Chapter 4: Exponential and Logarithmic Functions

Chaper 4: Eponenial and Logarihmic Funcions Secion 4.1 Eponenial Funcions... 15 Secion 4. Graphs of Eponenial Funcions... 3 Secion 4.3 Logarihmic Funcions... 4 Secion 4.4 Logarihmic Properies... 53 Secion 4.5 Graphs of Logarihmic Funcions... 6 Secion 4.6 Eponenial and Logarihmic Models... 70 Secion 4.7 Fiing Eponenials o Daa... 89 Secion 4.1 Eponenial Funcions India is he second mos populous counry in he world, wih a populaion in 008 of abou 1.14 billion people. The populaion is growing by abou 1.34% each year 1. We migh ask if we can find a formula o model he populaion, P, as a funcion of ime,, in years afer 008, if he populaion coninues o grow a his rae. In linear growh, we had a consan rae of change a consan number ha he oupu increased for each increase in inpu. For eample, in he equaion f ( ) = 3 + 4, he slope ells us he oupu increases by hree each ime he inpu increases by one. This populaion scenario is differen we have a percen rae of change raher han a consan number of people as our rae of change. To see he significance of his difference consider hese wo companies: Company A has 100 sores, and epands by opening 50 new sores a year Company B has 100 sores, and epands by increasing he number of sores by 50% of heir oal each year. Looking a a few years of growh for hese companies: Year Sores, company A Sores, company B 0 100 Saring wih 100 each 100 1 100 + 50 = 150 They boh grow by 50 sores in he firs year. 150 + 50 = 00 Sore A grows by 50, Sore B grows by 75 3 00 + 50 = 50 Sore A grows by 50, Sore B grows by 11.5 100 + 50% of 100 100 + 0.50(100) = 150 150 + 50% of 150 150 + 0.50(150) = 5 5 + 50% of 5 5 + 0.50(5) = 337.5 1 World Bank, World Developmen Indicaors, as repored on hp://www.google.com/publicdaa, rerieved Augus 0, 010 This chaper is par of Precalculus: An Invesigaion of Funcions Lippman & Rasmussen 011. This maerial is licensed under a Creaive Commons CC-BY-SA license.

16 Chaper 4 Noice ha wih he percen growh, each year he company is grows by 50% of he curren year s oal, so as he company grows larger, he number of sores added in a year grows as well. To ry o simplify he calculaions, noice ha afer 1 year he number of sores for company B was: 100 + 0.50(100) or equivalenly by facoring 100 (1 + 0.50) = 150 We can hink of his as he new number of sores is he original 100% plus anoher 50%. Afer years, he number of sores was: 150 + 0.50(150) or equivalenly by facoring 150 (1 + 0.50) now recall he 150 came from 100(1+0.50). Subsiuing ha, 100(1 + 0.50)(1 + 0.50) = 100(1 + 0.50) = 5 Afer 3 years, he number of sores was: 5 + 0.50(5) or equivalenly by facoring 5 (1 + 0.50) now recall he 5 came from 3 100(1 + 0.50) (1 + 0.50) = 100(1 + 0.50) = 337.5 100 (1 + 0.50). Subsiuing ha, From his, we can generalize, noicing ha o show a 50% increase, each year we muliply by a facor of (1+0.50), so afer n years, our equaion would be n B ( n) = 100(1 + 0.50) In his equaion, he 100 represened he iniial quaniy, and he 0.50 was he percen growh rae. Generalizing furher, we arrive a he general form of eponenial funcions. Eponenial Funcion An eponenial growh or decay funcion is a funcion ha grows or shrinks a a consan percen growh rae. The equaion can be wrien in he form f ( ) = a(1 + r) or f ( ) = ab where b = 1+r Where a is he iniial or saring value of he funcion r is he percen growh or decay rae, wrien as a decimal b is he growh facor or growh muliplier. Since powers of negaive numbers behave srangely, we limi b o posiive values. To see more clearly he difference beween eponenial and linear growh, compare he wo ables and graphs below, which illusrae he growh of company A and B described above over a longer ime frame if he growh paerns were o coninue

Secion 4.1 Eponenial Funcions 17 years Company A Company B 00 5 4 300 506 6 400 1139 8 500 563 10 600 5767 B A Eample 1 Wrie an eponenial funcion for India s populaion, and use i o predic he populaion in 00. A he beginning of he chaper we were given India s populaion of 1.14 billion in he year 008 and a percen growh rae of 1.34%. Using 008 as our saring ime ( = 0), our iniial populaion will be 1.14 billion. Since he percen growh rae was 1.34%, our value for r is 0.0134. Using he basic formula for eponenial growh f ( ) = a(1 + r) we can wrie he formula, f ( ) = 1.14(1 + 0.0134) To esimae he populaion in 00, we evaluae he funcion a = 1, since 00 is 1 years afer 008. f (1) = 1.14(1 + 0.0134) 1 1.337 billion people in 00 Try i Now 1. Given he hree saemens below, idenify which represen eponenial funcions. A. The cos of living allowance for sae employees increases salaries by 3.1% each year. B. Sae employees can epec a $300 raise each year hey work for he sae. C. Tuiion coss have increased by.8% each year for he las 3 years. Eample A cerificae of deposi (CD) is a ype of savings accoun offered by banks, ypically offering a higher ineres rae in reurn for a fied lengh of ime you will leave your money invesed. If a bank offers a 4 monh CD wih an annual ineres rae of 1.% compounded monhly, how much will a $1000 invesmen grow o over hose 4 monhs? Firs, we mus noice ha he ineres rae is an annual rae, bu is compounded monhly, meaning ineres is calculaed and added o he accoun monhly. To find he monhly ineres rae, we divide he annual rae of 1.% by 1 since here are 1 monhs in a

18 Chaper 4 year: 1.%/1 = 0.1%. Each monh we will earn 0.1% ineres. From his, we can se up an eponenial funcion, wih our iniial amoun of $1000 and a growh rae of r = 0.001, and our inpu m measured in monhs. f m.01 ( ) = 1000 1 + 1 f ( m) = 1000(1 + 0.001) m m Afer 4 monhs, he accoun will have grown o 4 f (4) = 1000(1 + 0.001) = $104.8 Try i Now. Looking a hese wo equaions ha represen he balance in wo differen savings accouns, which accoun is growing faser, and which accoun will have a higher balance afer 3 years? A( ) = 1000 1. 05 B( ) = 900( 1. 075) ( ) In all he preceding eamples, we saw eponenial growh. Eponenial funcions can also be used o model quaniies ha are decreasing a a consan percen rae. An eample of his is radioacive decay, a process in which radioacive isoopes of cerain aoms ransform o an aom of a differen ype, causing a percenage decrease of he original maerial over ime. Eample 3 Bismuh-10 is an isoope ha radioacively decays by abou 13% each day, meaning 13% of he remaining Bismuh-10 ransforms ino anoher aom (polonium-10 in his case) each day. If you begin wih 100 mg of Bismuh-10, how much remains afer one week? Wih radioacive decay, insead of he quaniy increasing a a percen rae, he quaniy is decreasing a a percen rae. Our iniial quaniy is a = 100 mg, and our growh rae will be negaive 13%, since we are decreasing: r = -0.13. This gives he equaion: d d Q ( d) = 100(1 0.13) = 100(0.87) This can also be eplained by recognizing ha if 13% decays, hen 87 % remains. Afer one week, 7 days, he quaniy remaining would be Q(7) = 100(0.87) 7 = 37.73 mg of Bismuh-10 remains. Try i Now 3. A populaion of 1000 is decreasing 3% each year. Find he populaion in 30 years.

Secion 4.1 Eponenial Funcions 19 Eample 4 T(q) represens he oal number of Android smar phone conracs, in housands, held by a cerain Verizon sore region measured quarerly since January 1, 010, Inerpre all of he pars of he equaion T () = 86(1.64) = 31. 3056. Inerpreing his from he basic eponenial form, we know ha 86 is our iniial value. This means ha on Jan. 1, 010 his region had 86,000 Android smar phone conracs. Since b = 1 + r = 1.64, we know ha every quarer he number of smar phone conracs grows by 64%. T() = 31.3056 means ha in he nd quarer (or a he end of he second quarer) here were approimaely 31,305 Android smar phone conracs. Finding Equaions of Eponenial Funcions In he previous eamples, we were able o wrie equaions for eponenial funcions since we knew he iniial quaniy and he growh rae. If we do no know he growh rae, bu insead know only some inpu and oupu pairs of values, we can sill consruc an eponenial funcion. Eample 5 In 00, 80 deer were reinroduced ino a wildlife refuge area from which he populaion had previously been huned o eliminaion. By 008, he populaion had grown o 180 deer. If his populaion grows eponenially, find a formula for he funcion. By defining our inpu variable o be, years afer 00, he informaion lised can be wrien as wo inpu-oupu pairs: (0,80) and (6,180). Noice ha by choosing our inpu variable o be measured as years afer he firs year value provided, we have effecively given ourselves he iniial value for he funcion: a = 80. This gives us an equaion of he form f ( ) = 80b. Subsiuing in our second inpu-oupu pair allows us o solve for b: 180 80b 6 = Divide by 80 6 180 9 b = = Take he 6 h roo of boh sides. 80 4 b = 9 6 = 1.1447 4 This gives us our equaion for he populaion: f ( ) = 80(1.1447) Recall ha since b = 1+r, we can inerpre his o mean ha he populaion growh rae is r = 0.1447, and so he populaion is growing by abou 14.47% each year. In his eample, you could also have used (9/4)^(1/6) o evaluae he 6 h roo if your calculaor doesn have an n h roo buon.

0 Chaper 4 In he previous eample, we chose o use he ( form of he eponenial f ) = ab funcion raher han he f ( ) = a(1 + r) form. This choice was enirely arbirary eiher form would be fine o use. When finding equaions, he value for b or r will usually have o be rounded o be wrien easily. To preserve accuracy, i is imporan o no over-round hese values. Typically, you wan o be sure o preserve a leas 3 significan digis in he growh rae. For eample, if your value for b was 1.00317643, you would wan o round his no furher han o 1.00318. In he previous eample, we were able o give ourselves he iniial value by clever definiion of our inpu variable. Ne we consider a siuaion where we can do his. Eample 6 Find a formula for an eponenial funcion passing hrough he poins (-,6) and (,1). Since we don have he iniial value, we will ake a general approach ha will work for any funcion form wih unknown parameers: we will subsiue in boh given inpuoupu pairs in he funcion form f ( ) = ab and solve for he unknown values, a and b. Subsiuing in (-, 6) gives 6 = ab Subsiuing in (, 1) gives 1 = ab We now solve hese as a sysem of equaions. To do so, we could ry a subsiuion approach, solving one equaion for a variable, hen subsiuing ha epression ino he second equaion. Solving 6 = ab for a: 6 a = = 6b b In he second equaion, 1 = ab, we subsiue he epression above for a: 1 = (6b ) b 4 1 = 6b 1 4 = b 6 1 b = 4 0.6389 6 Going back o he equaion a = 6b les us find a: a = 6b = 6(0.6389) =.449 Puing his ogeher gives he equaion f ( ) =.449(0.6389)

Secion 4.1 Eponenial Funcions 1 Try i Now 4. Given he wo poins (1, 3) and (, 4.5) find he equaion of an eponenial funcion ha passes hrough hese wo poins. Eample 7 Find an equaion for he eponenial funcion graphed below. The iniial value for he funcion is no clear in his graph, so we will insead work using wo clearer poins. There are hree fairly clear poins: (-1, 1), (1, ), and (3, 4). As we saw in he las eample, wo poins are sufficien o find he equaion for a sandard eponenial, so we will use he laer wo poins. 1 Subsiuing in (1,) gives = ab 3 Subsiuing in (3,4) gives 4 = ab Solving he firs equaion for a gives a =. b Subsiuing his epression for a ino he second equaion: 3 4 = ab 3 3 b 4 = b = Simplify he righ-hand side b b 4 = b = b b = ± Since we resric ourselves o posiive values of b, we will use b =. We can hen go back and find a: a = = = b This gives us a final equaion of f ( ) ( ) =.

Chaper 4 Compound Ineres In he bank cerificae of deposi (CD) eample earlier in he secion, we encounered compound ineres. Typically bank accouns and oher savings insrumens in which earnings are reinvesed, such as muual funds and reiremen accouns, uilize compound ineres. The erm compounding comes from he behavior ha ineres is earned no on he original value, bu on he accumulaed value of he accoun. In he eample from earlier, he ineres was compounded monhly, so we ook he annual ineres rae, usually called he nominal rae or annual percenage rae (APR) and divided by 1, he number of compounds in a year, o find he monhly ineres. The eponen was hen measured in monhs. Generalizing his, we can form a general formula for compound ineres. If he APR is wrien in decimal form as r, and here are k compounding periods per year, hen he ineres per compounding period will be r/k. Likewise, if we are ineresed in he value afer years, hen here will be k compounding periods in ha ime. Compound Ineres Formula Compound Ineres can be calculaed using he formula k r A ( ) = a 1+ k Where A() is he accoun value is measured in years a is he saring amoun of he accoun, ofen called he principal r is he annual percenage rae (APR), also called he nominal rae k is he number of compounding periods in one year Eample 8 If you inves $3,000 in an invesmen accoun paying 3% ineres compounded quarerly, how much will he accoun be worh in 10 years? Since we are saring wih $3000, a = 3000 Our ineres rae is 3%, so r = 0.03 Since we are compounding quarerly, we are compounding 4 imes per year, so k = 4 We wan o know he value of he accoun in 10 years, so we are looking for A(10), he value when = 10. 0.03 A (10) = 3000 1 + 4 4(10) = $4045.05 The accoun will be worh $4045.05 in 10 years.

Secion 4.1 Eponenial Funcions 3 Eample 9 A 59 plan is a college savings plan in which a relaive can inves money o pay for a child s laer college uiion, and he accoun grows a free. If Lily wans o se up a 59 accoun for her new granddaugher, wans he accoun o grow o $40,000 over 18 years, and she believes he accoun will earn 6% compounded semi-annually (wice a year), how much will Lily need o inves in he accoun now? Since he accoun is earning 6%, r = 0.06 Since ineres is compounded wice a year, k = In his problem, we don know how much we are saring wih, so we will be solving for a, he iniial amoun needed. We do know we wan he end amoun o be $40,000, so we will be looking for he value of a so ha A(18) = 40,000. 0.06 40,000 = A(18) = a 1 + 40,000 = a(.8983) a = 40,000.8983 $13,801 (18) Lily will need o inves $13,801 o have $40,000 in 18 years. Try i now 5. Recalculae eample from above wih quarerly compounding. Because of compounding hroughou he year, wih compound ineres he acual increase in a year is more han he annual percenage rae. If $1,000 were invesed a 10%, he able below shows he value afer 1 year a differen compounding frequencies: Frequency Value afer 1 year Annually $1100 Semiannually $110.50 Quarerly $1103.81 Monhly $1104.71 Daily $1105.16 If we were o compue he acual percenage increase for he daily compounding, here was an increase of $105.16 from an original amoun of $1,000, for a percenage increase 105.16 of = 0. 10516 = 10.516% increase. This quaniy is called he annual percenage 1000 yield (APY).

4 Chaper 4 Noice ha given any saring amoun, he amoun afer 1 year would be k r A ( 1) = a 1+. To find he oal change, we would subrac he original amoun, hen k o find he percenage change we would divide ha by he original amoun: a 1 + k r k a a = 1 + r k k 1 Annual Percenage Yield The annual percenage yield is he acual percen a quaniy increases in one year. I can be calculaed as k r APY = 1 + 1 k Noice his is equivalen o finding he value of $1 afer 1 year, and subracing he original dollar. Eample 10 Bank A offers an accoun paying 1.% compounded quarerly. Bank B offers an accoun paying 1.1% compounded monhly. Which is offering a beer rae? We can compare hese raes using he annual percenage yield he acual percen increase in a year. 4 0.01 Bank A: APY = 1 + 1 = 0. 01054 = 1.054% 4 1 0.011 Bank B: APY = 1 + 1 = 0. 011056 = 1.1056% 1 Bank B s monhly compounding is no enough o cach up wih Bank A s beer APR. Bank A offers a beer rae. A Limi o Compounding As we saw earlier, he amoun we earn increases as we increase he compounding frequency. The able, hough, shows ha he increase from annual o semi-annual compounding is larger han he increase from monhly o daily compounding. This migh lead us o believe ha alhough increasing he frequency of compounding will increase our resul, here is an upper limi o his process.

Secion 4.1 Eponenial Funcions 5 To see his, le us eamine he value of $1 invesed a 100% ineres for 1 year. Frequency Value Annual $ Semiannually $.5 Quarerly $.441406 Monhly $.613035 Daily $.714567 Hourly $.71817 Once per minue $.71879 Once per second $.7188 These values do indeed appear o be approaching an upper limi. This value ends up being so imporan ha i ges represened by is own leer, much like how π represens a number. Euler s Number: e e is he leer used o represen he value ha e.7188 k 1 k 1 + approaches as k ges big. Because e is ofen used as he base of an eponenial, mos scienific and graphing calculaors have a buon ha can calculae powers of e, usually labeled e. Some compuer sofware insead defines a funcion ep(), where ep() = e. Because e arises when he ime beween compounds becomes very small, e allows us o define coninuous growh and allows us o define a new oolki funcion, f( ) = e. Coninuous Growh Formula Coninuous Growh can be calculaed using he formula r f ( ) = ae where a is he saring amoun r is he coninuous growh rae This ype of equaion is commonly used when describing quaniies ha change more or less coninuously, like chemical reacions, growh of large populaions, and radioacive decay.

6 Chaper 4 Eample 11 Radon- decays a a coninuous rae of 17.3% per day. How much will 100mg of Radon- decay o in 3 days? Since we are given a coninuous decay rae, we use he coninuous growh formula. Since he subsance is decaying, we know he growh rae will be negaive: r = -0.173 0.173(3) f (3) = 100e 59.51 mg of Radon- will remain. Try i Now 0.1 6. Inerpre he following: S ( ) = 0e if S() represens he growh of a subsance in grams, and ime is measured in days. Coninuous growh is also ofen applied o compound ineres, allowing us o alk abou coninuous compounding. Eample 1 If $1000 is invesed in an accoun earning 10% compounded coninuously, find he value afer 1 year. Here, he coninuous growh rae is 10%, so r = 0.10. We sar wih $1000, so a = 1000. To find he value afer 1 year, 0.10(1) f (1) = 1000e $1105.17 Noice his is a $105.17 increase for he year. As a percen increase, his is 105.17 = 0.10517 = 10.517% increase over he original $1000. 1000 Noice ha his value is slighly larger han he amoun generaed by daily compounding in he able compued earlier. The coninuous growh rae is like he nominal growh rae (or APR) i reflecs he growh rae before compounding akes effec. This is differen han he annual growh rae used in he formula f ( ) = a(1 + r), which is like he annual percenage yield i reflecs he acual amoun he oupu grows in a year. While he coninuous growh rae in he eample above was 10%, he acual annual yield was 10.517%. This means we could wrie wo differen looking bu equivalen formulas for his accoun s growh: 0.10 f( ) = 1000e using he 10% coninuous growh rae f( ) = 1000(1.10517) using he 10.517% acual annual yield rae.

Secion 4.1 Eponenial Funcions 7 Imporan Topics of his Secion Percen growh Eponenial funcions Finding formulas Inerpreing equaions Graphs Eponenial Growh & Decay Compound ineres Annual Percen Yield Coninuous Growh Try i Now Answers 1. A & C are eponenial funcions, hey grow by a % no a consan number.. B() is growing faser, bu afer 3 years A() sill has a higher accoun balance 3. 1000(0.97) 30 = 401. 0071 4. f ( ) = ( 1. 5) 5. $104.5 6. An iniial subsance weighing 0g is growing a a coninuous rae of 1% per day.

8 Chaper 4 Secion 4.1 Eercises For each able below, could he able represen a funcion ha is linear, eponenial, or neiher? 1. 1 3 4 f() 70 40 10-0 3. 1 3 4 h() 70 49 34.3 4.01 5. 1 3 4 m() 80 61 4.9 5.61. 1 3 4 g() 40 3 6 4. 1 3 4 k() 90 80 70 60 6. 1 3 4 n() 90 81 7.9 65.61 7. A populaion numbers 11,000 organisms iniially and grows by 8.5% each year. Wrie an eponenial model for he populaion. 8. A populaion is currenly 6,000 and has been increasing by 1.% each day. Wrie an eponenial model for he populaion. 9. The fo populaion in a cerain region has an annual growh rae of 9 percen per year. I is esimaed ha he populaion in he year 010 was 3,900. Esimae he fo populaion in he year 018. 10. The amoun of area covered by blackberry bushes in a park has been growing by 1% each year. I is esimaed ha he area covered in 009 was 4,500 square fee. Esimae he area ha will be covered in 00. 11. A vehicle purchased for $3,500 depreciaes a a consan rae of 5% each year. Deermine he approimae value of he vehicle 1 years afer purchase. 1. A business purchases $15,000 of office furniure which depreciaes a a consan rae of 1% each year. Find he residual value of he furniure 6 years afer purchase.

Secion 4.1 Eponenial Funcions 9 Find a formula for an eponenial funcion passing hrough he wo poins. 0, 3, (, 75) 13. ( 0, 6 ), (3, 750) 14. ( ) 15. ( 0, 000 ), (, 0) 16. ( 0, 9000 ), (3, 7) 3 17. 1,,( 3, 4) 5 18. 1,,( 1,10) 19. (, 6 ),( 3,1) 0. ( ) 3,4, (3, ) 1. ( 3,1 ), (5, 4). (,5 ), (6, 9) 3. A radioacive subsance decays eponenially. A scienis begins wih 100 milligrams of a radioacive subsance. Afer 35 hours, 50 mg of he subsance remains. How many milligrams will remain afer 54 hours? 4. A radioacive subsance decays eponenially. A scienis begins wih 110 milligrams of a radioacive subsance. Afer 31 hours, 55 mg of he subsance remains. How many milligrams will remain afer 4 hours? 5. A house was valued a $110,000 in he year 1985. The value appreciaed o $145,000 by he year 005. Wha was he annual growh rae beween 1985 and 005? Assume ha he house value coninues o grow by he same percenage. Wha did he value equal in he year 010? 6. An invesmen was valued a $11,000 in he year 1995. The value appreciaed o $14,000 by he year 008. Wha was he annual growh rae beween 1995 and 008? Assume ha he value coninues o grow by he same percenage. Wha did he value equal in he year 01? 7. A car was valued a $38,000 in he year 003. The value depreciaed o $11,000 by he year 009. Assume ha he car value coninues o drop by he same percenage. Wha will he value be in he year 013? 8. A car was valued a $4,000 in he year 006. The value depreciaed o $0,000 by he year 009. Assume ha he car value coninues o drop by he same percenage. Wha will he value be in he year 014? 9. If $4,000 is invesed in a bank accoun a an ineres rae of 7 per cen per year, find he amoun in he bank afer 9 years if ineres is compounded annually, quarerly, monhly, and coninuously.

30 Chaper 4 30. If $6,000 is invesed in a bank accoun a an ineres rae of 9 per cen per year, find he amoun in he bank afer 5 years if ineres is compounded annually, quarerly, monhly, and coninuously. 31. Find he annual percenage yield (APY) for a savings accoun wih annual percenage rae of 3% compounded quarerly. 3. Find he annual percenage yield (APY) for a savings accoun wih annual percenage rae of 5% compounded monhly. 0.1 33. A populaion of baceria is growing according o he equaion P ( ) = 1600e, wih measured in years. Esimae when he populaion will eceed 7569. 0.17 34. A populaion of baceria is growing according o he equaion P ( ) = 100e, wih measured in years. Esimae when he populaion will eceed 3443. 35. In 1968, he U.S. minimum wage was $1.60 per hour. In 1976, he minimum wage was $.30 per hour. Assume he minimum wage grows according o an eponenial model w (), where represens he ime in years afer 1960. [UW] a. Find a formula for w (). b. Wha does he model predic for he minimum wage in 1960? c. If he minimum wage was $5.15 in 1996, is his above, below or equal o wha he model predics? 36. In 1989, research scieniss published a model for predicing he cumulaive number 3 1980 of AIDS cases (in housands) repored in he Unied Saes: a( ) = 155 10, where is he year. This paper was considered a relief, since here was a fear he correc model would be of eponenial ype. Pick wo daa poins prediced by he research model a () o consruc a new eponenial model b () for he number of cumulaive AIDS cases. Discuss how he wo models differ and eplain he use of he word relief. [UW]

Secion 4.1 Eponenial Funcions 31 37. You have a chess board as picured, wih squares numbered 1 hrough 64. You also have a huge change jar wih an unlimied number of dimes. On he firs square you place one dime. On he second square you sack dimes. Then you coninue, always doubling he number from he previous square. [UW] a. How many dimes will you have sacked on he 10h square? b. How many dimes will you have sacked on he nh square? c. How many dimes will you have sacked on he 64h square? d. Assuming a dime is 1 mm hick, how high will his las pile be? e. The disance from he earh o he sun is approimaely 150 million km. Relae he heigh of he las pile of dimes o his disance.

3 Chaper 4 Secion 4. Graphs of Eponenial Funcions Like wih linear funcions, he graph of an eponenial funcion is deermined by he values for he parameers in he funcion s formula. To ge a sense for he behavior of eponenials, le us begin by looking more closely a he funcion f ( ) =. Lising a able of values for his funcion: -3 - -1 0 1 3 1 1 1 f() 1 4 8 8 4 Noice ha: 1) This funcion is posiive for all values of. ) As increases, he funcion grows faser and faser (he rae of change increases). 3) As decreases, he funcion values grow smaller, approaching zero. 4) This is an eample of eponenial growh. 1 Looking a he funcion g( ) = -3 - -1 0 1 3 1 1 1 g() 8 4 1 4 8 Noe his funcion is also posiive for all values of, bu in his case grows as decreases, and decreases owards zero as increases. This is an eample of eponenial decay. You may noice from he able ha his funcion appears o be he horizonal reflecion of he f ( ) = able. This is in fac he case: f ( ) = = ( ) 1 1 = = g( ) Looking a he graphs also confirms his relaionship:

Secion 4. Graphs of Eponenial Funcions 33 Consider a funcion for he form f ( ) = ab. Since a, which we called he iniial value in he las secion, is he funcion value a an inpu of zero, a will give us he verical inercep of he graph. From he graphs above, we can see ha an eponenial graph will have a horizonal asympoe on one side of he graph, and can eiher increase or decrease, depending upon he growh facor. This horizonal asympoe will also help us deermine he long run behavior and is easy o deermine from he graph. The graph will grow when he growh rae is posiive, which will make he growh facor b larger han one. When i s negaive, he growh facor will be less han one. Graphical Feaures of Eponenial Funcions Graphically, in he funcion f ( ) = ab a is he verical inercep of he graph b deermines he rae a which he graph grows he funcion will increase if b > 1 he funcion will decrease if 0 < b < 1 The graph will have a horizonal asympoe a y = 0 The graph will be concave up if a > 0; concave down if a < 0. The domain of he funcion is all real numbers The range of he funcion is (0, ) When skeching he graph of an eponenial funcion, i can be helpful o remember ha he graph will pass hrough he poins (0, a) and (1, ab). The value b will deermine he funcion s long run behavior: If b > 1, as, f () and as, f ( ) 0. If 0 < b < 1, as, f ( ) 0 and as, f (). Eample 1 Skech a graph of 1 f ( ) = 4 3 This graph will have a verical inercep a (0,4), and 4 pass hrough he poin 1,. Since b < 1, he graph 3 will be decreasing owards zero. Since a > 0, he graph will be concave up. We can also see from he graph he long run behavior: as, f ( ) 0 and as, f ().

34 Chaper 4 To ge a beer feeling for he effec of a and b on he graph, eamine he ses of graphs below. The firs se shows various graphs, where a remains he same and we only change he value for b. ( 1 3 ) 3 ( 1 ) 1.5 0.9 Noice ha he closer he value of b is o 1, he less seep he graph will be. In he ne se of graphs, a is alered and our value for b remains he same. 4( 1. ) 3( 1. ) ( 1. ) 1. 0.5( 1. ) Noice ha changing he value for a changes he verical inercep. Since a is muliplying he b erm, a acs as a verical srech facor, no as a shif. Noice also ha he long run behavior for all of hese funcions is he same because he growh facor did no change and none of hese a values inroduced a verical flip.

Secion 4. Graphs of Eponenial Funcions 35 Eample Mach each equaion wih is graph. f ( ) = (1.3) g( ) = (1.8) h( ) = 4(1.3) k( ) = 4(0.7) The graph of k() is he easies o idenify, since i is he only equaion wih a growh facor less han one, which will produce a decreasing graph. The graph of h() can be idenified as he only growing eponenial funcion wih a verical inercep a (0,4). The graphs of f() and g() boh have a verical inercep a (0,), bu since g() has a larger growh facor, we can idenify i as he graph increasing faser. g() f() h() k() Try i Now 1. Graph he following funcions on he same ais: h ( ) = (1/ ). f ) ( ) = ( ; g ) ( ) = ( ; Transformaions of Eponenial Graphs While eponenial funcions can be ransformed following he same rules as any funcion, here are a few ineresing feaures of ransformaions ha can be idenified. The firs was seen a he beginning of he secion ha a horizonal reflecion is equivalen o a change in he growh facor. Likewise, since a is iself a srech facor, a verical srech of an eponenial corresponds wih a change in he iniial value of he funcion.

36 Chaper 4 Ne consider he effec of a horizonal shif on an eponenial funcion. Shifing he + 4 funcion f ( ) = 3() four unis o he lef would give f ( + 4) = 3(). Employing eponen rules, we could rewrie his: + 4 4 f ( + 4) = 3() = 3() ( ) = 48() Ineresingly, i urns ou ha a horizonal shif of an eponenial funcion corresponds wih a change in iniial value of he funcion. Lasly, consider he effec of a verical shif on an eponenial funcion. Shifing f ( ) = 3() down 4 unis would give he equaion f ( ) = 3() 4, yielding he graph Noice ha his graph is subsanially differen han he basic eponenial graph. Unlike a basic eponenial, his graph does no have a horizonal asympoe a y = 0; due o he verical shif, he horizonal asympoe has also shifed o y = -4. We can see ha as, f( ) and as, f( ) 4. We have deermined ha a verical shif is he only ransformaion of an eponenial funcion ha changes he graph in a way ha canno be achieved by alering he parameers a and b in he basic eponenial funcion f ( ) = ab. Transformaions of Eponenials Any ransformed eponenial can be wrien in he form f ( ) = ab + c where y = c is he horizonal asympoe. Noe ha, due o he shif, he verical inercep is shifed o (0, a+c). Try i Now. Wrie he equaion and graph he eponenial funcion described as follows: f ( ) = e is verically sreched by a facor of, flipped across he y ais and shifed up 4 unis.

Secion 4. Graphs of Eponenial Funcions 37 Eample 3 1 Skech a graph of f ( ) = 3 + 4. Noice ha in his eponenial funcion, he negaive in he srech facor -3 will cause a verical reflecion, and he verical shif up 4 will move he horizonal asympoe o y = 4. Skeching his as a ransformaion of The basic 1 g( ) =, 1 g( ) = Verically refleced and sreched by 3 Verically shifed up four unis Noice ha while he domain of his funcion is unchanged, due o he reflecion and,4. shif, he range of his funcion is ( ) As, f ( ) 4 and as, f( ). Funcions leading o graphs like he one above are common as models for learning and models of growh approaching a limi.

38 Chaper 4 Eample 4 Find an equaion for he graph skeched below. Looking a his graph, i appears o have a horizonal asympoe a y = 5, suggesing an equaion of he form f ( ) = ab + 5. To find values for a and b, we can idenify wo oher poins on he graph. I appears he graph passes hrough (0,) and (-1,3), so we can use hose poins. Subsiuing in (0,) allows us o solve for a 0 = ab + 5 = a + 5 a = 3 Subsiuing in (-1,3) allows us o solve for b 1 3 = 3b + 5 3 = b b = 3 b = 3 = 1.5 The final formula for our funcion is f ( ) = 3(1.5) + 5. Try i Now 3. Given he graph of he ransformed eponenial funcion, find a formula and describe he long run behavior.

Secion 4. Graphs of Eponenial Funcions 39 Imporan Topics of his Secion Graphs of eponenial funcions Inercep Growh facor Eponenial Growh Eponenial Decay Horizonal inerceps Long run behavior Transformaions Try i Now Answers 1 h ( ) = g= ( ) ( ) f( ) = 1.. f ( ) = e + 4 ; 3. f ( ) = 3(.5) 1 or f( ) = 3( ) 1; As, f () and as, f ( ) 1

40 Chaper 4 Secion 4. Eercises Mach each funcion wih one of he graphs below. 1. f ( ) = ( 0.69). f ( ) = ( 1.8) 3. f ( ) = ( 0.81) 4. f ( ) = 4( 1.8) 5. f ( ) = ( 1.59) 6. f ( ) = 4( 0.69) A B C D E F If all he graphs o he righ have equaions wih form f = ab, ( ) 7. Which graph has he larges value for b? A B C D E 8. Which graph has he smalles value for b? F 9. Which graph has he larges value for a? 10. Which graph has he smalles value for a? Skech a graph of each of he following ransformaions of f ( ) = 11. f ( ) = 1. g( ) = 13. h( ) = + 3 14. f ( ) = 4 k 15. f ( ) 3 = 16. ( ) Saring wih he graph of f ( ) = 4, find a formula for he funcion ha resuls from 17. Shifing f( ) 4 unis upwards 18. Shifing f( ) 3 unis downwards 19. Shifing f( ) unis lef 0. Shifing f( ) 5 unis righ 1. Reflecing f( ) abou he -ais. Reflecing f( ) abou he y-ais =

Secion 4. Graphs of Eponenial Funcions 41 Describe he long run behavior, as and of each funcion 54 1 f = 3 + 3. f ( ) = ( ) 4. ( ) ( ) 5. f ( ) 1 = 3 6. f ( ) 1 = 4 + 1 4 7. f ( ) = 34 ( ) + 8. f ( ) ( ) = 3 1 Find a formula for each funcion graphed as a ransformaion of f ( ) =. 9. 30. 31. 3. Find an equaion for he eponenial funcion graphed. 33. 34. 35. 36.

4 Chaper 4 Secion 4.3 Logarihmic Funcions A populaion of 50 flies is epeced o double every week, leading o a funcion of he form f ( ) = 50(), where represens he number of weeks ha have passed. When will his populaion reach 500? Trying o solve his problem leads o: 500 = 50() Dividing boh sides by 50 o isolae he eponenial 10 = While we have se up eponenial models and used hem o make predicions, you may have noiced ha solving eponenial equaions has no ye been menioned. The reason is simple: none of he algebraic ools discussed so far are sufficien o solve eponenial equaions. Consider he equaion = 10 above. We know ha 3 = 8 and 4 = 16, so i is clear ha mus be some value beween 3 and 4 since g= ( ) is increasing. We could use echnology o creae a able of values or graph o beer esimae he soluion. From he graph, we could beer esimae he soluion o be around 3.3. This resul is sill fairly unsaisfacory, and since he eponenial funcion is one-o-one, i would be grea o have an inverse funcion. None of he funcions we have already discussed would serve as an inverse funcion and so we mus inroduce a new funcion, named log as he inverse of an eponenial funcion. Since eponenial funcions have differen bases, we will define corresponding logarihms of differen bases as well. Logarihm The logarihm (base b) funcion, wrien log ( ) funcion (base b), b. b, is he inverse of he eponenial Since he logarihm and eponenial are inverses, i follows ha: Properies of Logs: Inverse Properies log b ( ) b = log b = b

Secion 4.3 Logarihmic Funcions 43 Recall also from he definiion of an inverse funcion ha if Applying his o he eponenial and logarihmic funcions: 1 f ( a) = c, hen f ( c) = a. Logarihm Equivalen o an Eponenial The saemen b a = c is equivalen o he saemen log ( c) a. b = a Alernaively, we could show his by saring wih he eponenial funcion c= b, hen a aking he log base b of boh sides, giving log b( c) = logbb. Using he inverse propery of logs we see ha log ( c b ) = a. Since log is a funcion, i is mos correcly wrien as logb ( c), using parenheses o denoe funcion evaluaion, jus as we would wih f(c). However, when he inpu is a single variable or number, i is common o see he parenheses dropped and he epression wrien as log c. b Eample 1 Wrie hese eponenial equaions as logarihmic equaions: 3 = 8 5 = 5 10 4 = 1 10000 3 = 8 is equivalen o log (8) = 3 5 = 5 is equivalen o log 5 (5) = 10 1 = is equivalen o log10 = 4 10000 10000 4 1 Eample Wrie hese logarihmic equaions as eponenial equaions: 1 log 6 ( 6) = log 3 ( 9) = ( 6) 1 log 6 = is equivalen o 6 1 / = 6 log 3 = is equivalen o 3 = 9 ( 9) Try i Now Wrie he eponenial equaion 4 = 16 as a logarihmic equaion.

44 Chaper 4 By esablishing he relaionship beween eponenial and logarihmic funcions, we can now solve basic logarihmic and eponenial equaions by rewriing. Eample 3 Solve log 4 ( ) = for. By rewriing his epression as an eponenial, 4 =, so = 16 Eample 4 Solve = 10 for. By rewriing his epression as a logarihm, we ge = log (10) While his does define a soluion, and an eac soluion a ha, you may find i somewha unsaisfying since i is difficul o compare his epression o he decimal esimae we made earlier. Also, giving an eac epression for a soluion is no always useful ofen we really need a decimal approimaion o he soluion. Luckily, his is a ask calculaors and compuers are quie adep a. Unluckily for us, mos calculaors and compuers will only evaluae logarihms of wo bases. Happily, his ends up no being a problem, as we ll see briefly. Common and Naural Logarihms The common log is he logarihm wih base 10, and is ypically wrien log( ). The naural log is he logarihm wih base e, and is ypically wrien ln( ). Eample 5 Evaluae log( 1000) using he definiion of he common log. To evaluae log( 1000), we can say = log(1000), hen rewrie ino eponenial form using he common log base of 10. 10 = 1000 From his, we migh recognize ha 1000 is he cube of 10, so = 3. We also can use he inverse propery of logs o 3 wrie log ( 10 ) 3 10 = Values of he common log number number as log(number) eponenial 1000 10 3 3 100 10 10 10 1 1 1 10 0 0 0.1 10-1 -1 0.01 10 - - 0.001 10-3 -3

Secion 4.3 Logarihmic Funcions 45 Try i Now. Evaluae log( 1000000). Eample 6 Evaluae ln ( e ). We can rewrie ( e ) ln as ln( e 1/ ) propery for logs: ln( 1/ ) = log ( e 1/ ). Since ln is a log base e, we can use he inverse 1 e e =. Eample 7 Evaluae log(500) using your calculaor or compuer. Using a compuer, we can evaluae log( 500). 69897 To uilize he common or naural logarihm funcions o evaluae epressions like log (10), we need o esablish some addiional properies. Properies of Logs: Eponen Propery r log A = r log A b ( ) ( ) b To show why his is rue, we offer a proof. Since he logarihmic and eponenial funcions are inverses, r log A So ( ) r b A = b p Uilizing he eponenial rule ha saes ( ) q pq =, r log r b A r logb A A = ( b ) = b r r logb A So hen log ( A ) = log ( b ) b b b log b A = Again uilizing he inverse propery on he righ side yields he resul r log A = r log b ( ) A b A. Eample 8 log 3 5 using he eponen propery for logs. Rewrie ( ) Since 5 = 5, log 5 = log 5 log3 ( ) ( ) 5 3 3 =

46 Chaper 4 Eample 9 Rewrie 4ln( ) using he eponen propery for logs. 4 Using he propery in reverse, 4ln( ) = ln( ) Try i Now 1 3. Rewrie using he eponen propery for logs: ln. The eponen propery allows us o find a mehod for changing he base of a logarihmic epression. Properies of Logs: Change of Base log c ( A) logb ( A) = log ( b) c Proof: Le log b ( A) =. Rewriing as an eponenial gives b = A. Taking he log base c of boh sides of his equaion gives log c b = log c A Now uilizing he eponen propery for logs on he lef side, log c b = log c A Dividing, we obain log c A log c A = or replacing our epression for, log b A = log b log b c Wih his change of base formula, we can finally find a good decimal approimaion o our quesion from he beginning of he secion. c Eample 10 Evaluae log (10) using he change of base formula. According o he change of base formula, we can rewrie he log base as a logarihm of any oher base. Since our calculaors can evaluae he naural log, we migh choose o use he naural logarihm, which is he log base e: log e 10 ln10 log 10 = = log e ln Using our calculaors o evaluae his,

Secion 4.3 Logarihmic Funcions 47 ln10 ln.3059 3.319 0.69315 This finally allows us o answer our original quesion he populaion of flies we discussed a he beginning of he secion will ake 3.3 weeks o grow o 500. Eample 11 Evaluae log 5 (100) using he change of base formula. We can rewrie his epression using any oher base. If our calculaors are able o evaluae he common logarihm, we could rewrie using he common log, base 10. log10 100 log (100) = log 5 0.69897 5 = 10.861 While we were able o solve he basic eponenial equaion = 10 by rewriing in logarihmic form and hen using he change of base formula o evaluae he logarihm, he proof of he change of base formula illuminaes an alernaive approach o solving eponenial equaions. Solving eponenial equaions: 1. Isolae he eponenial epressions when possible. Take he logarihm of boh sides 3. Uilize he eponen propery for logarihms o pull he variable ou of he eponen 4. Use algebra o solve for he variable. Eample 1 Solve = 10 for. Using his alernaive approach, raher han rewrie his eponenial ino logarihmic form, we will ake he logarihm of boh sides of he equaion. Since we ofen wish o evaluae he resul o a decimal answer, we will usually uilize eiher he common log or naural log. For his eample, we ll use he naural log: ln ( ) = ln(10) Uilizing he eponen propery for logs, ln ( ) = ln(10) Now dividing by ln(), ln(10) =.861 ln ( ) Noice ha his resul maches he resul we found using he change of base formula.

48 Chaper 4 Eample 13 In he firs secion, we prediced he populaion (in billions) of India years afer 008 by using he funcion f ( ) = 1.14(1 + 0.0134). If he populaion coninues following his rend, when will he populaion reach billion? We need o solve for he so ha f() = = 1.14(1.0134) Divide by 1.14 o isolae he eponenial epression = Take he logarihm of boh sides of he equaion 1.14 1.0134 ln = ln( 1.0134 ) 1.14 Apply he eponen propery on he righ side ln = ln( 1.0134) 1.14 Divide boh sides by ln(1.0134) ln 1.14 = 4.3 years ln 1.0134 ( ) If his growh rae coninues, he model predics he populaion of India will reach billion abou 4 years afer 008, or approimaely in he year 050. Try i Now 4. Solve 5 (0.93) = 10. In addiion o solving eponenial equaions, logarihmic epressions are common in many physical siuaions. Eample 14 In chemisry, ph is a measure of he acidiy or basiciy of a liquid. The ph is relaed o he concenraion of hydrogen ions, [H + ], measured in moles per lier, by he equaion ph = H +. log ( ) If a liquid has concenraion of 0.0001 moles per liber, deermine he ph. Deermine he hydrogen ion concenraion of a liquid wih ph of 7. To answer he firs quesion, we evaluae he epression log( 0.0001). While we could use our calculaors for his, we do no really need hem here, since we can use he inverse propery of logs: 4 log 0.0001 = log 10 = ( 4) = ( ) ( ) 4

Secion 4.3 Logarihmic Funcions 49 To answer he second quesion, we need o solve he equaion 7 log ( H + ) =. Begin by isolaing he logarihm on one side of he equaion by muliplying boh sides by -1: ( ) 7 = log H + Rewriing ino eponenial form yields he answer 7 H + = 10 = 0.0000001 moles per lier. Logarihms also provide us a mechanism for finding coninuous growh models for eponenial growh given wo daa poins. Eample 15 A populaion grows from 100 o 130 in weeks. Find he coninuous growh rae. Measuring in weeks, we are looking for an equaion P() = 130. Using he firs pair of values, r 0 100 = ae, so a = 100. P = r ( ) ae so ha P(0) = 100 and Using he second pair of values, 130 = 100e r Divide by 100 130 r = e Take he naural log of boh sides 100 r ln(1.3) = ln( e ) Use he inverse propery of logs ln(1.3) = r ln(1.3) r = 0.131 This populaion is growing a a coninuous rae of 13.1% per week. In general, we can relae he sandard form of an eponenial wih he coninuous growh form by noing (using k o represen he coninuous growh rae o avoid he confusion of using r in wo differen ways in he same formula): k a ( 1+ r) = ae k ( 1+ r ) = e k 1 + r = e Using his, we see ha i is always possible o conver from he coninuous growh form of an eponenial o he sandard form and vice versa. Remember ha he coninuous growh rae k represens he nominal growh rae before accouning for he effecs of coninuous compounding, while r represens he acual percen increase in one ime uni (one week, one year, ec.).

50 Chaper 4 Eample 16 A company s sales can be modeled by he funcion years. Find he annual growh rae. S 0.1 ( ) = 5000e, wih measured in k 0.1 Noing ha 1 + r = e, hen r = e 1 = 0. 175, so he annual growh rae is 1.75%. The sales funcion could also be wrien in he form S ( ) = 5000(1 + 0.175). Imporan Topics of his Secion The Logarihmic funcion as he inverse of he eponenial funcion Wriing logarihmic & eponenial epressions Properies of logs Inverse properies Eponenial properies Change of base Common log Naural log Solving eponenial equaions Try i Now Answers log 4 16 = = log 4 4 = log 4. 6 3. ln( ) ln() 4. 9. 5513 ln(0.93) 1. ( ) 4

Secion 4.3 Logarihmic Funcions 51 Secion 4.3 Eercises Rewrie each equaion in eponenial form 1. log 4( q) = m. log 3( ) = k 3. log ( b a ) = c 4. log ( z p ) = u 5. log ( v) = 6. log ( r) = s 7. ln ( w) = n 8. ln ( ) = y Rewrie each equaion in logarihmic form. 9. 4 = y 10. 5 y = 11. 13. 10 a = b 14. 10 p = v 15. d c k e = k 1. = h 16. z n y e = L = Solve for. 17. log3 ( ) = 18. log 4( ) = 3 19. log ( ) = 3 0. log 5( ) = 1 1. log ( ) = 3. log ( ) = 5 3. ln ( ) = 4. ( ) ln = Simplify each epression using logarihm properies. 1 5. log5 ( 5 ) 6. log ( 8 ) 7. log3 7 3 9. log6 ( 6 ) 30. 5 ( ) 33. log ( 0.001 ) 34. ( ) 1 8. log6 36 log 5 31. log ( 10,000 ) 3. log ( 100 ) log 0.00001 35. ln ( e 3 ) 36. ln ( e ) Evaluae using your calculaor. 37. log ( 0.04 ) 38. log ( 1045 ) 39. ln ( 15 ) 40. ln ( 0.0 ) Solve each equaion for he variable. 41. 5 = 14 4. 3 = 3 43. 45. 5 e = 17 46. 3 e = 1 47. 1 7 = 44. 3 15 4 5 3 = 38 48. 49. 1000( 1.03) = 5000 50. ( ) 51. ( ) 3 3 1.04 8 = 5. ( ) 4 00 1.06 = 550 1.08 = 7 = 1 4 3 4 = 44 53. e 0.03 = 54. 10e = 4 0.1 50 10 1 55. 10 8 = 5 1 56. 100 100 = 70 4

5 Chaper 4 f = ae. Conver he equaion ino coninuous growh form, ( ) k 57. f ( ) = 300( 0.91) 58. f ( ) = 10( 0.07) 59. f ( ) = 10( 1.04) 60. f ( ) = 1400( 1.1) f = ab. Conver he equaion ino annual growh form, ( ) 0.06 0.1 61. f ( ) = 1 5 0 e 6. f ( ) = 100e 0.01 63. f ( ) = 50e 0.85 64. f ( ) = 80e 65. The populaion of Kenya was 39.8 million in 009 and has been growing by abou.6% each year. If his rend coninues, when will he populaion eceed 45 million? 66. The populaion of Algeria was 34.9 million in 009 and has been growing by abou 1.5% each year. If his rend coninues, when will he populaion eceed 45 million? 67. The populaion of Seale grew from 563,374 in 000 o 608,660 in 010. If he populaion coninues o grow eponenially a he same rae, when will he populaion eceed 1 million people? 68. The median household income (adjused for inflaion) in Seale grew from $4,948 in 1990 o $45,736 in 000. If i coninues o grow eponenially a he same rae, when will median income eceed $50,000? 69. A scienis begins wih 100 mg of a radioacive subsance. Afer 4 hours, i has decayed o 80 mg. How long afer he process began will i ake o decay o 15 mg? 70. A scienis begins wih 100 mg of a radioacive subsance. Afer 6 days, i has decayed o 60 mg. How long afer he process began will i ake o decay o 10 mg? 71. If $1000 is invesed in an accoun earning 3% compounded monhly, how long will i ake he accoun o grow in value o $1500? 7. If $1000 is invesed in an accoun earning % compounded quarerly, how long will i ake he accoun o grow in value o $1300?

Secion 4.4 Logarihmic Properies 53 Secion 4.4 Logarihmic Properies In he previous secion, we derived wo imporan properies of logarihms, which allowed us o solve some basic eponenial and logarihmic equaions. Properies of Logs Inverse Properies: log b b ( ) b = log b = Eponenial Propery: r log A = r log A b ( ) ( ) Change of Base: log c ( A) logb ( A) = log ( b) c b While hese properies allow us o solve a large number of problems, hey are no sufficien o solve all problems involving eponenial and logarihmic equaions. Properies of Logs Sum of Logs Propery: log A + log C log ( AC ( ) ( ) ) b b = Difference of Logs Propery: A log b ( A) logb ( C) = logb C b I s jus as imporan o know wha properies logarihms do no saisfy as o memorize he valid properies lised above. In paricular, he logarihm is no a linear funcion, which means ha i does no disribue: log(a + B) log(a) + log(b). To help in his process we offer a proof o help solidify our new rules and show how hey follow from properies you ve already seen. Le a = log ( A) and c log ( C) b =, so by definiion of he logarihm, b a = A and b c = C b

54 Chaper 4 Using hese epressions, AC = b a b c a+ c Using eponen rules on he righ, AC = b Taking he log of boh sides, and uilizing he inverse propery of logs, a+ c log b ( AC) = logb ( b ) = a + c Replacing a and c wih heir definiion esablishes he resul log AC = log A + log b ( ) C b b The proof for he difference propery is very similar. Wih hese properies, we can rewrie epressions involving muliple logs as a single log, or break an epression involving a single log ino epressions involving muliple logs. Eample 1 Wrie log3( 5) + log3( 8) log3( ) as a single logarihm. Using he sum of logs propery on he firs wo erms, log3 ( 5) + log3( 8) = log3( 5 8) = log3( 40) This reduces our original epression o ( 40) log ( ) Then using he difference of logs propery, 40 log3 ( 40) log3( ) = log3 = log3( 0) log 3 3 Eample log 5 + log 4 wihou a calculaor by firs rewriing as a single logarihm. Evaluae ( ) ( ) On he firs erm, we can use he eponen propery of logs o wrie log 5 = log 5 = log 5 ( ) ( ) ( ) Wih he epression reduced o a sum of wo logs, ( 5) log( 4) sum of logs propery log 5 + log 4 = log(4 5) = log(100 ( ) ( ) ) Since 100 = 10, we can evaluae his log wihou a calculaor: log(100) = log 10 = ( ) log +, we can uilize he Try i Now 1. Wihou a calculaor evaluae by firs rewriing as a single logarihm: log ( 8) + log ( 4)

Secion 4.4 Logarihmic Properies 55 Eample 3 Rewrie 4 y ln 7 as a sum or difference of logs Firs, noicing we have a quoien of wo epressions, we can uilize he difference propery of logs o wrie 4 y 4 ln = ln( ) ln(7) 7 y Then seeing he produc in he firs erm, we use he sum propery 4 4 ln y ln(7) = ln + ln( y) ln(7 ( ) ( ) ) Finally, we could use he eponen propery on he firs erm ln 4 + ln( y) ln(7) = 4ln( ) + ln( y) ln(7 ( ) ) Ineresingly, solving eponenial equaions was no he reason logarihms were originally developed. Hisorically, up unil he adven of calculaors and compuers, he power of logarihms was ha hese log properies reduced muliplicaion, division, roos, or powers o be evaluaed using addiion, subracion, division and muliplicaion, respecively, which are much easier o compue wihou a calculaor. Large books were published lising he logarihms of numbers, such as in he able o he righ. To find he produc of wo numbers, he sum of log propery was used. Suppose for eample we didn know he value of imes 3. Using he sum propery of logs: log( 3) = log() + log(3) value log(value) 1 0.0000000 0.3010300 3 0.477113 4 0.600600 5 0.6989700 6 0.7781513 7 0.8450980 8 0.9030900 9 0.95445 10 1.0000000 Using he log able, log( 3) = log() + log(3) = 0.3010300 + 0.477113 = 0.7781513 We can hen use he able again in reverse, looking for 0.7781513 as an oupu of he logarihm. From ha we can deermine: log( 3) = 0.7781513 = log(6). By doing addiion and he able of logs, we were able o deermine 3 = 6. Likewise, o compue a cube roo like 3 8 1/ 3 1 1 log( 8) == log 8 = log(8) = (0.9030900) = 0.3010300 3 3 3 So 8 =. ( ) log() 3 =

56 Chaper 4 Alhough hese calculaions are simple and insignifican hey illusrae he same idea ha was used for hundreds of years as an efficien way o calculae he produc, quoien, roos, and powers of large and complicaed numbers, eiher using ables of logarihms or mechanical ools called slide rules. These properies sill have oher pracical applicaions for inerpreing changes in eponenial and logarihmic relaionships. Eample 4 Recall ha in chemisry, ph log ( H + ) =. If he concenraion of hydrogen ions in a liquid is doubled, wha is he affec on ph? Suppose C is he original concenraion of hydrogen ions, and P is he original ph of he liquid, so P = log( C). If he concenraion is doubled, he new concenraion is C. Then he ph of he new liquid is ph = log C ( ) Using he sum propery of logs, ph = log C = log() + log( C) Since P log( C) ( ) ( ) = log() log( C) =, he new ph is ph = P log( ) = P 0.301 When he concenraion of hydrogen ions is doubled, he ph decreases by 0.301. Log properies in solving equaions The logarihm properies ofen arise when solving problems involving logarihms. Eample 5 Solve log( 50 + 5) log( ) =. In order o rewrie in eponenial form, we need a single logarihmic epression on he lef side of he equaion. Using he difference propery of logs, we can rewrie he lef side: 50 + 5 log = Rewriing in eponenial form reduces his o an algebraic equaion: 50 + 5 = 10 = 100

Secion 4.4 Logarihmic Properies 57 Solving, 50 + 5 = 100 5 = 50 = 5 50 = 1 Checking his answer in he original equaion, we can verify here are no domain issues, and his answer is correc. Try i Now. Solve log( 4) = 1+ log( + ). More comple eponenial equaions can ofen be solved in more han one way. In he following eample, we will solve he same problem in wo ways one using logarihm properies, and he oher using eponenial properies. Eample 6a In 008, he populaion of Kenya was approimaely 38.8 million, and was growing by.64% each year, while he populaion of Sudan was approimaely 41.3 million and growing by.4% each year. If hese rends coninue, when will he populaion of Kenya mach ha of Sudan? We sar by wriing an equaion for each populaion in erms of, he number of years afer 008. Kenya( ) = 38.8(1 + 0.064) Sudan( ) = 41.3(1 + 0.04) To find when he populaions will be equal, we can se he equaions equal 38.8(1.064) = 41.3(1.04) For our firs approach, we ake he log of boh sides of he equaion log 38.8(1.064) = log 41.3(1.04) ( ) ( ) Uilizing he sum propery of logs, we can rewrie each side, log(38.8) log 1.064 + = log(41.3) + log 1.04 ( ) ( ) Then uilizing he eponen propery, we can pull he variables ou of he eponen World Bank, World Developmen Indicaors, as repored on hp://www.google.com/publicdaa, rerieved Augus 4, 010

58 Chaper 4 ( ) ( ) log(38.8) + log 1.064 = log(41.3) + log 1.04 Moving all he erms involving o one side of he equaion and he res of he erms o he oher side, log 1.064 log 1.04 = log(41.3) log(38.8) ( ) ( ) Facoring ou he on he lef, log 1.064 log 1.04 = log(41.3) log(38.8) ( ( ) ( )) Dividing o solve for log(41.3) log(38.8) = 15.991years unil he populaions will be equal. log 1.064 log 1.04 ( ) ( ) Eample 6b Solve he problem above by rewriing before aking he log. Saring a he equaion 38.8(1.064) = 41.3(1.04) Divide o move he eponenial erms o one side of he equaion and he consans o he oher side 1.064 41.3 = 1.04 38.8 Using eponen rules o group on he lef, 1.064 41.3 = 1.04 38.8 Taking he log of boh sides 1.064 41.3 log = log 1.04 38.8 Uilizing he eponen propery on he lef, 1.064 41.3 log = log 1.04 38.8 Dividing gives 41.3 log 38.8 = 15.991 years 1.064 log 1.04

Secion 4.4 Logarihmic Properies 59 While he answer does no immediaely appear idenical o ha produced using he previous mehod, noe ha by using he difference propery of logs, he answer could be rewrien: 41.3 log 38.8 log(41.3) log(38.8) = = 1.064 log(1.064) log(1.04) log 1.04 While boh mehods work equally well, i ofen requires fewer seps o uilize algebra before aking logs, raher han relying solely on log properies. Try i Now 3. Tank A conains 10 liers of waer, and 35% of he waer evaporaes each week. Tank B conains 30 liers of waer, and 50% of he waer evaporaes each week. In how many weeks will he anks conain he same amoun of waer? Imporan Topics of his Secion Inverse Eponenial Change of base Sum of logs propery Difference of logs propery Solving equaions using log rules Try i Now Answers 1. 5. 1 3. 4.1874 weeks

60 Chaper 4 Secion 4.4 Eercises Simplify o a single logarihm, using logarihm properies. log 8 log 7 log 3 log 4 1. ( ) ( ). ( ) ( ) 3 3 3 3 1 3. log3 7 1 + 10 5. log log ( 50) 3 3 1 4. log4 5 6. ( ) log 3 + log (7) 4 4 1 log 7 8 3 7. ( ) 1 log 5 36 8. ( ) 4 5 9. log ( ) + log ( 3 ) 3 10. ln ( 4 ) + ln ( 3 ) 9 11. ln ( 6 ) ln ( 3 ) 4 1. log ( 1 ) log ( 4) + + 14. 3log ( ) + log ( ) 13. log ( ) 3log ( 1) 1 15. log ( ) log ( y) + 3log ( z) 16. log ( ) + log ( y) log ( z) 1 3 Use logarihm properies o epand each epression. 15 13 17. log y 19 18. log ab 5 z c a 19. ln 4 5 b c 0. 3 ln a b 5 c 3 3 4 1. log ( y ) 3. log ( y ) 3. ln y y 1 y 4. ln 1 3 5 5. log ( y ) 3 y 6. ( ) 3 4 7 log y y 3 9

Secion 4.4 Logarihmic Properies 61 Solve each equaion for he variable. 4 7 9 6 7. 4 = 3 8. = 7 5 3 7 9. 17( 1.14) = 19( 1.16) 30. 0( 1.07) = 8( 1.13) 31. 0.1 0.08 5e = 10e 3. 3e = e 0.09 0.14 log 7 + 6 = 3 34. log 3( + 4) = 33. ( ) 35. ln ( 3) + 3 = 1 36. ( ) 4ln 5 + 5 = 3 5 37. log ( ) = 38. ( ) log = 3 39. log ( ) + log ( + 3) = 3 40. ( ) ( ) log + 4 + log = 9 41. log ( + 4) log ( + 3) = 1 4. ( ) ( ) log + 5 log + = 43. ( ) log log ( + 1) = 1 44. 6 6 log ( ) log ( + ) = 5 3 3 45. log ( + 1) = log ( ) + log ( 1) 46. log ( + 15) = log ( ) + log ( 15) 47. ln ( ) + ln ( 3) = ln ( 7) 48. ln ( ) + ln ( 6) = ln ( 6)

6 Chaper 4 Secion 4.5 Graphs of Logarihmic Funcions Recall ha he eponenial funcion f ( ) = produces his able of values -3 - -1 0 1 3 1 1 1 f() 1 4 8 8 4 Since he logarihmic funcion is an inverse of he eponenial, g ( ) = log ( ) produces he able of values 1 1 1 1 4 8 8 4 g() -3 - -1 0 1 3 In his second able, noice ha 1) As he inpu increases, he oupu increases. ) As inpu increases, he oupu increases more slowly. 3) Since he eponenial funcion only oupus posiive values, he logarihm can only accep posiive values as inpus, so he domain of he log funcion is ( 0, ). 4) Since he eponenial funcion can accep all real numbers as inpus, he logarihm can oupu any real number, so he range is all real numbers or (, ). Skeching he graph, noice ha as he inpu approaches zero from he righ, he oupu of he funcion grows very large in he negaive direcion, indicaing a verical asympoe a = 0. In symbolic noaion we wrie as 0 +, f ( ), and as, f ( ) Graphical Feaures of he Logarihm Graphically, in he funcion g ( ) = log b( ) The graph has a horizonal inercep a (1, 0) The graph has a verical asympoe a = 0 The graph is increasing and concave down The domain of he funcion is > 0, or ( 0, ) The range of he funcion is all real numbers, or (, ) When skeching a general logarihm wih base b, i can be helpful o remember ha he graph will pass hrough he poins (1, 0) and (b, 1).

Secion 4.5 Graphs of Logarihmic Funcions 63 To ge a feeling for how he base affecs he shape of he graph, eamine he graphs below. log ( ) ln( ) log( ) Noice ha he larger he base, he slower he graph grows. For eample, he common log graph, while i grows wihou bound, i does so very slowly. For eample, o reach an oupu of 8, he inpu mus be 100,000,000. Anoher imporan observaion made was he domain of he logarihm. Like he reciprocal and square roo funcions, he logarihm has a resriced domain which mus be considered when finding he domain of a composiion involving a log. Eample 1 Find he domain of he funcion f ( ) = log(5 ) The logarihm is only defined wih he inpu is posiive, so his funcion will only be defined when 5 > 0. Solving his inequaliy, > 5 5 < The domain of his funcion is 5 5 <, or in inerval noaion,, Try i Now 1. Find he domain of he funcion f ( ) = log( 5) + ; before solving his as an inequaliy, consider how he funcion has been ransformed.

64 Chaper 4 Transformaions of he Logarihmic Funcion Transformaions can be applied o a logarihmic funcion using he basic ransformaion echniques, bu as wih eponenial funcions, several ransformaions resul in ineresing relaionships. log c 1 Firs recall he change of base propery ells us ha log b = = log c log c b log c b From his, we can see ha log b is a verical srech or compression of he graph of he log c graph. This ells us ha a verical srech or compression is equivalen o a change of base. For his reason, we ypically represen all graphs of logarihmic funcions in erms of he common or naural log funcions. Ne, consider he effec of a horizonal compression on he graph of a logarihmic funcion. Considering f ( ) = log( c), we can use he sum propery o see f ( ) = log( c) = log( c) + log( ) Since log(c) is a consan, he effec of a horizonal compression is he same as he effec of a verical shif. Eample Skech f ( ) = ln( ) and g ( ) = ln( ) +. Graphing hese, g ( ) = ln( ) + f ( ) = ln( ) Noe ha, his verical shif could also be wrien as a horizonal compression: g ( ) = ln( ) + = ln( ) + ln( e ) = ln( e ). While a horizonal srech or compression can be wrien as a verical shif, a horizonal reflecion is unique and separae from verical shifing. Finally, we will consider he effec of a horizonal shif on he graph of a logarihm

Secion 4.5 Graphs of Logarihmic Funcions 65 Eample 3 Skech a graph of f ( ) = ln( + ). This is a horizonal shif o he lef by unis. Noice ha none of our logarihm rules allow us rewrie his in anoher form, so he effec of his ransformaion is unique. Shifing he graph, Noice ha due o he horizonal shif, he verical asympoe shifed o = -, and he domain shifed o (, ). Combining hese ransformaions, Eample 4 Skech a graph of f ( ) = 5log( + ). Facoring he inside as f ( ) = 5log( ( )) reveals ha his graph is ha of he common logarihm, horizonally refleced, verically sreched by a facor of 5, and shifed o he righ by unis. The verical asympoe will be shifed o =, and he graph will have domain (, ). A rough skech can be creaed by using he verical asympoe along wih a couple poins on he graph, such as f (1) = 5log( 1+ ) = 5log(1) = 0 f ( 8) = 5log( ( 8) + ) = 5log(10) = 5 Try i Now. Skech a graph of he funcion f ( ) = 3log( ) + 1.

66 Chaper 4 Transformaions of Logs Any ransformed logarihmic funcion can be wrien in he form f( ) alog( b) k f = a b + k if horizonally refleced, ( ) log ( ) = +, or ( ) where = b is he verical asympoe. Eample 5 Find an equaion for he logarihmic funcion graphed below. This graph has a verical asympoe a = and has been verically refleced. We do no know ye he verical shif (equivalen o horizonal srech) or he verical srech (equivalen o a change of base). We know so far ha he equaion will have form f ( ) = a log( + ) + k I appears he graph passes hrough he poins ( 1, 1) and (, 1). Subsiuing in ( 1, 1), 1 = a log( 1+ ) + k 1 = a log(1) + k 1 = k Ne, subsiuing in (, 1), 1 = a log( + ) + 1 = a log(4) a = log(4) This gives us he equaion f ( ) = log( + ) + 1. log(4) This could also be wrien as f( ) = log 4( + ) + 1. Flashback 3. Wrie he domain and range of he funcion graphed in Eample 5, and describe is long run behavior.

Secion 4.5 Graphs of Logarihmic Funcions 67 Imporan Topics of his Secion Graph of he logarihmic funcion (domain and range) Transformaion of logarihmic funcions Creaing graphs from equaions Creaing equaions from graphs Try i Now Answers 1. Domain: { > 5}. Flashback Answers 3. Domain: { >-}, Range: all real numbers; As +, f ( ) and as, f ( ).

68 Chaper 4 Secion 4.5 Eercises For each funcion, find he domain and he verical asympoe. f log 5 f = log + 1. ( ) = ( ). ( ) ( ) 3. f ( ) = ln ( 3 ) 4. f ( ) = ln ( 5 ) 5. f ( ) = log ( 3+ 1) 6. f ( ) = log ( + 5) 7. f ( ) = 3log ( ) + 8. f ( ) ( ) = log + 1 Skech a graph of each pair of funcion. f log, g ln 9. ( ) = ( ) ( ) = ( ) 10. f ( ) = log ( ), g( ) = log ( ) Skech each ransformaion. f log 11. ( ) = ( ) 1. f ( ) = 3ln ( ) 13. f ( ) = ln ( ) 14. f ( ) = log ( ) 4 15. f ( ) = log ( + ) 16. f ( ) = log ( + 4) 3 Find a formula for he ransformed logarihm graph shown. 17. 18. 19. 0.

Secion 4.5 Graphs of Logarihmic Funcions 69 Find a formula for he ransformed logarihm graph shown. 1.. 3. 4.

70 Chaper 4 Secion 4.6 Eponenial and Logarihmic Models While we have eplored some basic applicaions of eponenial and logarihmic funcions, in his secion we eplore some imporan applicaions in more deph. Radioacive Decay In an earlier secion, we discussed radioacive decay he idea ha radioacive isoopes change over ime. One of he common erms associaed wih radioacive decay is halflife. Half Life The half-life of a radioacive isoope is he ime i akes for half he subsance o decay. Given he basic eponenial growh/decay equaion (, half-life can be found by h ) = ab solving for when half he original amoun remains; by solving simply 1 = b 1 a a( b) =, or more. Noice how he iniial amoun is irrelevan when solving for half-life. Eample 1 Bismuh-10 is an isoope ha decays by abou 13% each day. Wha is he half-life of Bismuh-10? We were no given a saring quaniy, so we could eiher make up a value or use an unknown consan o represen he saring amoun. To show ha saring quaniy does no affec he resul, le us denoe he iniial quaniy by he consan a. Then he decay d of Bismuh-10 can be described by he equaion Q ( d) = a(0.87). To find he half-life, we wan o deermine when he remaining quaniy is half he original: 1 a. Solving, 1 d a = a(0.87) Dividing by a, 1 = d 0.87 Take he log of boh sides 1 d log = log( 0.87 ) Use he eponen propery of logs 1 log = d log( 0.87) Divide o solve for d

Secion 4.6 Eponenial and Logarihmic Models 71 1 log d = 4.977 days log 0.87 ( ) This ells us ha he half-life of Bismuh-10 is approimaely 5 days. Eample Cesium-137 has a half-life of abou 30 years. If you begin wih 00mg of cesium-137, how much will remain afer 30 years? 60 years? 90 years? Since he half-life is 30 years, afer 30 years, half he original amoun, 100mg, will remain. Afer 60 years, anoher 30 years have passed, so during ha second 30 years, anoher half of he subsance will decay, leaving 50mg. Afer 90 years, anoher 30 years have passed, so anoher half of he subsance will decay, leaving 5mg. Eample 3 Cesium-137 has a half-life of abou 30 years. Find he annual decay rae. Since we are looking for an annual decay rae, we will use an equaion of he form Q ( ) = a(1 + r). We know ha afer 30 years, half he original amoun will remain. Using his informaion 1 30 a = a(1 + r) Dividing by a 1 30 = (1 + r) Taking he 30 h roo of boh sides 1 30 = 1+ r Subracing one from boh sides, 1 r = 30 1 0.084 This ells us cesium-137 is decaying a an annual rae of.84% per year. Try i Now Chlorine-36 is eliminaed from he body wih a biological half-life of 10 days 3. Find he daily decay rae. 3 hp://www.ead.anl.gov/pub/doc/chlorine.pdf

7 Chaper 4 Eample 4 Carbon-14 is a radioacive isoope ha is presen in organic maerials, and is commonly used for daing hisorical arifacs. Carbon-14 has a half-life of 5730 years. If a bone fragmen is found ha conains 0% of is original carbon-14, how old is he bone? To find how old he bone is, we firs will need o find an equaion for he decay of he carbon-14. We could eiher use a coninuous or annual decay formula, bu op o use he coninuous decay formula since i is more common in scienific es. The half life ells us ha afer 5730 years, half he original subsance remains. Solving for he rae, 1 r5730 a = ae Dividing by a 1 r5730 = e Taking he naural log of boh sides 1 r5730 ln = ln( e ) Use he inverse propery of logs on he righ side 1 ln = 5730r Divide by 5730 1 ln r = 0.00011 5730 0.00011 Now we know he decay will follow he equaion Q( ) = ae. To find how old he bone fragmen is ha conains 0% of he original amoun, we solve for so ha Q() = 0.0a. 0.00011 0.0a = ae 0.0 = e 0. 00011 ln(0.0) = ln( e 0. 00011 ) ln( 0.0) = 0.00011 ln(0.0) = 13301 years 0.00011 The bone fragmen is abou 13,300 years old. Try i Now. In Eample, we learned ha Cesium-137 has a half-life of abou 30 years. If you begin wih 00mg of cesium-137, will i ake more or less han 30 years unil only 1 milligram remains?

Secion 4.6 Eponenial and Logarihmic Models 73 Doubling Time For decaying quaniies, we asked how long i akes for half he subsance o decay. For growing quaniies we migh ask how long i akes for he quaniy o double. Doubling Time The doubling ime of a growing quaniy is he ime i akes for he quaniy o double. Given he basic eponenial growh equaion h = solving for when he original quaniy has doubled; by solving ( ) ab, doubling ime can be found by a = a( b), or more simply = b. Again noice how he iniial amoun is irrelevan when solving for doubling ime. Eample 5 Cancer cells someimes increase eponenially. If a cancerous growh conained 300 cells las monh and 360 cells his monh, how long will i ake for he number of cancer cells o double? Defining o be ime in monhs, wih = 0 corresponding o his monh, we are given wo pieces of daa: his monh, (0, 360), and las monh, (-1, 300). From his daa, we can find an equaion for he growh. Using he form C ( ) = ab, we know immediaely a = 360, giving C( ) = 360b. Subsiuing in (-1, 300), 1 300 = 360b 360 300 = b 360 b = = 1. 300 This gives us he equaion C ( ) = 360(1.) To find he doubling ime, we look for he ime unil we have wice he original amoun, so when C() = a. a = a(1.) = (1.) log ( ) = log( 1. ) log( ) = log( 1.) log( ) = 3.80 monhs. log( 1.) I will ake abou 3.8 monhs for he number of cancer cells o double.

74 Chaper 4 Eample 6 Use of a new social neworking websie has been growing eponenially, wih he number of new members doubling every 5 monhs. If he sie currenly has 10,000 users and his rend coninues, how many users will he sie have in 1 year? We can use he doubling ime o find a funcion ha models he number of sie users, and hen use ha equaion o answer he quesion. While we could use an arbirary a as we have before for he iniial amoun, in his case, we know he iniial amoun was 10,000. r If we use a coninuous growh equaion, i would look like N( ) = 10e, measured in housands of users afer monhs. Based on he doubling ime, here would be 40 housand users afer 5 monhs. This allows us o solve for he coninuous growh rae: r5 40 = 10e r5 = e ln = 5r ln r = 0.1386 5 0.1386 Now ha we have an equaion, N( ) = 10e, we can predic he number of users afer 1 monhs: 0.1386(1) N (1) = 10e = 633.140 housand users. So afer 1 year, we would epec he sie o have around 633,140 users. Try i Now 3. If uiion a a college is increasing by 6.6% each year, how many years will i ake for uiion o double? Newon s Law of Cooling When a ho objec is lef in surrounding air ha is a a lower emperaure, he objec s emperaure will decrease eponenially, leveling off owards he surrounding air emperaure. This "leveling off" will correspond o a horizonal asympoe in he graph of he emperaure funcion. Unless he room emperaure is zero, his will correspond o a verical shif of he generic eponenial decay funcion.

Secion 4.6 Eponenial and Logarihmic Models 75 Newon s Law of Cooling The emperaure of an objec, T, in surrounding air wih emperaure T s will behave according o he formula k T ( ) = ae + Ts Where is ime a is a consan deermined by he iniial emperaure of he objec k is a consan, he coninuous rae of cooling of he objec While an equaion of he form is more common. T ) = ab + ( T could be used, he coninuous growh form s Eample 7 A cheesecake is aken ou of he oven wih an ideal inernal emperaure of 165 degrees Fahrenhei, and is placed ino a 35 degree refrigeraor. Afer 10 minues, he cheesecake has cooled o 150 degrees. If you mus wai unil he cheesecake has cooled o 70 degrees before you ea i, how long will you have o wai? Since he surrounding air emperaure in he refrigeraor is 35 degrees, he cheesecake s emperaure will decay eponenially owards 35, following he equaion k T ( ) = ae + 35 We know he iniial emperaure was 165, so T ( 0) = 165. Subsiuing in hese values, 0 165 = ae k + 35 165 = a + 35 a = 130 We were given anoher pair of daa, T ( 10) = 150, which we can use o solve for k k 150 = 130e 10 + 35 k10 115 = 130e 115 10k = e 130 115 ln = 10k 130 115 ln 130 k = = 0.013 10 0.013 Togeher his gives us he equaion for cooling: T ( ) = 130e + 35.

76 Chaper 4 Now we can solve for he ime i will ake for he emperaure o cool o 70 degrees. 0.013 70 = 130e + 35 35 = 130e 0. 013 35 = e 0. 013 130 35 ln = 0.013 130 35 ln 130 = 106.68 0.013 I will ake abou 107 minues, or one hour and 47 minues, for he cheesecake o cool. Of course, if you like your cheesecake served chilled, you d have o wai a bi longer. Try i Now 4. A picher of waer a 40 degrees Fahrenhei is placed ino a 70 degree room. One hour laer he emperaure has risen o 45 degrees. How long will i ake for he emperaure o rise o 60 degrees? Logarihmic Scales For quaniies ha vary grealy in magniude, a sandard scale of measuremen is no always effecive, and uilizing logarihms can make he values more manageable. For eample, if he average disances from he sun o he major bodies in our solar sysem are lised, you see hey vary grealy. Plane Disance (millions of km) Mercury 58 Venus 108 Earh 150 Mars 8 Jupier 779 Saurn 1430 Uranus 880 Nepune 4500 Placed on a linear scale one wih equally spaced values hese values ge bunched up. Mercury Venus Earh Mars Jupier Saurn Uranus Nepune 0 500 1000 1500 000 500 3000 3500 4000 4500 disance

Secion 4.6 Eponenial and Logarihmic Models 77 However, compuing he logarihm of each value and ploing hese new values on a number line resuls in a more manageable graph, and makes he relaive disances more apparen. 4 Plane Disance (millions of km) log(disance) Mercury 58 1.76 Venus 108.03 Earh 150.18 Mars 8.36 Jupier 779.89 Saurn 1430 3.16 Uranus 880 3.46 Nepune 4500 3.65 Mercury Venus Earh Mars Jupier Saurn Uranus Nepune 1.5 1.75.5.5.75 3 3.5 3.5 3.75 4 log(disance) 10 =100 10 3 =100 10 4 =10000 Someimes, as shown above, he scale on a logarihmic number line will show he log values, bu more commonly he original values are lised as powers of 10, as shown below. P A B C D 10-10 -1 10 0 10 1 10 10 3 10 4 10 5 10 6 10 7 Eample 8 Esimae he value of poin P on he log scale above The poin P appears o be half way beween - and -1 in log value, so if V is he value of his poin, log( V ) 1.5 Rewriing in eponenial form, V 10 1. 5 = 0.0316 4 I is ineresing o noe he large gap beween Mars and Jupier on he log number line. The aseroid bel, which scieniss believe consiss of he remnans of an ancien plane, is locaed here.

78 Chaper 4 Eample 9 Place he number 6000 on a logarihmic scale. Since log( 6000) 3. 8, his poin would belong on he log scale abou here: 6000 10-10 -1 10 0 10 1 10 10 3 10 4 10 5 10 6 10 7 Try i Now 5. Plo he daa in he able below on a logarihmic scale 5. Source of Sound/Noise Approimae Sound Pressure in µpa (micro Pascals) Launching of he Space Shule,000,000,000 Full Symphony Orchesra,000,000 Diesel Freigh Train a High Speed a 5 m 00,000 Normal Conversaion 0,000 Sof Whispering a m in Library,000 Unoccupied Broadcas Sudio 00 Sofes Sound a human can hear 0 Noice ha on he log scale above Eample 8, he visual disance on he scale beween poins A and B and beween C and D is he same. When looking a he values hese poins correspond o, noice B is en imes he value of A, and D is en imes he value of C. A visual linear difference beween poins corresponds o a relaive (raio) change beween he corresponding values. Logarihms are useful for showing hese relaive changes. For eample, comparing $1,000,000 o $10,000, he firs is 100 imes larger han he second. 1,000,000 = 100 = 10 10,000 Likewise, comparing $1000 o $10, he firs is 100 imes larger han he second. 1,000 = 100 = 10 10 When one quaniy is roughly en imes larger han anoher, we say i is one order of magniude larger. In boh cases described above, he firs number was wo orders of magniude larger han he second. 5 From hp://www.epd.gov.hk/epd/noise_educaion/web/eng_epd_html/m1/inro_5.hml, rerieved Oc, 010

Secion 4.6 Eponenial and Logarihmic Models 79 Noice ha he order of magniude can be found as he common logarihm of he raio of he quaniies. On he log scale above, B is one order of magniude larger han A, and D is one order of magniude larger han C. Orders of Magniude Given wo values A and B, o deermine how many orders of magniude A is greaer han B, A Difference in orders of magniude = log B Eample 10 On he log scale above Eample 8, how many orders of magniude larger is C han B? The value B corresponds o 10 = 100 The value C corresponds o 10 5 = 100, 000 5 100,000 10 3 The relaive change is = 1000 = = 10. The log of his value is 3. 100 10 C is hree orders of magniude greaer han B, which can be seen on he log scale by he visual difference beween he poins on he scale. Try i Now 6. Using he able from Try i Now #5, wha is he difference of order of magniude beween he sofes sound a human can hear and he launching of he space shule? An eample of a logarihmic scale is he Momen Magniude Scale (MMS) used for earhquakes. This scale is commonly and misakenly called he Richer Scale, which was a very similar scale succeeded by he MMS. Momen Magniude Scale For an earhquake wih seismic momen S, a measuremen of earh movemen, he MMS value, or magniude of he earhquake, is S M = log 3 S0 16 Where S 0 = 10 is a baseline measure for he seismic momen.

80 Chaper 4 Eample 11 If one earhquake has a MMS magniude of 6.0, and anoher has a magniude of 8.0, how much more powerful (in erms of earh movemen) is he second earhquake? Since he firs earhquake has magniude 6.0, we can find he amoun of earh movemen. The value of S 0 is no paricularly relevan, so we will no replace i wih is value. S 6.0 = log 3 S0 3 6.0 = log S 9 = log S0 S 9 = 10 S 0 S = 9 10 S0 S S 0 Doing he same wih he second earhquake wih a magniude of 8.0, S 8.0 = log 3 S0 1 S = 10 S 0 From his, we can see ha his second value s earh movemen is 1000 imes as large as he firs earhquake. Eample 1 One earhquake has magniude of 3.0. If a second earhquake has wice as much earh movemen as he firs earhquake, find he magniude of he second quake. Since he firs quake has magniude 3.0,

Secion 4.6 Eponenial and Logarihmic Models 81 0 4.5 0 4.5 0 0 0 10 10 log 4.5 log 3 3.0 log 3 3.0 S S S S S S S S S S = = = = = Since he second earhquake has wice as much earh movemen, for he second quake, 0 4.5 10 S S = Finding he magniude, = 0 0 4.5 10 log 3 S S M ( ) 01 3. 10 log 3 5 4. = M The second earhquake wih wice as much earh movemen will have a magniude of abou 3.. In fac, using log properies, we could show ha whenever he earh movemen doubles, he magniude will increase by abou 0.01: = = 0 0 log 3 log 3 S S S S M + = 0 log log() 3 S S M + = 0 log 3 log() 3 S S M + = 0 log 3 0.01 S S M This illusraes he mos imporan feaure of a log scale: ha muliplying he quaniy being considered will add o he scale value, and vice versa.

8 Chaper 4 Imporan Topics of his Secion Radioacive decay Half life Doubling ime Newon s law of cooling Logarihmic Scales Orders of Magniude Momen Magniude scale Try i Now Answers 1 1. r = 10 1 0. 067 or 6.7% is he daily rae of decay.. Less han 30 years, 9.3157 o be eac 3. I will ake 10.845 years, or approimaely 11 years, for uiion o double. 4. 6.06 hours 5. Broadcas Conversaion Sofes room Sof Sound Whisper Train Symphony Space Shule 10 1 10 10 3 10 4 10 5 10 6 10 7 10 8 10 9 10 10 9 10 8 6. = 10 The sound pressure in µpa creaed by launching he space shule is 8 1 10 orders of magniude greaer han he sound pressure in µpa creaed by he sofes sound a human ear can hear.

Secion 4.6 Eponenial and Logarihmic Models 83 Secion 4.6 Eercises 1. You go o he docor and he injecs you wih 13 milligrams of radioacive dye. Afer 1 minues, 4.75 milligrams of dye remain in your sysem. To leave he docor's office, you mus pass hrough a radiaion deecor wihou sounding he alarm. If he deecor will sound he alarm whenever more han milligrams of he dye are in your sysem, how long will your visi o he docor ake, assuming you were given he dye as soon as you arrived and he amoun of dye decays eponenially?. You ake 00 milligrams of a headache medicine, and afer 4 hours, 10 milligrams remain in your sysem. If he effecs of he medicine wear off when less han 80 milligrams remain, when will you need o ake a second dose, assuming he amoun of medicine in your sysem decays eponenially? 3. The half-life of Radium-6 is 1590 years. If a sample iniially conains 00 mg, how many milligrams will remain afer 1000 years? 4. The half-life of Fermium-53 is 3 days. If a sample iniially conains 100 mg, how many milligrams will remain afer 1 week? 5. The half-life of Erbium-165 is 10.4 hours. Afer 4 hours a sample sill conains mg. Wha was he iniial mass of he sample, and how much will remain afer anoher 3 days? 6. The half-life of Nobelium-59 is 58 minues. Afer 3 hours a sample sill conains10 mg. Wha was he iniial mass of he sample, and how much will remain afer anoher 8 hours? 7. A scienis begins wih 50 grams of a radioacive subsance. Afer 5 minues, he sample has decayed o 3 grams. Find he half-life of his subsance. 8. A scienis begins wih 0 grams of a radioacive subsance. Afer 7 days, he sample has decayed o 17 grams. Find he half-life of his subsance. 9. A wooden arifac from an archeological dig conains 60 percen of he carbon-14 ha is presen in living rees. How long ago was he arifac made? (The half-life of carbon-14 is 5730 years.)

84 Chaper 4 10. A wooden arifac from an archeological dig conains 15 percen of he carbon-14 ha is presen in living rees. How long ago was he arifac made? (The half-life of carbon-14 is 5730 years.) 11. A baceria culure iniially conains 1500 baceria and doubles in size every half hour. Find he size of he populaion afer: a) hours b) 100 minues 1. A baceria culure iniially conains 000 baceria and doubles in size every half hour. Find he size of he populaion afer: a) 3 hours b) 80 minues 13. The coun of baceria in a culure was 800 afer 10 minues and 1800 afer 40 minues. a. Wha was he iniial size of he culure? b. Find he doubling ime. c. Find he populaion afer 105 minues. d. When will he populaion reach 11000? 14. The coun of baceria in a culure was 600 afer 0 minues and 000 afer 35 minues. a. Wha was he iniial size of he culure? b. Find he doubling ime. c. Find he populaion afer 170 minues. d. When will he populaion reach 1000? 15. Find he ime required for an invesmen o double in value if invesed in an accoun paying 3% compounded quarerly. 16. Find he ime required for an invesmen o double in value if invesed in an accoun paying 4% compounded monhly = 0e. 17. The number of crysals ha have formed afer hours is given by ( ) 0.013 How long does i ake he number of crysals o double? 18. The number of building permis in Pasco years afer 199 roughly followed he 0.143 equaion n ( ) = 400e. Wha is he doubling ime? n

Secion 4.6 Eponenial and Logarihmic Models 85 19. A urkey is pulled from he oven when he inernal emperaure is 165 Fahrenhei, and is allowed o cool in a 75 room. If he emperaure of he urkey is 145 afer half an hour, a. Wha will he emperaure be afer 50 minues? b. How long will i ake he urkey o cool o 110? 0. A cup of coffee is poured a 190 Fahrenhei, and is allowed o cool in a 70 room. If he emperaure of he coffee is 170 afer half an hour, a. Wha will he emperaure be afer 70 minues? b. How long will i ake he coffee o cool o 10? 1. The populaion of fish in a farm-socked lake afer years could be modeled by he 1000 equaion P( ) =. 0.6 1 + 9e a. Skech a graph of his equaion. b. Wha is he iniial populaion of fish? c. Wha will he populaion be afer years? d. How long will i ake for he populaion o reach 900?. The number of people in a own who have heard a rumor afer days can be modeled 500 by he equaion N( ) = 0.7 1 49e. + a. Skech a graph of his equaion. b. How many people sared he rumor? c. How many people have heard he rumor afer 3 days? d. How long will i ake unil 300 people have heard he rumor? Find he value of he number shown on each logarihmic scale 3. 4. 5. 6. Plo each se of approimae values on a logarihmic scale. 10 4 7. Inensiy of sounds: Whisper: 10 W / m, Vacuum: 10 W / m, Je: 10 W / m 8. Mass: Amoeba: 5 10 g, Human: 5 10 g, Saue of Libery: 8 10 g

86 Chaper 4 9. The 1906 San Francisco earhquake had a magniude of 7.9 on he MMS scale. Laer here was an earhquake wih magniude 4.7 ha caused only minor damage. How many imes more inense was he San Francisco earhquake han he second one? 30. The 1906 San Francisco earhquake had a magniude of 7.9 on he MMS scale. Laer here was an earhquake wih magniude 6.5 ha caused less damage. How many imes more inense was he San Francisco earhquake han he second one? 31. One earhquake has magniude 3.9 on he MMS scale. If a second earhquake has 750 imes as much energy as he firs, find he magniude of he second quake. 3. One earhquake has magniude 4.8 on he MMS scale. If a second earhquake has 100 imes as much energy as he firs, find he magniude of he second quake. 33. A colony of yeas cells is esimaed o conain 10 6 cells a ime = 0. Afer collecing eperimenal daa in he lab, you decide ha he oal populaion of cells a ime 6 0.495105 hours is given by he funcion f ( ) = 10 e. [UW] a. How many cells are presen afer one hour? b. How long does i ake of he populaion o double?. c. Cherie, anoher member of your lab, looks a your noebook and says: Tha formula is wrong, my calculaions predic he formula for he number of yeas 6 cells is given by he funcion. ( ) 10 (.0477 ) 0.693147 f =. Should you be worried by Cherie s remark? d. Anja, a hird member of your lab working wih he same yeas cells, ook 6 6 hese wo measuremens: 7.46 10 cells afer 4 hours; 16.504 10 cells afer 6 hours. Should you be worried by Anja s resuls? If Anja s measuremens are correc, does your model over esimae or under esimae he number of yeas cells a ime? 34. As ligh from he surface peneraes waer, is inensiy is diminished. In he clear waers of he Caribbean, he inensiy is decreased by 15 percen for every 3 meers of deph. Thus, he inensiy will have he form of a general eponenial funcion. [UW] a. If he inensiy of ligh a he waer s surface is I 0, find a formula for Id, ( ) he inensiy of ligh a a deph of d meers. Your formula should depend on I0 and d. b. A wha deph will he ligh inensiy be decreased o 1% of is surface inensiy?

Secion 4.6 Eponenial and Logarihmic Models 87 35. Myoglobin and hemoglobin are oygen-carrying molecules in he human body. Hemoglobin is found inside red blood cells, which flow from he lungs o he muscles hrough he bloodsream. Myoglobin is found in muscle cells. The funcion p Y = M ( p) = calculaes he fracion of myoglobin sauraed wih oygen a a 1 + p given pressure p Torrs. For eample, a a pressure of 1 Torr, M(1) = 0.5, which means half of he myoglobin (i.e. 50%) is oygen sauraed. (Noe: More precisely, you need o use somehing called he parial pressure, bu he disincion is no imporan for.8 p his problem.) Likewise, he funcion Y = H( p) = calculaes he fracion.8.8 6 + p of hemoglobin sauraed wih oygen a a given pressure p. [UW] a. The graphs of M( p ) and H( p ) are given here on he domain 0 p 100; which is which? b. If he pressure in he lungs is 100 Torrs, wha is he level of oygen sauraion of he hemoglobin in he lungs? c. The pressure in an acive muscle is 0 Torrs. Wha is he level of oygen sauraion of myoglobin in an acive muscle? Wha is he level of hemoglobin in an acive muscle? d. Define he efficiency of oygen ranspor a a given pressure p o be M( p) H( p). Wha is he oygen ranspor efficiency a 0 Torrs? A 40 Torrs? A 60 Torrs? Skech he graph of M( p) H( p) ; are here condiions under which ranspor efficiency is maimized (eplain)? 36. The lengh of some fish are modeled by a von Beralanffy growh funcion. For 0.18 L = 00 1 0.957e where L () is Pacific halibu, his funcion has he form ( ) ( ) he lengh (in cenimeers) of a fish years old. [UW] a. Wha is he lengh of a newborn halibu a birh? b. Use he formula o esimae he lengh of a 6 year old halibu. c. A wha age would you epec he halibu o be 10 cm long? d. Wha is he pracical (physical) significance of he number 00 in he formula for L ()?

88 Chaper 4 37. A cancer cell lacks normal biological growh regulaion and can divide coninuously. Suppose a single mouse skin cell is cancerous and is mioic cell cycle (he ime for he cell o divide once) is 0 hours. The number of cells a ime grows according o an eponenial model. [UW] a. Find a formula C () for he number of cancerous skin cells afer hours. b. Assume a ypical mouse skin cell is spherical of radius 50 10 4 cm. Find he combined volume of all cancerous skin cells afer hours. When will he volume of cancerous cells be 1 cm 3? 38. A ship embarked on a long voyage. A he sar of he voyage, here were 500 ans in he cargo hold of he ship. One week ino he voyage, here were 800 ans. Suppose he populaion of ans is an eponenial funcion of ime. [UW] a. How long did i ake he populaion o double? b. How long did i ake he populaion o riple? c. When were here be 10,000 ans on board? d. There also was an eponenially growing populaion of aneaers on board. A he sar of he voyage here were 17 aneaers, and he populaion of aneaers doubled every.8 weeks. How long ino he voyage were here 00 ans per aneaer? 39. The populaions of ermies and spiders in a cerain house are growing eponenially. The house conains 100 ermies he day you move in. Afer 4 days, he house conains 00 ermies. Three days afer moving in, here are wo imes as many ermies as spiders. Eigh days afer moving in, here were four imes as many ermies as spiders. How long (in days) does i ake he populaion of spiders o riple? [UW]

Secion 4.7 Fiing Eponenial Models o Daa 89 Secion 4.7 Fiing Eponenial Models o Daa In he previous secion, we saw number lines using logarihmic scales. I is also common o see wo dimensional graphs wih one or boh aes using a logarihmic scale. One common use of a logarihmic scale on he verical ais is o graph quaniies ha are changing eponenially, since i helps reveal relaive differences. This is commonly used in sock chars, since values hisorically have grown eponenially over ime. Boh sock chars below show he Dow Jones Indusrial Average, from 198 o 010. Boh chars have a linear horizonal scale, bu he firs graph has a linear verical scale, while he second has a logarihmic verical scale. The firs scale is he one we are more familiar wih, and shows wha appears o be a srong eponenial rend, a leas up unil he year 000.

90 Chaper 4 Eample 1 There were sock marke drops in 199 and 008. Which was larger? In he firs graph, he sock marke drop around 008 looks very large, and in erms of dollar values, i was indeed a large drop. However he second graph shows relaive changes, and he drop in 009 seems less major on his graph, and in fac he drop saring in 199 was, percenage-wise, much more significan. Specifically, in 008, he Dow value dropped from abou 14,000 o 8,000, a drop of 6,000. This is obviously a large value drop, and amouns o abou a 43% drop. In 199, he Dow value dropped from a high of around 380 o a low of 4 by July of 193. While value-wise his drop of 338 is much smaller han he 008 drop, i corresponds o a 89% drop, a much larger relaive drop han in 008. The logarihmic scale shows hese relaive changes. The second graph above, in which one ais uses a linear scale and he oher ais uses a logarihmic scale, is an eample of a semi-log graph. Semi-log and Log-log Graphs A semi-log graph is a graph wih one ais using a linear scale and one ais using a logarihmic scale. A log-log graph is a graph wih boh aes using logarihmic scales. Eample Plo 5 poins on he graph of on he verical ais. ( = on a semi-log graph wih a logarihmic scale f ) 3() To do his, we need o find 5 poins on he graph, hen calculae he logarihm of he oupu value. Arbirarily choosing 5 inpu values, f() log(f()) -3 3 3() = 8-0.46-1 3 3() = 0.176 0 3() 0 = 3 0.477 3() = 1 1.079 5 3() 5 = 96 1.98

Secion 4.7 Fiing Eponenial Models o Daa 91 Ploing hese values on a semi-log graph, log(f()) 3 1 0-4 -3 - -1 0 1 3 4 5 6-1 Noice ha on his semi-log scale, values from he eponenial funcion appear linear. We can show his behavior is epeced by uilizing logarihmic properies. For he funcion f ( ) = ab, finding log(f()) gives ( f ( ) ) log( ab ) ( f ( ) ) log( a) log( b ) log = Uilizing he sum propery of logs, log = + Now uilizing he eponen propery, log f ( ) = log a + log b ( ) ( ) ( ) This relaionship is linear, wih log(a) as he verical inercep, and log(b) as he slope. This relaionship can also be uilized in reverse. Eample 3 An eponenial graph is ploed on a semi-log graph below. Find a formula for he eponenial funcion g() ha generaed his graph. log(g()) 5 4 3 1 0-4 -3 - -1-1 0 1 3 4 - -3 The graph is linear, wih verical inercep a (0, 1). Looking a he change beween he poins (0, 1) and (4, 4), we can deermine he slope of he line is 4 3. Since he oupu is 3 =. 4 log(g()), his leads o he equaion log ( g( ) ) 1+

9 Chaper 4 We can solve his formula for g() by rewriing in eponenial form and simplifying: 3 log ( g( ) ) = 1+ Rewriing as an eponenial, 4 3 1+ 4 g( ) 10 = Breaking his apar using eponen rules, 3 4 g( ) 10 1 10 = Using eponen rules o group he second facor, 3 1 g 4 ( ) = 10 10 Evaluaing he powers of 10, ( 5. ) g( ) = 10 63 Try i Now 1. An eponenial graph is ploed on a semi-log graph below. Find a formula for he eponenial funcion g() ha generaed his graph. log(g()) 5 4 3 1 0-4 -3 - -1-1 0 1 3 4 Fiing Eponenial Funcions o Daa Some echnology opions provide dedicaed funcions for finding eponenial funcions ha fi daa, bu many only provide funcions for fiing linear funcions o daa. The semi-log scale provides us wih a mehod o fi an eponenial funcion o daa by building upon he echniques we have for fiing linear funcions o daa. To fi an eponenial funcion o a se of daa using linearizaion 1. Find he log of he daa oupu values. Find he linear equaion ha fis he (inpu, log(oupu)) pairs. This equaion will be of he form log(f()) = b + m 3. Solve his equaion for he eponenial funcion f()

Secion 4.7 Fiing Eponenial Models o Daa 93 Eample 4 The able below shows he cos in dollars per megabye of sorage space on compuer hard drives from 1980 o 004 6, and he daa is shown on a sandard graph o he righ, wih he inpu changed o years afer 1980 Year Cos per MB 1980 19.31 1984 87.86 1988 15.98 199 4 1996 0.173 000 0.006849 004 0.001149 This daa appears o be decreasing eponenially. To find a funcion ha models his decay, we would sar by finding he log of he coss. Year Cos per MB log(cos) 1980 19.31.8400 1984 87.86 1.943791 1988 15.98 1.03577 199 4 0.6006 1996 0.173-0.76195 000 0.006849 -.16437 004 0.001149 -.9395 As epeced, he graph of he log of coss appears fairly linear, suggesing an eponenial funcion will fi he original daa will fi reasonably well. Using echnology, we can find a linear equaion o fi he log(cos) values. Using as years afer 1980, linear regression gives he equaion: log( C( )) =.794 0. 31 Solving for C(),.794 C( ) = 10 C( ) = 10 C( ) = 10.794.794 0.31 10 0.31 0.31 ( 10 ) ( 0. ) C( ) = 6 5877 0 0 4 8 1 16 0 4 This equaion suggess ha he cos per megabye for sorage on compuer hard drives is decreasing by abou 41% each year. 50 00 150 100 - -3-4 50 3 1 0-1 0 4 8 1 16 0 4 6 Seleced values from hp://www.swivel.com/workbooks/6190-cos-per-megabye-of-hard-drive- Space, rerieved Aug 6, 010

94 Chaper 4 Using his funcion, we could predic he cos of sorage in he fuure. Predicing he cos in he year 00 ( = 40): C(40) = 6 ( 0.5877) 40 0. 000000364 dollars per megabye, a really small number. Tha is equivalen o $0.36 per erabye of hard drive sorage. Comparing he values prediced by his model o he acual daa, we see he model maches he original daa in order of magniude, bu he specific values appear quie differen. This is, unforunaely, he bes eponenial model ha can fi he daa. I is possible ha a non-eponenial model would fi he daa beer, or here could jus be wide enough variabiliy in he daa ha no relaively simple model would fi he daa any beer. Year Acual Cos per MB Cos prediced by model 1980 19.31 6.3 1984 87.86 74.3 1988 15.98 8.9 199 4 1.1 1996 0.173 0.13 000 0.006849 0.015 004 0.001149 0.0018 Try i Now. The able below shows he value V, in billions of dollars, of US impors from China years afer 000. year 000 001 00 003 004 005 0 1 3 4 5 V 100 10.3 15. 15.4 196.7 43.5 This daa appears o be growing eponenially. Linearize his daa and build a model o predic how many billions of dollars of impors were epeced in 011. Imporan Topics of his Secion Semi-log graph Log-log graph Linearizing eponenial funcions Fiing an eponenial equaion o daa Try i Now Answers 1. f ( ) = 100(0.316). V ( ) = 90.545(1.078). Predicing in 011, V ( 11) = 7. 45billion dollars

Secion 4.7 Fiing Eponenial Models o Daa 95 Secion 4.7 Eercises Graph each funcion on a semi-log scale, hen find a formula for he linearized funcion in he log f = m + b. ( ) form ( ) 1. f ( ) = 4( 1.3). f ( ) = ( 1.5) 3. f ( ) = 10( 0.) 4. f ( ) = ( ) 30 0.7 The graph below is on a semi-log scale, as indicaed. Find a formula for he eponenial funcion y. ( ) 5. 6. 7. 8. Use regression o find an eponenial funcion ha bes fis he daa given. 9. 1 3 4 5 6 y 115 1495 310 394 4650 6361 10. 1 3 4 5 6 y 643 89 90 1073 1330 1631 11. 1 3 4 5 6 y 555 383 307 10 158 1

96 Chaper 4 1. 1 3 4 5 6 y 699 701 695 668 683 71 13. Toal ependiures (in billions of dollars) in he US for nursing home care are shown below. Use regression o find an eponenial funcion ha models he daa. Wha does he model predic ependiures will be in 015? Year 1990 1995 000 003 005 008 Ependiure 53 74 95 110 11 138 14. Ligh inensiy as i passes hrough waer decreases eponenially wih deph. The daa below shows he ligh inensiy (in lumens) a various dephs. Use regression o find an funcion ha models he daa. Wha does he model predic he inensiy will be a 5 fee? Deph (f) 3 6 9 1 15 18 Lumen 11.5 8.6 6.7 5. 3.8.9 15. The average price of elecriciy (in cens per kilowa hour) from 1990 hrough 008 is given below. Deermine if a linear or eponenial model beer fis he daa, and use he beer model o predic he price of elecriciy in 014. Year 1990 199 1994 1996 1998 000 00 004 006 008 Cos 7.83 8.1 8.38 8.36 8.6 8.4 8.44 8.95 10.40 11.6 16. The average cos of a loaf of whie bread from 1986 hrough 008 is given below. Deermine if a linear or eponenial model beer fis he daa, and use he beer model o predic he price of a loaf of bread in 016. Year 1986 1988 1990 1995 1997 000 00 004 006 008 Cos 0.57 0.66 0.70 0.84 0.88 0.99 1.03 0.97 1.14 1.4