Chapter 4: Exponential and Logarithmic Functions


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1 Chaper 4: Eponenial and Logarihmic Funcions Secion 4.1 Eponenial Funcions Secion 4. Graphs of Eponenial Funcions... 3 Secion 4.3 Logarihmic Funcions... 4 Secion 4.4 Logarihmic Properies Secion 4.5 Graphs of Logarihmic Funcions... 6 Secion 4.6 Eponenial and Logarihmic Models Secion 4.7 Fiing Eponenials o Daa 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 Saring wih 100 each = 150 They boh grow by 50 sores in he firs year = 00 Sore A grows by 50, Sore B grows by = 50 Sore A grows by 50, Sore B grows by % of (100) = % of (150) = % of (5) = 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 CCBYSA license.
2 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) or equivalenly by facoring 100 ( ) = 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) or equivalenly by facoring 150 ( ) now recall he 150 came from 100(1+0.50). Subsiuing ha, 100( )( ) = 100( ) = 5 Afer 3 years, he number of sores was: (5) or equivalenly by facoring 5 ( ) now recall he 5 came from 3 100( ) ( ) = 100( ) = ( ). 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( ) 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
3 Secion 4.1 Eponenial Funcions 17 years Company A Company B 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 Using he basic formula for eponenial growh f ( ) = a(1 + r) we can wrie he formula, f ( ) = 1.14( ) To esimae he populaion in 00, we evaluae he funcion a = 1, since 00 is 1 years afer 008. f (1) = 1.14( ) 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
4 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 ( ) = f ( m) = 1000( ) m m Afer 4 monhs, he accoun will have grown o 4 f (4) = 1000( ) = $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( ) = B( ) = 900( ) ( ) 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 Bismuh10 is an isoope ha radioacively decays by abou 13% each day, meaning 13% of he remaining Bismuh10 ransforms ino anoher aom (polonium10 in his case) each day. If you begin wih 100 mg of Bismuh10, 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 = 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 = mg of Bismuh10 remains. Try i Now 3. A populaion of 1000 is decreasing 3% each year. Find he populaion in 30 years.
5 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) = 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() = 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 inpuoupu 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 inpuoupu pair allows us o solve for b: b 6 = Divide by b = = Take he 6 h roo of boh sides b = 9 6 = 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 = , 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.
6 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 overround 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 , you would wan o round his no furher han o 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 = 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)
7 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 righhand 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 ( ) ( ) =.
8 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 = A (10) = (10) = $ The accoun will be worh $ in 10 years.
9 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 semiannually (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 = A(18) = a ,000 = a(.8983) a = 40, $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 $ Quarerly $ Monhly $ Daily $ If we were o compue he acual percenage increase for he daily compounding, here was an increase of $ from an original amoun of $1,000, for a percenage increase of = = % increase. This quaniy is called he annual percenage 1000 yield (APY).
10 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 = 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 Bank A: APY = = = 1.054% Bank B: APY = = = % 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 semiannual 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.
11 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 $ Monhly $ Daily $ Hourly $ Once per minue $ 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.
12 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 = (3) f (3) = 100e mg of Radon will remain. Try i Now 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 = We sar wih $1000, so a = To find he value afer 1 year, 0.10(1) f (1) = 1000e $ Noice his is a $ increase for he year. As a percen increase, his is = = % increase over he original $ 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 %. 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( ) using he % acual annual yield rae.
13 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 (0.97) 30 = f ( ) = ( 1. 5) 5. $ An iniial subsance weighing 0g is growing a a coninuous rae of 1% per day.
14 8 Chaper 4 Secion 4.1 Eercises For each able below, could he able represen a funcion ha is linear, eponenial, or neiher? f() h() m() g() k() n() 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 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 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.
15 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, 4) ,,( 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 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 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.
16 30 Chaper 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 A populaion of baceria is growing according o he equaion P ( ) = 1600e, wih measured in years. Esimae when he populaion will eceed A populaion of baceria is growing according o he equaion P ( ) = 100e, wih measured in years. Esimae when he populaion will eceed 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 [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 of AIDS cases (in housands) repored in he Unied Saes: a( ) = , 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]
17 Secion 4.1 Eponenial Funcions 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.
18 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: f() 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( ) = g() 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:
19 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 ().
20 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 ) 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. ) 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.
21 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.
22 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: 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.
23 Secion 4. Graphs of Eponenial Funcions 37 Eample 3 1 Skech a graph of f ( ) = 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.
24 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 = 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.
25 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
26 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( ) = 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 yais =
27 Secion 4. Graphs of Eponenial Funcions 41 Describe he long run behavior, as and of each funcion 54 1 f = f ( ) = ( ) 4. ( ) ( ) 5. f ( ) 1 = 3 6. f ( ) 1 = f ( ) = 34 ( ) + 8. f ( ) ( ) = 3 1 Find a formula for each funcion graphed as a ransformaion of f ( ) = Find an equaion for he eponenial funcion graphed
28 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 oneoone, 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
29 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 = = = 8 is equivalen o log (8) = 3 5 = 5 is equivalen o log 5 (5) = 10 1 = is equivalen o log10 = 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.
30 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 = 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
31 Secion 4.3 Logarihmic Funcions 45 Try i Now. Evaluae log( ). 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) 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 ( ) ( ) =
32 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,
33 Secion 4.3 Logarihmic Funcions 47 ln10 ln 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. log log (100) = log = 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.
34 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( ). 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 ln = ln( ) 1.14 Apply he eponen propery on he righ side ln = ln( ) 1.14 Divide boh sides by ln(1.0134) ln 1.14 = 4.3 years ln ( ) 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 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( ). 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 = log 10 = ( 4) = ( ) ( ) 4
35 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 = 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 = ae, so a = 100. P = r ( ) ae so ha P(0) = 100 and Using he second pair of values, 130 = 100e r Divide by 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 = 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.).
36 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 = , so he annual growh rae is 1.75%. The sales funcion could also be wrien in he form S ( ) = 5000( ). 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 ln( ) ln() ln(0.93) 1. ( ) 4
37 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 = y y = a = b p = v 15. d c k e = k 1. = h 16. z n y e = L = Solve for. 17. log3 ( ) = 18. log 4( ) = log ( ) = 3 0. log 5( ) = 1 1. log ( ) = 3. log ( ) = 5 3. ln ( ) = 4. ( ) ln = Simplify each epression using logarihm properies log5 ( 5 ) 6. log ( 8 ) 7. log log6 ( 6 ) ( ) 33. log ( ) 34. ( ) 1 8. log6 36 log log ( 10,000 ) 3. log ( 100 ) log 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 = = e = e = = = ( 1.03) = ( ) 51. ( ) = 5. ( ) = = 7 = = e 0.03 = e = = = 70 4
38 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, ( ) f ( ) = e 6. f ( ) = 100e f ( ) = 50e 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?
39 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
40 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)
41 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) Using he log able, log( 3) = log() + log(3) = = We can hen use he able again in reverse, looking for as an oupu of he logarihm. From ha we can deermine: log( 3) = = 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/ log( 8) == log 8 = log(8) = ( ) = So 8 =. ( ) log() 3 =
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