6.3 Exponential Equations and Inequalities

Transcription

1 448 Exponntial and Logarithmic Functions 6.3 Exponntial Equations and Inqualitis In this sction w will dvlop tchniqus for solving quations involving xponntial functions. Suppos, for instanc, w wantd to solv th quation x = 8. Aftr a momnt s calculation, w find 8 = 7, so w hav x = 7. Th on-to-on proprty of xponntial functions, dtaild in Thorm 6.4, tlls us that x = 7 if and only if x = 7. This mans that not only is x = 7 a solution to x = 7, it is th only solution. Now suppos w chang th problm vr so slightly to x = 9. W could us on of th invrs proprtis of xponntials and logarithms listd in Thorm 6.3 to writ 9 = log (9. W d thn hav x = log (9, which mans our solution is x = log (9. This maks sns bcaus, aftr all, th dfinition of log (9 is th xponnt w put on to gt 9. Indd w could hav obtaind this solution dirctly by rwriting th quation x = 9 in its logarithmic form log (9 = x. Eithr way, in ordr to gt a rasonabl dcimal approximation to this numbr, w d us th chang of bas formula, Thorm 6.7, to giv us somthing mor calculator frindly, say log (9 = ln(9 ln(. Anothr way to arriv at this answr is as follows x = 9 ln ( x = ln(9 Tak th natural log of both sids. x ln( = ln(9 Powr Rul x = ln(9 ln( Taking th natural log of both sids is akin to squaring both sids: sinc f(x = ln(x is a function, as long as two quantitis ar qual, thir natural logs ar qual. Also not that w trat ln( as any othr non-zro ral numbr and divid it through 3 to isolat th variabl x. W summariz blow th two common ways to solv xponntial quations, motivatd by our xampls. Stps for Solving an Equation involving Exponntial Functions. Isolat th xponntial function.. (a If convnint, xprss both sids with a common bas and quat th xponnts. (b Othrwis, tak th natural log of both sids of th quation and us th Powr Rul. Exampl Solv th following quations. Chck your answr graphically using a calculator.. 3x = 6 x. 000 = t x = 7 x = t 5. 5 x = 5 x x x = 5 Solution. You can us natural logs or common logs. W choos natural logs. (In Calculus, you ll larn ths ar th most mathy of th logarithms. This is also th if part of th statmnt log b (u = log b (w if and only if u = w in Thorm Plas rsist th tmptation to divid both sids by ln instad of ln(. Just lik it wouldn t mak sns to divid both sids by th squar root symbol whn solving x = 5, it maks no sns to divid by ln.

2 6.3 Exponntial Equations and Inqualitis 449. Sinc 6 is a powr of, w can rwrit 3x = 6 x as 3x = ( 4 x. Using proprtis of xponnts, w gt 3x = 4( x. Using th on-to-on proprty of xponntial functions, w gt 3x = 4( x which givs x = 4 7. To chck graphically, w st f(x = 3x and g(x = 6 x and s that thy intrsct at x = W bgin solving 000 = t by dividing both sids by 000 to isolat th xponntial which yilds 3 0.t =. Sinc it is inconvnint to writ as a powr of 3, w us th natural log to gt ln ( 3 0.t = ln(. Using th Powr Rul, w gt 0.t ln(3 = ln(, so w divid both sids by 0. ln(3 to gt t = ln( 0 ln( 0. ln(3 = ln(3. On th calculator, w graph f(x = 000 and g(x = x and find that thy intrsct at x = ln( ln(3 y = f(x = 3x and y = g(x = 6 x y = f(x = 000 and y = g(x = x 3. W first not that w can rwrit th quation 9 3 x = 7 x as 3 3 x = 7 x to obtain 3 x+ = 7 x. Sinc it is not convnint to xprss both sids as a powr of 3 (or 7 for that mattr w us th natural log: ln ( 3 x+ = ln ( 7 x. Th powr rul givs (x + ln(3 = x ln(7. Evn though this quation appars vry complicatd, kp in mind that ln(3 and ln(7 ar just constants. Th quation (x + ln(3 = x ln(7 is actually a linar quation and as such w gathr all of th trms with x on on sid, and th constants on th othr. W thn divid both sids by th cofficint of x, which w obtain by factoring. (x + ln(3 = x ln(7 x ln(3 + ln(3 = x ln(7 ln(3 = x ln(7 x ln(3 ln(3 = x( ln(7 ln(3 Factor. x = ln(3 ln(7 ln(3 Graphing f(x = 9 3 x and g(x = 7 x on th calculator, w s that ths two graphs intrsct at x = ln(3 ln(7 ln( Our objctiv in solving 75 = 00 is to first isolat th xponntial. To that nd, w +3 t clar dnominators and gt 75 ( + 3 t = 00. From this w gt t = 00, which lads to 5 t = 5, and finally, t = 9. Taking th natural log of both sids

3 450 Exponntial and Logarithmic Functions givs ln ( t = ln ( ( 9. Sinc natural log is log bas, ln t = t. W can also us th Powr Rul to writ ln ( 9 = ln(9. Putting ths two stps togthr, w simplify ln ( t = ln ( 9 to t = ln(9. W arriv at our solution, t = ln(9 which simplifis to t = ln(3. (Can you xplain why? Th calculator confirms th graphs of f(x = 75 and g(x = 00 intrsct at x = ln( x y = f(x = 9 3 x and y = f(x = 75 and y = g(x = 7 x y = g(x = x 5. W start solving 5 x = 5 x + 6 by rwriting 5 = 5 so that w hav ( 5 x = 5 x + 6, or 5 x = 5 x + 6. Evn though w hav a common bas, having two trms on th right hand sid of th quation foils our plan of quating xponnts or taking logs. If w star at this long nough, w notic that w hav thr trms with th xponnt on on trm xactly twic that of anothr. To our surpris and dlight, w hav a quadratic in disguis. Ltting u = 5 x, w hav u = (5 x = 5 x so th quation 5 x = 5 x + 6 bcoms u = u + 6. Solving this as u u 6 = 0 givs u = or u = 3. Sinc u = 5 x, w hav 5 x = or 5 x = 3. Sinc 5 x = has no ral solution, (Why not? w focus on 5 x = 3. Sinc it isn t convnint to xprss 3 as a powr of 5, w tak natural logs and gt ln (5 x = ln(3 so that x ln(5 = ln(3 or x = ln(3 ln(5. On th calculator, w s th graphs of f(x = 5x and g(x = 5 x + 6 intrsct at x = ln(3 ln( At first, it s unclar how to procd with x x = 5, bsids claring th dnominator to obtain x x = 0. Of cours, if w rwrit x =, w s w hav anothr dnominator x lurking in th problm: x = 0. Claring this dnominator givs us x = 0 x, x and onc again, w hav an quation with thr trms whr th xponnt on on trm is xactly twic that of anothr - a quadratic in disguis. If w lt u = x, thn u = x so th quation x = 0 x can b viwd as u = 0u. Solving u 0u = 0, w obtain by th quadratic formula u = 5 ± 6. From this, w hav x = 5 ± 6. Sinc 5 6 < 0, w gt no ral solution to x = 5 6, but for x = 5 + 6, w tak natural logs to obtain x = ln ( If w graph f(x = x x and g(x = 5, w s that th graphs intrsct at x = ln (

4 6.3 Exponntial Equations and Inqualitis 45 y = f(x = 5 x and y = f(x = x x and y = g(x = 5 x + 6 y = g(x = 5 Th authors would b rmiss not to mntion that Exampl 6.3. still holds grat ducational valu. Much can b larnd about logarithms and xponntials by vrifying th solutions obtaind in Exampl 6.3. analytically. For xampl, to vrify our solution to 000 = t, w substitut t = and obtain 0 ln( ln( (? = ? = ln( ln(3? 0 ln( ln(3 = log 3 ( Chang of Bas? = 000 Invrs Proprty 000 = 000 Th othr solutions can b vrifid by using a combination of log and invrs proprtis. Som fall out quit quickly, whil othrs ar mor involvd. W lav thm to th radr. Sinc xponntial functions ar continuous on thir domains, th Intrmdiat Valu Thorm 3. applis. As with th algbraic functions in Sction 5.3, this allows us to solv inqualitis using sign diagrams as dmonstratd blow. Exampl Solv th following inqualitis. Chck your answr graphically using a calculator.. x 3x 6 0. Solution. x x x x < 4x. Sinc w alrady hav 0 on on sid of th inquality, w st r(x = x 3x 6. Th domain of r is all ral numbrs, so in ordr to construct our sign diagram, w sd to find th zros of r. Stting r(x = 0 givs x 3x 6 = 0 or x 3x = 6. Sinc 6 = 4 w hav x 3x = 4, so by th on-to-on proprty of xponntial functions, x 3x = 4. Solving x 3x 4 = 0 givs x = 4 and x =. From th sign diagram, w s r(x 0 on (, ] [4,, which corrsponds to whr th graph of y = r(x = x 3x 6, is on or abov th x-axis.

5 45 Exponntial and Logarithmic Functions (+ 0 ( 0 (+ 4 y = r(x = x 3x 6. Th first stp w nd to tak to solv x x 4 3 is to gt 0 on on sid of th inquality. To that nd, w subtract 3 from both sids and gt a common dnominator x x 4 3 x x x x 4 3 (x 4 x 4 0 Common dnomintors. x x 4 0 W st r(x = x x 4 and w not that r is undfind whn its dnominator x 4 = 0, or whn x = 4. Solving this givs x = ln(4, so th domain of r is (, ln(4 (ln(4,. To find th zros of r, w solv r(x = 0 and obtain x = 0. Solving for x, w find x = 6, or x = ln(6. Whn w build our sign diagram, finding tst valus may b a littl tricky sinc w nd to chck valus around ln(4 and ln(6. Rcall that th function ln(x is incrasing 4 which mans ln(3 < ln(4 < ln(5 < ln(6 < ln(7. Whil th prospct of dtrmining th sign of r (ln(3 may b vry unsttling, rmmbr that ln(3 = 3, so r (ln(3 = ln(3 ln(3 4 = (3 3 4 = 6 W dtrmin th signs of r (ln(5 and r (ln(7 similarly. 5 From th sign diagram, w find our answr to b (, ln(4 [ln(6,. Using th calculator, w s th graph of f(x = x x 4 is blow th graph of g(x = 3 on (, ln(4 (ln(6,, and thy intrsct at x = ln( This is bcaus th bas of ln(x is >. If th bas b wr in th intrval 0 < b <, thn log b (x would dcrasing. 5 W could, of cours, us th calculator, but what fun would that b?

6 6.3 Exponntial Equations and Inqualitis 453 ( (+ 0 ( ln(4 ln(6 y = f(x = x x 4 y = g(x = 3 3. As bfor, w start solving x x < 4x by gtting 0 on on sid of th inquality, x x 4x < 0. W st r(x = x x 4x and sinc thr ar no dnominators, vn-indxd radicals, or logs, th domain of r is all ral numbrs. Stting r(x = 0 producs x x 4x = 0. W factor to gt x ( x 4 = 0 which givs x = 0 or x 4 = 0. To solv th lattr, w isolat th xponntial and tak logs to gt x = ln(4, or x = ln(4 = ln(. (Can you xplain th last quality using proprtis of logs? As in th prvious xampl, w nd to b carful about choosing tst valus. Sinc ln( = 0, w choos ln ( (, ln 3 and ln(3. Evaluating, 6 w gt r ( ln ( ( = ln ln( ( 4 ln = ln ( ln( 4 ln ( = ln ( ln( ( 4 4 ln = 4 ln ( ( 4 ln = 5 4 ln ( Powr Rul Sinc <, ln ( < 0 and w gt r(ln ( is (+, so r(x < 0 on (0, ln(. Th calculator confirms that th graph of f(x = x x is blow th graph of g(x = 4x on ths intrvals. 7 (+ 0 ( 0 (+ 0 ln( y = f(x = x x and y = g(x = 4x 6 A calculator can b usd at this point. As usual, w procd without apologis, with th analytical mthod. 7 Not: ln(

7 454 Exponntial and Logarithmic Functions Exampl Rcall from Exampl 6.. that th tmpratur of coff T (in dgrs Fahrnhit t minuts aftr it is srvd can b modld by T (t = t. Whn will th coff b warmr than 00 F? Solution. W nd to find whn T (t > 00, or in othr words, w nd to solv th inquality t > 00. Gtting 0 on on sid of th inquality, w hav 90 0.t 30 > 0, and w st r(t = 90 0.t 30. Th domain of r is artificially rstrictd du to th contxt of th problm to [0,, so w procd to find th zros of r. Solving 90 0.t 30 = 0 rsults in 0.t = 3 so that t = 0 ln ( 3 which, aftr a quick application of th Powr Rul lavs us with t = 0 ln(3. If w wish to avoid using th calculator to choos tst valus, w not that sinc < 3, 0 = ln( < ln(3 so that 0 ln(3 > 0. So w choos t = 0 as a tst valu in [0, 0 ln(3. Sinc 3 < 4, 0 ln(3 < 0 ln(4, so th lattr is our choic of a tst valu for th intrval (0 ln(3,. Our sign diagram is blow, and nxt to it is our graph of t = T (t from Exampl 6.. with th horizontal lin y = 00. y (+ 0 ( 40 0 y = ln( H.A. y = t y = T (t In ordr to intrprt what this mans in th contxt of th ral world, w nd a rasonabl approximation of th numbr 0 ln( This mans it taks approximatly minuts for th coff to cool to 00 F. Until thn, th coff is warmr than that. 8 W clos this sction by finding th invrs of a function which is a composition of a rational function with an xponntial function. Exampl Th function f(x = 5x x + is on-to-on. Find a formula for f (x and chck your answr graphically using your calculator. Solution. W start by writing y = f(x, and intrchang th rols of x and y. To solv for y, w first clar dnominators and thn isolat th xponntial function. 8 Critics may point out that sinc w ndd to us th calculator to intrprt our answr anyway, why not us it arlir to simplify th computations? It is a fair qustion which w answr unfairly: it s our book.

8 6.3 Exponntial Equations and Inqualitis 455 y = x = x ( y + = 5 y x y + x = 5 y 5 x x + 5 y y + x = 5 y x y Switch x and y x = y (5 x y x = 5 x ( x ln ( y = ln 5 x ( x y = ln 5 x ( W claim f (x = ln x 5 x. To vrify this analytically, w would nd to vrify th compositions ( f f (x = x for all x in th domain of f and that ( f f (x = x for all x in th domain of f. W lav this to th radr. To vrify our solution graphically, w graph y = f(x = 5x ( and y = g(x = ln x 5 x on th sam st of axs and obsrv th symmtry about th lin y = x. Not th domain of f is th rang of g and vic-vrsa. x + ( y = f(x = 5x x + and y = g(x = ln x 5 x