Unit 6: Solving Exponential Equations and More

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Randall Hall
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1 Habrman MTH 111 Sction II: Eonntial and Logarithmic Functions Unit 6: Solving Eonntial Equations and Mor EXAMPLE: Solv th quation for. Obtain an act solution. This quation is so asy to solv that most of us can do it in our hads, without showing any work. (Clarly, sinc ) But it might b hlful to notic what work would hav hld us solv this quation, so that w can larn how to solv onntial quations that w can t do in our hads. If it is ossibl, th asist way to solv onntial quations is to writ both sids of th quation as a owr of th sam bas: (writ both sids of th quation as a owr of 10) Sinc onntial functions ar on-to-on, w can now assum that sinc both sids of th quation ar rssd as owrs of th sam bas, th onnts must b qual, which allows us to conclud that. But this tchniqu rquirs us to b abl to writ both sids of th quation in trms of th sam bas. In ordr to larn how to solv onntial quations involving rssions that don t allow us to do this so asily, lt s invstigat anothr mthod of solving onntial quations. Rcall that th log-of-owrs rorty allows us to mov owrs out of th onnt. Thus, lt s utiliz a logarithmic function to bring th out of th onnt: log( 10 ) log( 10 ) log(10) log( 100) (tak th common logarithm of both sids of quation) (sinc log(10) 1 and log(100) ) EXAMPLE: Solv th quation 8 16 for. Obtain an act solution. Sinc both 8 and 16 can b rssd as owrs of th sam bas (i.., ), w can solv th quation using th sam bas mthod.

2 8 16 (sinc onntial functions ar on-to-on, w can st th onnts qual) But, again, w ar lucky to b abl to us this mthod, sinc both sids of th quation can b rssd as owrs of th sam bas. Anothr way to solv this quation is to us logarithms. W us th natural logarithm in this cas sinc it is asist to writ (only two lttrs!) and it is th usually asist logarithm to stimat on calculators (8) (16) (8) (8) (16) (8) (16) (8) () () (this valu could b calculatd on a calclator but w'll simlify th rssion using th log-of-owrs rorty) In gnral, whn solving onntial quations w ar not usually abl to us th sam bas mthod but w can ALWAYS us th mthod involving logarithms. Th nt fw amls w NEED logarithms to solv th quations. W will us th natural logarithm hr for th sam rasons givn in th aml abov. Although ANY logarithm can b usd, th natural logarithm (or common logarithm) is rcommndd.

3 EXAMPLE: Solv th quation 19 5 for. Obtain an act solution (5) (1) (1) (5) (first, isolat th onntial rssion) (aly th natural logarithm to both sids) Although our solution isn t vry usr frindly (most of us don t hav much intuition about th numbr (1) ), w wr askd to find and act solution, and w can t rss this (5) numbr in a simlr way. So w nd to lav this as our solution. If you want to aroimat it, that is fin, but th aroimation must b givn in addition to th act solution, NOT instad of th act solution. EXAMPLE: Solv th quation 5 1 for. Obtain an act solution (onc w hav th onntial rssion isolatd, w aly th natural logarithm to both sids) (this st isn't ncssary but allows us to ractic th log-of-quotints rorty)

4 EXAMPLE: Solv th quation t 1 8 for t. Obtain an act solution. t 1 8 t 1 t 1 t 1 8 (t 1) t 1 t () () () () 1 () t 1 () 1 (onc w hav th onntial rssion isolatd, w aly th natural logarithm to both sids) EXAMPLE: Solv th quation 5 7 for. Obtain an act solution (7) () (this st isn't ncssary but allows us to (5) () ractic th log-of-quotints rorty)

5 5 Alications Involving Eonntial Functions EXAMPLE: Suos that a oulation grows according to th modl t rrsnts yars aftr January 1, 018. P() t whr SOLUTIONS: a. What is th doubling-tim for th oulation? b. How long will it tak for th oulation to incras 10%? a. In ordr to find th doubling-tim, w nd to dtrmin how long it taks for th oulation to doubl. Sinc th oulation has an initial siz of C ol, w nd to find t such that P( t) C. (Although w don t know what numbr C rrsnts, w can still dtrmin whn th oulation is C.) P( t) C C () t Thrfor, th doubling-tim is about 9.9 yars. (On thing worth king in mind about th doubling-tim is that it rrsnts th tim it taks for th oulation to doubl no mattr whn you start counting. So it taks th sam 9.9 yars for th oulation to go from C to C or from 5C to 10C.) b. If it grows 10%. thn th oulation will b C 0.10C 1.10C, so w nd to find th tim t such that P( t) 1.10C : P( t) 1.10C 1.10C t Thrfor, it taks about 1.6 yars for th oulation to grow by 10%.

6 6 EXAMPLE: Iodin-11 was on of th radioactiv substancs rlasd into th atmoshr during th Chrnobyl disastr in th formr Sovit Union in If iodin-11 dcays at th continuous daily rat of 8.66%, find its half-lif. If you hav C grams of iodin-11 thn th amount rmaining aftr t days is givn by th following function: I() t t To find th half-lif, w nd to dtrmin how long it taks for th amount of iodin to shrink by 50%, so w nd to solv I( t) 0.5C for t t I( t) 0.5C t 0.5C t t 0.5 So th half-lif of iodin-11 is about 8 days. 0.5 t EXAMPLE: Th half-lif of radium-6 is 160 yars. What is th continuous annual rat of dcay for radium-6? In ordr to answr th qustion, w nd to solv th quation blow for k. k k 0.5C k k k So th continuous annual dcay rat for radium-6 is about 0.0%.

y = 2xe x + x 2 e x at (0, 3). solution: Since y is implicitly related to x we have to use implicit differentiation: 3 6y = 0 y = 1 2 x ln(b) ln(b)

y = 2xe x + x 2 e x at (0, 3). solution: Since y is implicitly related to x we have to use implicit differentiation: 3 6y = 0 y = 1 2 x ln(b) ln(b) 4. y = y = + 5. Find th quation of th tangnt lin for th function y = ( + ) 3 whn = 0. solution: First not that whn = 0, y = (1 + 1) 3 = 8, so th lin gos through (0, 8) and thrfor its y-intrcpt is 8. y