Exponential and Logarithmic Functions

Transcription

1 5 Chpter Eponentil nd Logrithmic Functions 5. Eponentil Functions nd Their Grphs 5. Applictions of Eponentil Functions 5. Logrithmic Functions nd Their Grphs 5. Properties nd Applictions of Logrithms 5.5 Eponentil nd Logrithmic Equtions Chpter 5 Project Chpter 5 Summr Chpter 5 Review Chpter 5 Test B the end of this chpter ou should be ble to: Wht if ou hd to predict the mount of time it will tke for the erth s popultion to rech 0 billion (ignoring the effect of limited resources on popultion growth)? How would ou perform this clcultion? B the end of this chpter, ou ll be ble to ppl the skills regrding eponentil nd logrithmic functions to estimte the behvior of growing popultions, interest-erning ccounts, nd decing elements. On pge 8 ou ll encounter problem like the popultion problem given bove. You ll mster this tpe of problem using tools such s the Summr of Logrithmic Properties, found on pge.

2 9 Chpter 5 Introduction This chpter introduces two entirel new clsses of functions, both of which re of enormous importnce in mn nturl nd mn-mde contets. As we will see, the two clsses of functions re inverses of one nother, though historicll eponentil nd logrithmic functions were developed independentl nd for unrelted resons. We will begin with stud of eponentil functions. These re functions in which the vrible ppers in the eponent while the bse is constnt, just the opposite of wht we hve seen so often in the individul terms of polnomils. As with mn mthemticl concepts, the rgument cn be mde tht eponentil functions eist in the nturl world independentl of mnkind nd tht consequentl mthemticins hve done nothing more thn observe (nd formlize) wht there is to be seen. Eponentil behvior is ehibited, for emple, in the rte t which rdioctive substnces dec, in how the temperture of n object chnges when plced in n environment held t constnt temperture, nd in the fundmentl principles of popultion growth. But eponentil functions lso rise in discussing such mn-mde phenomen s the growth of investment funds. In fct, we will use the formul for compound interest to motivte the introduction of the most fmous nd useful bse for eponentil functions, the irrtionl constnt e (the first few digits of which re ). The Swiss mthemticin Leonhrd Euler (707 78), who identified mn of this number s unique properties (such s the fct tht e = ), ws one of the first to recognize the fundmentl importnce of e nd in fct is responsible for the choice of the letter e s its smbol. The constnt e lso rises ver nturll in the contet of clculus, but tht discussion must wit for lter course. Npier Logrithms re inverses, in the function sense, of eponentils, but historicll their development ws for ver different resons. Much of the development of logrithms is due to John Npier (550 67), Scottish writer of politicl nd religious trcts nd n mteur mthemticin nd scientist. It ws the gol of simplifing computtions of lrge numbers tht led him to devise erl versions of logrithmic functions, to construct wht tod would be clled tbles of logrithms, nd to design prototpe of wht would eventull become slide rule. Of course, tod it is not necessr to resort to logrithms in order to crr out difficult computtions, but the properties of logrithms mke them invluble in solving certin equtions. Further, the fct tht the re inverses of eponentil functions mens the hve just s much plce in the nturl world.

3 5. Eponentil Functions nd Their Grphs Section Eponentil Functions nd Their Grphs TOPICS. Definition nd clssifiction of eponentil functions. Grphing eponentil functions. Solving elementr eponentil equtions T. Inputting eponents, grphing eponentil functions TOPIC Definition nd Clssifiction of Eponentil Functions We hve studied mn functions in which vrible is rised to constnt power, including polnomil functions such s f = +, rdicl functions like g=, nd rtionl functions such s h =. An eponentil function is function in which constnt is rised to vrible power. Eponentil functions re etremel importnt becuse of the lrge number of nturl situtions in which the rise. Emples include rdioctive dec, popultion growth, compound interest, spred of epidemics, nd rtes of temperture chnge. DEFINITION Eponentil Functions Let be fied, positive rel number not equl to. The eponentil function with bse is the function f =. Wh do we hve the restrictions > 0,? The bse of the eponent cn t be negtive, since would not be rel for mn vlues of. For emple, if = nd =, is not rel. If we let =, we don t hve problem producing rel numbers. Insted, the eponentil function turns out to be constnt. Recll tht for ll vlues of, =. For this reson, is considered constnt function, not n eponentil function, nd should lws be written in its simplified form,. Note tht for n positive constnt, is defined for ll rel numbers. Consequentl, the domin of f = is the set of rel numbers. Wht bout the rnge of f? Since is positive, we know tht must be positive. We will see tht the rnge of ll eponentil functions is the set of ll positive rel numbers.

4 96 Chpter 5 Recll tht if is n nonzero number, 0 is defined to be. This mens the -intercept of n eponentil function, regrdless of the bse, is the point ( 0, ). Beond this, eponentil functions fll into two clsses, depending on whether lies between 0 nd or if is lrger thn. Consider the following clcultions for two smple eponentil functions: f = g = f ( )= 9 g( )= f ( )= g( )= 0 f 0 g 0 f f 9 = = ()= g()= = g= Note tht the vlues of f decrese s increses while the vlues of g do just the opposite. We s tht f is n emple of decresing function nd g is n emple of n incresing function. If we plot these points nd then fill in the gps with smooth curve, we get the grphs of f nd g tht pper in Figure. (, ) ( 0, ), (, ),, 0, Figure : Two Eponentil Functions nd tht The bove grphs suggest tht the rnge of n eponentil function is 0,, is indeed the cse. Note tht the bse does not mtter: the rnge of is the positive rel numbers for n llowble bse (tht is, for n positive not equl to ).

5 Eponentil Functions nd Their Grphs Section PROPERTIES Behvior of Eponentil Functions Given positive rel number not equl to, the function f = is: decresing function if 0 < <, with f s nd f 0 s. n incresing function if >, with f 0 s nd f s. lies on the grph of f, the domin of f is the set of rel In either cse, the point 0, numbers, nd the rnge of f is the set of positive rel numbers. TOPIC Grphing Eponentil Functions Given tht ll eponentil functions tke one of two bsic shpes (depending on whether is less thn or greter thn ), the re reltivel es to grph. Plotting few points, including the -intercept of 0,, EXAMPLE will provide n ccurte sketch. Grphing Eponentil Functions Note: Plugging in =, =0, nd = will produce good ide of the shpe of the grph. Sketch the grphs of the following eponentil functions:. f = b. g= Solutions: In both cses, we plot points b plugging in =, = 0, nd =. We then connect the points with smooth curve.. b., (, ) ( 0, ) (, ) 0,,

6 98 Chpter 5 An eponentil function, like n function, cn be trnsformed in ws tht result in the grph being shifted, reflected, stretched, or compressed. It often helps to grph the bse function before tring to grph the trnsformed one. EXAMPLE Grphing Eponentil Functions Note: For ech emple, first grph the bse function, then ppl n trnsformtions. Sketch the grphs of ech of the following functions.. f = Solutions:. +,, ( 0, ) (, ),, b. g= + c. h = First, drw the grph of the function, s in Emple b. Then, since hs been replced b +, shift the grph to the left b units. b. Begin with the grph of shown in green., 8 9, 9 (, ) 0, ( 00, ) (, ) The effect of multipling function b is to reflect the grph with respect to the -is. Following this, the second trnsformtion of dding to function cuses verticl shift of the grph. The purple curve t left is the grph of g() = - +. c. (, ),, ( 0, ), Begin b grphing the bse function, shown in green. For h, hs been replced b -, so we reflect the grph of cross the -is to obtin the grph of h, shown in purple. Note tht this grph is lso the grph of. Using properties of eponents is nother w to think bout this problem, s = =.

7 Eponentil Functions nd Their Grphs Section TOPIC Solving Elementr Eponentil Equtions As ou might epect, n eqution in which the vrible ppers s n eponent is clled n eponentil eqution. We re not et red to tckle eponentil equtions in full generlit. Even something s simple s = 5 currentl stumps us. We know tht = nd = 8, so the nswer must be between nd, but beond this we don t hve method to proceed. This must wit until we hve discussed clss of functions clled logrithms (which will hppen in Section 5.). However, we re red to solve eponentil equtions tht cn be written in the form b =, where is n eponentil bse (positive nd not equl to ) nd b is constnt. b You might guess, just from the form of the eqution = tht the solution is = b. This guess is correct, but we need to investigte wh it is true. The reson tht the single vlue b is the solution of n eponentil eqution of the b form = is tht the eponentil function f = is one-to-one (its grph psses the horizontl line test). Recll tht if g is one-to-one function, then the onl w for g( ) to equl g( ) is if =. In the cse of the function f = b, the eqution = is equivlent to the f = f b, nd this implies = b, since f is one-to-one. sttement b An eponentil eqution m not pper in the simple form = initill. The procedure below describes the steps ou m need to tke to solve n elementr eponentil eqution. PROCEDURE Solving Elementr Eponentil Equtions To solve n elementr eponentil eqution, Step : Isolte the eponentil. Move the eponentil contining to one side of the eqution nd n constnts or other vribles in the epression to the other side. Simplif, if necessr. Step : Find bse tht cn be used to rewrite both sides of the eqution. Step : Equte the powers, nd solve the resulting eqution.

8 00 Chpter 5 EXAMPLE Solving Elementr Eponentil Equtions Note: As lws, it is good prctice to check our solution in the originl eqution. Solve the following eponentil equtions = 0 b. 8 Solutions:. 5 5 = 0 b. 5 = 5 = = 5 = = 8 = = = = = = = c. = Begin b isolting the term with the vrible on one side. We cn write both sides with the sme bse since 5 nd 5 re both powers of 5. Simplif using properties of eponents. We then equte the power, resulting in liner eqution which we cn esil solve. Agin, we need to rewrite both sides using the sme bse. We see tht 8 nd cn both be written s powers of. Simplif using properties of eponents. Set the eponents equl to ech other. 9 Solve the resulting liner eqution. c. = 9 = = = Sometimes, the choice of bse is not obvious. Initill, we write the right-hnd side s shown. Then, we cn mke the two bses equl b using properties of eponents. After mking both bses the sme, we equte the eponents to find the solution.

9 Eponentil Functions nd Their Grphs Section 5. 0 TOPIC T Inputting Eponents, Grphing Eponentil Functions To input eponents into grphing clcultor, we cn use one of two methods. If the eponent is, we cn tpe the bse nd then press. Otherwise, we need to. use the cret smbol,. For emple, to clculte 6, we would tpe the bse, 6, then followed b the eponent,. CAUTION!.. Regrdless of which method is used, be creful with our negtive signs. We know tht = 9 nd not 9 becuse negtive times negtive is positive. However, if we tpe into clcultor, the output is 9. The number tht we re squring is, not, so we need to tpe ( ) or ( ) ^ into the clcultor to get the correct nswer, 9.. To grph n eponentil function, where the eponent is the vrible, we use the technique tht incorportes the cret smbol. Consider the grph of the function f =. To grph this in clcultor, press nd tpe in the following: 5. Notice tht onl the frction is being rised to the eponent, so we put it in 6. prentheses. The grph looks like In the net section, we will lern bout the irrtionl number, e, which is used often in eponentil equtions. To clculte or input vlue such s e 05. into grphing clcultor, tpe LN. Then, tpe in the eponent, 0.5, nd close the prentheses b pressing. Then press ENTER: 9.

10 0 Chpter 5 0. Eercises Sketch the grphs of the following functions. See Emples nd... f =. g = ( 05).. s=. +. f = + 5. r= h= 7. f = 8. r= 9. g = 0. h=. s = ( 0).. f = +.. g=. r= 5. h = m= 7. p= 8. q= r= 9 0. p =. r= 5 Solve the following eponentil equtions. See Emple = 5. = 7. 9 = = = = = 7. = = =. 5 = = 9 +. = = = = = = = 5 =. 0 = = 6 7. = 5. 5 = ( e ) = ( e ) ( e ) 7 7. = 8

11 Eponentil Functions nd Their Grphs Section Mtch the grphs of the following functions to the pproprite eqution. 8. f = 9. h = g = ( ). 5. p = 5. f = = 6 5. r 5. m= g = h =. { } 57. s =. { }. { }. b. c. 5. { } 6. { } 7. { } 8. { 7 } 9. { } 0. { } d. e. f.. { }. { }. {, }. { } 5. { 7 } 6. { } g. h. i. 7. { } 8. { } 9. { 9 } 0. { }. { }. { 6 }. { }. { } 5. { } j. 7. No solution g 50. i 5. b 5. d 5. f 5. e 55. c 56. h 57. j

12 0 Chpter 5 Applictions of Eponentil Functions TOPICS 5... Models of popultion growth. Rdioctive dec Compound interest nd the number e TOPIC Models of Popultion Growth Eponentil functions rise nturll in wide rr of situtions. In this section, we will stud few such situtions in some detil, beginning with popultion models. Mn people working in such res s mthemtics, biolog, nd sociolog stud mthemticl models of popultion. The models represent mn tpes of popultions, such s people in given cit or countr, wolves in wildlife hbitt, or number of bcteri in Petri dish. While such models cn be quite comple, depending on fctors like vilbilit of food, spce constrints, nd effects of disese nd predtion, t their core, mn popultion models ssume tht popultion growth displs eponentil behvior. The reson for this is tht the growth of popultion usull depends to lrge etent on the number of members cpble of producing more members. This ssumes n bundnt food suppl nd no constrints on popultion from lck of spce, but t lest initill this is often the cse. In n sitution where the rte of growth of popultion is proportionl to the size of the popultion, the popultion will grow eponentill, so we cn write the function Pt ()= P t 0. This function tells us the size of popultion Pt () t time t. Wht do the other terms in the function represent? 0 Consider wht hppens if we substitute t = 0. P( 0)= P0, nd since 0 = for ll, we hve P= P. Thus, P 0 is the initil popultion (popultion t time 0). 0 0 Recll from our work in grphing eponentil equtions tht when >, the lrger the vlue of is, the fster the eponentil function grows. In situtions of popultion growth, is greter thn (otherwise we would hve popultion dec). For this reson, the vlue of is the growth rte of the popultion. Often, we will need to use informtion bout popultion to determine the vlues of P 0 nd. Once we hve the function for popultion growth, we cn nswer other questions bout the popultion. Emple illustrtes this with specific model of bcteril popultion growth.

13 Applictions of Eponentil Functions Section EXAMPLE Popultion Growth A biologist is culturing bcteri in Petri dish. She begins with 000 bcteri, nd supplies sufficient food so tht for the first five hours the bcteri popultion grows eponentill, doubling ever hour.. Find function tht models the popultion growth of this bcteri culture. b. Determine when the popultion reches 6,000 bcteri. c. Clculte the popultion two nd hlf hours fter the scientist begins. Solutions:. We know tht we seek function of the form Pt ()= P t 0. Since the scientist strts with 000 bcteri, the initil popultion, P 0 = 000. To solve for, we use the fct tht the popultion doubles ever hour. 000 P Substitute, using the fct tht the popultion 000 = = 000 fter one hour equls 000 bcteri. Simplif, then solve for. = Thus, function tht models the popultion growth of the bcteri culture is t Pt 000, where t is mesured in hours. = () ()= b. We wish to find the time t for which Pt ()= 6, 000. t, = t 6 = t = t = Substitute the desired popultion vlue. Divide both sides b 000. Rewrite both sides with the sme bse,. Equte the eponents nd solve. Thus, the bcteri culture reches popultion of 6,000 in hours. c. To clculte the popultion two nd hlf hours fter growth begins, we substitute t = 5. into the popultion model function. = P While we could rewrite this using rtionl eponents, nd tr to find n ect vlue, frequentl this will be ver tedious or impossible, so we use clcultor to evlute. P ( 5. ) 5657 bcteri.

14 06 Chpter 5 TOPIC Rdioctive Dec In contrst to popultions (t lest helth popultions), rdioctive substnces diminish with time. To be ect, the mss of rdioctive element decreses over time s it decs into other elements. Since eponentil functions with bse between 0 nd re decresing functions, this suggests tht rdioctive dec is modeled b At ()= A t 0, where A() t represents the mount of the substnce t time t, A 0 is the mount t time t = 0, nd is number between 0 nd. Rdioctive dec is so predictble tht rcheologists nd nthropologists cn use it to clock how long go n orgnism died. The method of rdiocrbon dting depends on the fct tht living orgnisms constntl bsorb molecules of the rdioctive substnce crbon- while live, but the intke of crbon- ceses once the orgnism dies. The percentge of crbon- on Erth (reltive to other isotopes of crbon) hs been reltivel constnt over time, so living orgnisms provide n ccurte reference. B compring the smller percentge of crbon- in ded orgnism to the percentge found in living tissue, n estimte of when the orgnism died cn then be mde. Frequentl, the dec rte of rdioctive substnce is described b how long it tkes for hlf of the mteril to dec into other elements. This is known s the hlf-life of the substnce. The hlf-life of rdioctive element rnges widel; Frncium ehibits hlf-life of bout minutes, while Bismuth hs hlf life of ers, bout one billion times the estimted ge of the universe! In the cse of crbon-, its hlf-life of 578 ers mens tht if n orgnism contined grms of crbon- t deth, it would contin 6 grms fter 578 ers, grms fter nother 578 ers, nd so on. EXAMPLE Rdioctive Dec Determine the bse,, so tht the function At ()= A t 0 ccurtel describes the dec of crbon- s function of t ers. Solution: = =, so A 0 represents the mount of crbon- t time 0 Note tht A 0 A0 A0 t = 0. Since we re seeking generl formul, we don t know the vlue of A 0, tht is, the vlue of A 0 will vr depending on the detils of the sitution. But we cn still determine the bse constnt,. Wht we know is tht hlf of the originl mount of crbon- decs over period of 578 ers, so A 578 will be hlf of A 0. This gives us the eqution A A = = A 0 ( ). To solve this eqution for, we cn first divide both sides b A 0 ; the fct tht A 0 then cncels from the eqution emphsizes tht its ect vlue is irrelevnt in determining the vlue of.

15 Applictions of Eponentil Functions Section We re left with the eqution 578 =. At this point we need clcultor, s we need to tke the 578 th root of both sides: 578 = t The function is thus: At ()= A0 ( ) (using our pproimte vlue for ). TOPIC Compound Interest nd the Number e One of the most commonl encountered pplictions of eponentil functions is in compounding interest. We run into compound interest when erning mone (b interest on svings ccount or investment) nd when spending mone (on cr lons, mortgges nd credit crds). The bsic compound interest formul cn be understood b considering wht hppens when mone is invested in svings ccount. Tpicll, svings ccount is set up to p interest t n nnul rte of r (which we will write in deciml form) compounded n times er. For instnce, bnk m offer n nnul interest rte of 5% (mening r = 005. ) compounded monthl (so n = ). Compounding is the ct of clculting the interest erned on n investment nd dding tht mount to the investment. An investment in monthl compounded ccount will hve interest dded to it twelve times over the course of er, once ech month. Suppose n mount of P (for principl) dollrs is invested in svings ccount t n nnul rte of r compounded n times per er. We wnt formul for the mount of mone At () in the ccount fter t ers. If we s tht period is the length of time between compoundings, interest is clculted t the rte of r per period (for instnce, if r = 005. nd n =, interest is erned t n rte of per month). Tble illustrtes how compounding increses the mount in the ccount over the course of severl periods.

16 08 Chpter 5 Period 0 A= P Amount r A= P + n r r r A= P + P n + n = + n r r r A= P + P n + n = + n k r r r k A= P + P n + n = + n k Tble : Effect of Compounding Interest on n Investment of P Dollrs Since there re nt compounding periods in t ers, we obtin the following formul. DEFINITION Compound Interest Formul An investment of P dollrs, compounded n times per er t n nnul interest rte of r, hs vlue fter t ers of nt r At ()= P + n. EXAMPLE Compound Interest Formul Snd invests $0,000 in svings ccount erning.5% nnul interest compounded qurterl. Wht is the vlue of her investment fter three nd hlf ers? Solution: We know tht P = 0, 000, r = (remember to epress the interest rte in deciml form), n = (since the ccount is compounded four times er), nd t = 5.. Now we substitute nd evlute. ( 5. ) A( 5. ) = 0, = 0, 000(. 05) $, Thus, fter three nd hlf ers, Snd s investment grows to $,695.5.

17 Applictions of Eponentil Functions Section The compound interest formul cn lso be used to determine the interest rte of n eisting svings ccount, s shown in Emple. EXAMPLE Compound Interest Formul Nine months fter depositing $50.00 in monthl compounded svings ccount, Frnk checks his blnce nd finds the ccount hs $58.8. Being the forgetful tpe, he cn t remember wht the nnul interest rte for his ccount is, nd sees the bnk is dvertising rte of.5% for new ccounts. Should he close out his eisting ccount nd open new one? Solution: As in Emple, we will begin b identifing the known quntities in the compound interest formul: P = 50, n = ( compoundings per er), nd t = 075. (nine months is three-qurters of er). Further, the mount in the ccount, A, t this time is $58.8. This gives us the eqution to solve for r, the nnul interest rte. r = 50 + ( 075. ) = r Simplif the eponent. 9 r. 07 = + Divide both sides b 50. r Tke the ninth root of both sides. r Simplif to solve for r r Thus Frnk s current svings ccount is ping n nnul interest rte of.5%, so he would gin slight dvntge b switching to new ccount. Even though the interest rte is divided b n, the number of periods per er, incresing the frequenc of compounding lws increses the totl interest erned on the investment. This is becuse interest is lws clculted on the current blnce. For emple, if ou invest $000 in n ccount with 5% interest, compounded once per er, ou ern $50 in interest. However, if the interest is compounded twice per er, ou ern.5%, or $5 for the first period, then.5% of the new blnce of $05. This brings the totl interest erned to $50.6! Tble shows the interest erned in one er on n investment of $000 in n ccount with 5% nnul interest rte, compounded n times per er.

18 0 Chpter 5 n Clcultion Vlue After Yer (nnull) A = 000( ) $ (binnull) A = (qurterl) A = (monthl) A = (weekl) A = (dil) A = (hourl) A = $050.6 $ $05.6 $05.5 $05.7 $05.7 Tble : Vlue of n Investment with Interest Compounded n Times per Yer Looking t the tble, we see tht the mount of interest keeps growing s we divide the er into more compounding periods, but tht this growth slows drmticll. It looks s if there is limit to how much interest we cn ern, nd this is in fct the cse. Let s emine the formul s n. In order to do this we need to perform some lgebric mnipultion of the formul: nt r At () = P + n rmt P = + m m = P + m rt Substitute m n r =, so =. r m n Bring together the instnces of the vrible m using the properties of eponents. Although the mnipultion m pper strnge, it hs ccomplished the importnt tsk of isolting the prt of the formul tht chnges s n. Looking t the chnge of vribles m = n, we see tht letting n mens tht m s well. Since ever r other quntit remins fied, we onl hve to understnd wht hppens to s m grows without bound. + m m

19 Applictions of Eponentil Functions Section 5. This is not trivil undertking. We might think tht letting m get lrger nd lrger would mke the epression grow lrger nd lrger, s m is the eponent in the epression nd the bse is lrger thn. But t the sme time, the bse pproches s m increses without bound, nd rised to n power is simpl. It turns out tht these two effects blnce one nother out, s we cn see in the tble below: m + m , , Tble : Vlues of + m s m Increses m As m gets lrger nd lrger, we find tht the epression pproches fied number: + m m m This vlue is ver importnt number in mthemtics, so importnt tht it gets its own smbol, the letter e (which highlights its connection to eponentil functions). The Number e DEFINITION The number e is defined s the vlue of + m m e s m. In most pplictions, the pproimtion e. 78 is sufficientl ccurte. Let s look bck t the compound interest formul from before. r At ()= P + P n = + m nt m rt This mens tht if P dollrs is invested in n ccount tht is compounded continuousl, tht is, with n, then the mount in the ccount is determined b the following formul, which results from substituting the definition of e. DEFINITION Continuous Compounding Formul An investment of P dollrs compounded continuousl t n nnul interest rte of r hs vlue fter t ers of: ()= Pe rt. At

20 Chpter 5 EXAMPLE 5 Continuous Compounding Formul If Snd (lst seen in Emple ) hs the option of investing her $0,000 in continuousl compounded ccount erning.5% nnul interest, wht will be the vlue of her ccount in three nd hlf ers? Solution: Agin, the solution boils down to substituting the correct vlues nd evluting the result. Here, P = 0, 000, r = 0. 05, nd t = 5.. ( 0. 05)( 5. ) A( 5. )= 0, 000e = 0, 000e $, This ccount erns $0.9 more thn the qurterl compounded ccount in Emple. As we will see in Section 5., ll eponentil functions cn be epressed with the bse e (or n other bse, for tht mtter). The bse e is so commonl used for eponentil functions tht it is often clled the nturl bse. For instnce, the formul for the rdioctive dec of crbon-, using the bse e, is t At Ae 0. You should verif tht this version of the dec formul does indeed give the sme vlues for At () s the version derived in Emple. ()= Eercises The following problems ll involve working with eponentil functions similr to those encountered in this section.. V 78 people. w 6.6 wtts. A new virus hs broken out in isolted prts of Afric nd is spreding eponentill through tribl villges. The growth of this new virus cn be mpped using the following formul where P stnds for the number of people in villge nd d stnds for the number of ds since the virus first ppered. According to this eqution, how mn people in tribe of 00 will be infected fter 5 ds? 08. d V = P e. A new hbrid cr is equipped with btter tht is ment to improve gs milege b mking the cr run on electric power s well s gsoline. The btteries in these new crs must be chnged out ever so often to ensure proper opertion. The power in the btter decreses ccording to the eponentil eqution below. After 0 ds (d), how much power in wtts (w) is left in the btter? d w( d)= 0e 006.

21 Applictions of Eponentil Functions Section 5.. C $ P 5,50 rbbits 5. W 9 computers 6. F prtments b grms c grms 8. Approimtel 6.%. A oung economics student hs come cross ver profitble investment scheme in which his mone will ccrue interest ccording to the eqution listed below. If this student invests $50 into this lucrtive endevor, how much mone will he hve fter months? I represents the investment nd m represents the number of months the mone hs been invested for. C = Ie 008. m. A fmil releses couple of pet rbbits into the wild. Upon being relesed the rbbits begin to reproduce t n eponentil rte, s shown in the formul below. After ers how lrge is the rbbit popultion where n stnds for the initil rbbit popultion () nd m stnds for the number of months? 05. P = ne m 5. Inside business network, n e-mil worm ws downloded b n emploee. This worm goes through the infected computer s ddress book nd sends itself to ll the listed e-mil ddresses. This worm ver rpidl works its w through the network following the eqution below where C is the number of computers in the network nd W is the number of computers infected h hours fter its discover. After onl 8 hours, how mn computers hs the worm infected if there re 50 computers in the network? 0. h W = C e 6. A construction crew hs been ssigned to build n prtment comple. The work of the crew cn be modeled using the eponentil formul below where A is the totl number of prtments to be built, w is the number of weeks, nd F is the number of finished prtments. Out of totl of 00 prtments, how mn prtments hve been finished fter weeks of work? 0. w F = A e 7. The hlf-life of rdium is pproimtel 600 ers.. Determine so tht A()= t A t 0 describes the mount of rdium left fter t ers, where A 0 is the mount t time t = 0. b. How much of -grm smple of rdium would remin fter 00 ers? c. How much of -grm smple of rdium would remin fter 000 ers? 8. The rdioctive element Polonium- 0 hs reltivel short hlf-life of 0 ds, nd one w to model the mount of Polonium-0 remining fter t ds is with the function At ()= Ae t 0, where A 0 is the mss t time t = 0 (note tht A0 A( 0) =.) Wht percentge of the originl mss of smple of Polonium-0 remins fter one er?

22 Chpter ers b. 9 ers 0. 97,959 people. 8 rbbits. Approimtel. 7. billion b. 5.8 billion. The bnk offering.75% nd monthl compounding.. 0 cents 5. Approimtel.8% 6. Approimtel.9% 7. $,9 9. A certin species of fish is to be introduced into new mn-mde lke, nd wildlife eperts estimte the popultion will grow ccording to ()= t P t 000, where t represents the number of ers from the time of introduction.. Wht is the doubling time for this popultion of fish? b. How long will it tke for the popultion to rech 8000 fish, ccording to this model? 0. The popultion of certin inner-cit re is estimted to be declining ccording to the model t P()= t 7 000e 0., 08, where t is the number of ers from the present. Wht does this model predict the popultion will be in ten ers?. In n effort to control vegettion overgrowth, 00 rbbits re relesed in n isolted re tht is free of predtors. After one er, it is estimted tht the rbbit popultion hs incresed to 500. Assuming eponentil popultion growth, wht will the popultion be fter nother 6 months?. Assuming current world popultion of 6 billion people, n nnul growth rte of.9% per er, nd worstcse scenrio of eponentil growth, wht will the world popultion be in:. 0 ers? b. 50 ers?. Mdih hs $500 tht she wnts to invest in simple svings ccount for two nd hlf ers, t which time she plns to close out the ccount nd use the mone s down pment on cr. She finds one locl bnk offering n nnul interest rte of.75% compounded monthl, nd nother bnk offering n nnul interest rte of.7% compounded dil (65 times per er). Which bnk should she choose?. Mdih, from the lst problem, does some more serching nd finds n online bnk offering n nnul rte of.75% compounded continuousl. How much more mone will she ern over two nd hlf ers if she chooses this bnk rther thn the locl bnk offering the sme rte compounded monthl? 5. Tom hopes to ern $000 in interest in three ers time from $0,000 tht he hs vilble to invest. To decide if it s fesible to do this b investing in simple monthl compounded svings ccount, he needs to determine the nnul interest rte such n ccount would hve to offer for him to meet his gol. Wht would the nnul rte of interest hve to be? 6. An investment firm clims tht its clients usull double their principl in five ers time. Wht nnul rte of interest would svings ccount, compounded monthl, hve to offer in order to mtch this clim? 7. The function C()= t C + r t models the rise in the cost of product tht hs cost of C tod, subject to n verge erl infltion rte of r for t ers. If the verge nnul rte of infltion over the net decde is ssumed to be %, wht will the infltion-djusted cost of $00,000 house be in 0 ers? Round our nswer to the nerest dollr.

23 Applictions of Eponentil Functions Section Approimtel.5% b. 790 people c. The function pproches 0,000 s time goes on. 0. Approimtel b kg c mg. Although the difference is minor in the short term, the ccount with the continuousl compounded rte of.5% will p more over time... $5.7 b. $7.7.. P( t) = t 75(. 06 ) b. 860 bcteri 5. $959.8; $ Given the infltion model C()= t C( + r) t (see the previous problem), nd given tht lof of bred tht currentl sells for $.60 sold for $.0 si ers go, wht hs the verge nnul rte of infltion been for the pst si ers? 0, The function N()= t + 999e t models the number of people in smll town who hve cught the flu t weeks fter the initil outbrek.. How mn people were ill initill? b. How mn people hve cught the flu fter 8 weeks? c. Describe wht hppens to the function N() t s t. 0. The concentrtion, C(), t of certin drug in the bloodstrem fter t minutes is given b the formul 0. t C()= t 005. ( e ). Wht is the concentrtion fter 0 minutes?. Crbon- hs rdioctive hlf-life of pproimtel 0 minutes, nd is useful s dignostic tool in certin medicl pplictions. Becuse of the reltivel short hlf-life, time is crucil fctor when conducting eperiments with this element.. Determine so tht A()= t A t 0 describes the mount of crbon- left fter t minutes, where A 0 is the mount t time t = 0. b. How much of -kg smple of crbon- would be left fter 0 minutes? c. How mn milligrms of -kg smple of crbon- would be left fter 6 hours?. Chrles hs recentl inherited $8000 tht he wnts to deposit into svings ccount. He hs determined tht his two best bets re n ccount tht compounds nnull t rte of.0% nd n ccount tht compounds continuousl t n nnul rte of.5%. Which ccount would p Chrles more interest?. Mrshll invests $50 in mutul fund which bosts 5.7% nnul return compounded seminnull (twice er). After three nd hlf ers, Mrshll decides to withdrw his mone.. How much is in his ccount? b. How much hs he mde in interest from his investment?. Adm is working in lb testing bcteri popultions. After strting out with popultion of 75 bcteri, he observes the chnge in popultion nd notices tht the popultion doubles ever 7 minutes. Find. the eqution for the popultion P in terms of time t in minutes, rounding to the nerest thousndth, nd b. the popultion fter hours. 5. Your credit union offers specil interest rte of 0% compounded monthl for the first er for student svings ccount opened in August if the student deposits $5000 or more. You received totl of $9000 for grdution, nd ou decide to deposit ll of it in this specil ccount. Assuming ou open our ccount in August nd mke no withdrwls for the first er, how much mone will ou hve in our ccount t the end of Februr (fter 6 months)? How much will ou hve t the end of the following Jul (fter one full er)?

24 6 Chpter $.87 b. $9.8 c. Second cse; eplntions will vr. 7. $ Yes; $0.0 remining 9. $0, $70,66.7 b. $. c. $6. d. $6,5.9. $78.7. $ $ b. $ c. Yes; eplntions will vr. 6. You hve svings ccount of $000 with n interest rte of 6.8%.. How much interest would be erned in ers if the interest is compounded nnull? b. How much interest would be erned in ers if the interest is compounded seminnull? c. In which cse do ou mke more mone on interest? Eplin wh this is so. 7. If $500 is invested in continuousl compounded certificte of deposit with n nnul interest rte of.%, wht would be the ccount blnce t the end of ers? 8. The new furniture store in town bosts specil in which ou cn bu n set of furniture in their store nd mke no monthl pments for the first er. However, the fine print ss tht the interest rte of 7.5% is compounded qurterl beginning when ou bu the furniture. You re considering buing set of living room furniture for $000, but know ou cnnot sve up more thn $500 in one er s time. Cn ou full p off our furniture on the one er nniversr of hving bought the furniture? If so, how much mone will ou hve left over? If not, how much more mone will ou need? 9. When Nicole ws born, her grndmother ws so ecited bout her birth tht she opened certificte of deposit in Nicole s honor to help send her to college. Now t ge 8, Nicole s ccount hs $8,6.9. How much did her grndmother originll invest if the interest rte hs been 8.% compounded nnull? 0. Infltion is reltive mesure of our purchsing power over time. The formul for infltion is the sme s the compound interest formul, but with n =. Given the current vlues below, wht will the vlues of the following items be 0 ers from now if infltion is t 6.%?. n SUV: $8,000 b. lof of bred: $.79 c. gllon of milk: $.0 d. our slr: $,000. Deprecition is the decrese of n item s vlue nd cn be determined using formul similr to tht for compound interest: V = P( r) t, where V is the new vlue. If the prticulr cr ou bu upon grdution from college costs $7,500 nd deprecites t rte of 6% per er, wht will the vlue of the cr be in 5 ers when ou p it off?. Assume the interest on our credit crd is compounded continuousl with n APR (nnul percentge rte) of 9.8%. If ou put our first term bill of $98 on our credit crd, but do not hve to mke pments until ou grdute ( ers lter), how much will ou owe when ou strt mking pments?. Suppose ou deposit $5000 in n ccount for 5 ers t rte of 8.5%.. Wht would be the ending ccount blnce if the interest is continuousl compounded? b. Wht would be the ending ccount blnce if the interest is compounded dil? c. Are these two nswers similr? Wh or wh not?

25 5. Logrithmic Functions nd Their Grphs Section 5. 7 Logrithmic Functions nd Their Grphs TOPICS. Definition of logrithmic functions. Grphing logrithmic functions. Evluting elementr logrithmic epressions. Common nd nturl logrithms T. Inputting logrithms, grphing logrithmic functions TOPIC Definition of Logrithmic Functions Currentl, we re onl ble to solve smll subset of possible eponentil equtions. Solvble We cn solve the eqution = 8 b writing 8 s nd equting eponents. This is n emple of n elementr eponentil eqution we lerned to solve in Section 5.. If we know A, P, n, nd t, we cn solve the compound interest eqution r A = P + n for the nnul interest rte r, s we did in Emple of Section 5.. nt Not Esil Solvble Yet We cnnot solve the eqution = 9 in the sme w, lthough this eqution is onl slightl different. All we cn s t the moment is tht must be bit lrger thn. If we know A, P, nd t, it is not so es to determine the nnul interest rte r from the eqution A = Pe rt, for clculting interest tht is continuousl compounded. In both of the first two equtions, the vrible ppers in the eponent (mking them eponentil equtions), nd we ren t ble to rewrite the eqution with the vrible outside the eponent. We cn onl solve the first eqution becuse we cn rewrite 8 s power of, nd even this doesn t remove the vrible from the eponent. In the second pir of equtions, we re ble to solve for r in the first cse, but it is difficult in the continuous-compounding cse: once gin the vrible is inconvenientl stuck in the eponent. We need w of undoing eponentition. This might seem fmilir. We hve undone functions before (whether we relized it t the time or not) b finding inverses of functions. Therefore, we need to find the inverse of the generl eponentil function f =. Fortuntel, we know before we

26 8 Chpter 5 strt tht f = hs n inverse, becuse the grph of f psses the horizontl line test (regrdless of whether 0 < < or > ). If we ppl the lgorithm for finding inverses of functions, we hve the following: f = = Rewrite the function s n eqution b replcing f with. = =? Then interchnge nd nd proceed to solve for. At this point, we re stuck gin. Wht is? No concept or nottion tht we hve encountered up to this point llows us to solve the eqution = for the vrible, nd this is the reson for introducing new clss of functions clled logrithms. DEFINITION Logrithmic Functions Let be fied positive rel number not equl to. The logrithmic function with bse is defined to be the inverse of the eponentil function with bse, nd is denoted log. In smbols, if f =, then f = log. In eqution form, the definition of logrithm mens tht the equtions = nd = log re equivlent. Note tht is the bse in both equtions: either the bse of the eponentil function or the bse of the logrithmic function. EXAMPLE Eponentil nd Logrithmic Equtions Use the definition of logrithmic functions to rewrite the following eponentil equtions s logrithmic equtions.. 8 = b. 5 = 65 c. 7 = z Then rewrite the following logrithmic equtions s eponentil equtions. d. log 9 = e. = log 8 5 f. = log Solutions:. = log 8 b. = log 5 65 c. = log 7 z d. = 9 e. 8 = 5 f. = Note tht in ech cse, the bse of the eponentil eqution is the bse of the logrithmic eqution.

27 Logrithmic Functions nd Their Grphs Section 5. 9 TOPIC Grphing Logrithmic Functions Since logrithmic functions re inverses of eponentil functions, we cn lern gret del bout the grphs of logrithmic functions b reclling the grphs of eponentil functions. For instnce, the domin of log (for n llowble ) is the positive rel numbers, becuse the positive rel numbers constitute the rnge of (the domin of f is the rnge of f ). Similrl, the rnge of log is the entire set of rel numbers, becuse the rel numbers mke up the domin of. Recll tht the grphs of function nd its inverse re reflections of one nother with respect to the line =. Since eponentil functions come in two forms bsed on the vlue of (0 < < nd > ), logrithmic functions lso fll into two ctegories. In both of the grphs in Figure, the green curve is the grph of n eponentil function nd the purple curve is the corresponding logrithmic function. = =,, ( 0, ) (, ) ( 0, ),, ( 0, ) (, ) (,) ( 0, ), ( ) < > Figure : The Two Clsses of Logrithmic Functions (, ), nd In ll cses, the points 0,,, lie on the grph of logrithmic function with bse. Frequentl, these points will be enough to get good ide of the shpe of the grph. Note tht the domin of ech of the logrithmic functions in Figure is indeed ( 0, ), nd tht the rnge in ech cse is (, ). Also note tht the -is is verticl smptote for both, nd tht neither hs horizontl smptote. CAUTION! While it m pper so on the grph, logrithmic functions do not hve horizontl smptote. The reson tht logrithmic functions often look like the hve horizontl smptote is becuse the re mong the slowest growing functions in mthemtics! Consider the function f= log. We hve f ( 0)= 0, f ( 08)=, nd f ( 096)=. While the function m increse its vlue ver slowl s increses, it never pproches n smptote.

28 0 Chpter 5 EXAMPLE Grphing Logrithmic Functions Sketch the grphs of the following logrithmic functions.. f= log b. g= log Note: Once gin, plotting few ke points will provide good ide of the shpe of the function. Solutions:. = 0 ( 0, ), (, ) b. = 0, ( 0, ), EXAMPLE Grphing Logrithmic Functions Sketch the grphs of the following functions.. f = log ( + ) + b. g= log c. h = log Note: Begin b grphing the bse function, then ppl n trnsformtions. Solutions:. = 5, 0 (, ) (, ) ( 0, ), (, ) Begin b grphing the bse function, which is log. This is shown in green. Since hs been replced b +, we shift the grph units to the left. To find the grph of f, we shift the result up unit, since hs been dded to the function. This grph is shown in purple. Note tht the smptote hs lso shifted to the left.

29 Logrithmic Functions nd Their Grphs Section 5. b. = (, ) ( 0, ), (, ) 0,, The bsic shpe of the grph of g is the sme s the shpe of = log, drwn in green. To obtin g from log, the vrible is replced with, which shifts the grph unit to the right, nd then is replced b, which reflects the grph with respect to the -is. c. = 0, ( 0, ), (, ) (, ), We begin with the green curve, which is the grph of log. We then shift the grph units down to obtin the grph of the function h. TOPIC Evluting Elementr Logrithmic Epressions Now tht we hve grphed logrithmic functions, we cn ugment our understnding of their behvior with few lgebric observtions. These will enble us to evlute some logrithmic epressions nd solve some elementr logrithmic equtions. is lws on the grph of First, our work in Emple suggests tht the point 0, log for n llowble bse, nd this is indeed the cse. A similr observtion is tht (,) is lws on the grph of log. These two fcts re ctull just resttements of two corresponding fcts bout eponentil functions, consequence of the definition of logrithms: log = 0, becuse 0 = log =, becuse = More generll, we cn use the fct tht the functions log nd re inverses of one nother to write: log log = = nd In Emple, we use the first sttement to evlute logrithmic epressions. Note the similrit to solving eponentil equtions; we rewrite the rgument of the logrithm s power of the bse, llowing us to simplif the epression.

30 Chpter 5 EXAMPLE Logrithmic Epressions Note: We write ech of the equivlent eponentil equtions s reference. Evlute the following logrithmic epressions.. log 5 5 b. log c. log π π d. log 7 e. log 6 f. log 0 00 Solutions:. log 5 = log = Rewrite 5 s power of 5. Equivlent eponentil eqution: 5 = 5 b. log = log = c. logπ π = log = d. log 7 = 0 π π Rewrite s power of. Equivlent eponentil eqution: = Rewrite the rdicl s rtionl eponent. Equivlent eponentil eqution: π = π Equivlent eponentil eqution: = 7 0 e. log6 = log6 6 = Rewrite s power of 6. Equivlent eponentil eqution: = 6 f. log 00 = log = Rewrite 00 s power of 0. Equivlent eponentil eqution: = 0 00 In Emple 5, we use both sttements to solve logrithmic nd eponentil equtions. Observe how converting between logrithmic nd eponentil form cn mke n eqution esier to solve.

31 Logrithmic Functions nd Their Grphs Section 5. EXAMPLE 5 Solving Logrithmic Equtions Solve the following equtions involving logrithms.. log6 = b. log = c. log 8 = 5 Solutions:. log6 = = 6 = 6 = log b. = log = log = = c. log 8 = 5 8 = = 5 5 = = 5 5 = 5 Convert the eqution to eponentil form. Simplif the eponent, then solve for. This time, we convert from the eponentil form to the logrithmic form. We cn equte the bses, just like we equte the eponents in n elementr eponentil eqution. Rewrite the eqution in eponentil form. Rewrite 8 s power of. Simplif using properties of eponents. Set the eponents equl to ech other, then solve. TOPIC Common nd Nturl Logrithms We hve lred mentioned the fct tht the number e, clled the nturl bse, pls fundmentl role in mn importnt rel-world situtions nd in higher mthemtics, so it is not surprising tht the logrithmic function with bse e is worth of specil ttention. For historicl resons, nmel the fct tht our number sstem is bsed on powers of 0, the logrithmic function with bse 0 is lso singled out. DEFINITION Common nd Nturl Logrithms The function log 0 is clled the common logrithm, nd is usull written log. The function log e is clled the nturl logrithm, nd is usull written ln.

32 Chpter 5 Another w in which these prticulr logrithms re specil is tht most clcultors, if the re cpble of clculting logrithms t ll, re onl equipped to evlute common nd nturl logrithms. Such clcultors normll hve button lbeled LOG for the common logrithm nd button lbeled LN for the nturl logrithm. Properties of Nturl Logrithms Properties ln = e = Properties Resons.. ln= 0 ln e = Rise e to the power 0 to get. Rise e to the power to get e.. ln e = Rise e to the power to get e. ln. e = ln is the power to which e must be rised to get. EXAMPLE 6 Evluting Logrithmic Epressions Evlute the following logrithmic epressions.. ln e c. ln 78. b. log000 d. log ( 0. 5) Solutions:. ln e = ln e = b. log000 = log0 = c. ln ( 78. ) 56. d. log ( 0. 5) 0. No clcultor is necessr for this problem, just n ppliction of n elementr propert of logrithms. Agin, no clcultor is required. This time, clcultor is needed, nd onl n pproimte nswer cn be given. Be sure to use the correct logrithm. Agin, we must use clcultor, though we cn s beforehnd tht the nswer should be onl slightl lrger thn, s log0 = nd 0.5 is onl slightl lrger thn 0. TOPIC T Inputting Logrithms, Grphing Logrithmic Functions Common nd nturl logrithms cn be entered into grphing clcultor using the LOG nd LN buttons, respectivel. For instnce, to grph the function f = log( ), press nd LOG. The opening prenthesis ppers utomticll, so we just hve to tpe in the rgument,, nd the right-hnd prenthesis.

33 Logrithmic Functions nd Their Grphs Section = log = log 6 6. = log 7. = log b. 5. = log. C 6. = log. V 7. = log 8. = log = log 0. = log e. e = log. e = log N The following grph should pper:. 8 = = = b = 7. 5 = b 8. 8 = 5 d 9. W = 5 0. T = 7 6. =. π=(. e =. e = 5 5. ) CAUTION! Notice tht the grph ppers to cut off round the point where =. We know, however, tht the grph hs verticl smptote t =, nd pproches tht smptote even though it does not show on the grphing clcultor. Eercises Write the following equtions in logrithmic terms.. 65 = 5. 6 = 6. = 7. b =. 5.. = C 6.. = V = 7. = 8. 6 = e e =. =. e = N Write the following logrithmic equtions s eponentil equtions. 6.. log 8 =. log 8 = 5. log b = 6. log 9 = 7. log 5 = b 8. log 5 8 = d log 5 W = 0. log 7 T = 6. log π =. log π=. ln =. ln 5 =

34 6 Chpter 5 8. Sketch the grphs of the following functions. See Emples nd f = log ( ) 6. g= log 5 ( + ) 7. r= log 8. p= log q= log ( ) 0. s= 5 log. h= log 7 ( ) +. m= log.. f = log ( 6 ). p= log( + ) 5. s= ( ) log 6. g= log 5. Mtch the grph of the pproprite eqution to the logrithmic function. 7. f = log 8. f = log ( ) 9. f = log 0. f = log ( ). f = log. f = log.. f = log ( ). f = log 5. f = log +. b. c.. 5. d. e. f e g. h. i. 8. g 9. b 0. f. h

35 . c. d. 5. i {} { } {} {} { } No solution 70. { 0} 7. {} { } { } Logrithmic Functions nd Their Grphs Section 5. 7 Evlute the following logrithmic epressions without the use of clcultor. See Emples nd log log 8. log log log 7 5. log ( log ) ln e log log 55. ln e ln e 56. log( log ( 00 )) 57. log 59. log log 8 log 000 Use the elementr properties of logrithms to solve the following equtions. See Emple log 6 = 6. log 6 log 6. log 5 5 = 65. log = 6. log6 = log b = = log log log = 68. log7 = 69. = log 0 = 0 7. log log = 7. 6 log e = e Solve the following logrithmic equtions, using clcultor if necessr to evlute the logrithms. See Emples 5 nd 6. Epress our nswer either s frction or deciml rounded to the nerest hundredth. = 7. = 7. log. log 75. ln( + ) = 76. ln = 77. ln e = ln( ln ) = log9 = 80. log e = log = 8. log log ( ) 8. log 9 7. ± {9.09} 76. {0.8} 77. {5.60} 78. ± = { } e or { ± 65. } { } { } log 00 = 9 8. { 97. 8} 8. { } 8. { 0, 000, 000, 00}

36 8 Chpter 5 Properties nd Applictions of Logrithms TOPICS 5... Properties of logrithms. The chnge of bse formul Applictions of logrithmic functions TOPIC Properties of Logrithms In Section 5., we introduced logrithmic functions nd studied some of their elementr properties. The motivtion ws our inbilit, t tht time, to solve certin eponentil equtions. Let us reconsider the two smple problems tht initited our discussion of logrithms nd see if we hve mde progress. We begin with the continuousl compounding interest problem. EXAMPLE Continuousl Compounded Interest Anne reds n d in the pper for new bnk in town. The bnk is dvertising continuousl compounded svings ccounts in n ttempt to ttrct customers, but fils to mention the nnul interest rte. Curious, she goes to the bnk nd is told b n ccount gent tht if she were to invest, $0,000 in n ccount, her mone would grow to $0,0.0 in one er s time. But, strngel, the gent lso refuses to divulge the erl interest rte. Wht rte is the bnk offering? Solution: We need to solve the eqution A = Pe rt for r, given tht A = 0, 0. 0, P = 0, 000, nd t =. 0, 0. 0 = 0, 000e. 000 = e ln (. 000)= r r r 00. r Substitute the given vlues. Divide both sides b 0,000. Convert to logrithmic form. Evlute using clcultor. Note tht we use the nturl logrithm since the bse of the eponentil function is e. While we must use clcultor, we cn now solve the eqution for r.

37 Properties nd Applictions of Logrithms Section 5. 9 EXAMPLE Solving Eponentil Equtions Solve the eqution = 9. Solution: We convert the eqution to logrithmic form to obtin the solution = log 9. Unfortuntel, this nswer still doesn t tell us nthing bout in deciml form, other thn tht it is bound to be slightl more thn. Further, we cn t use clcultor to evlute log 9 since the bse is neither 0 nor e. As Emple shows, our bilit to work with logrithms is still incomplete. In this section we will derive some importnt properties of logrithms tht llow us to solve more complicted equtions, s well s provide deciml pproimtion to the solution of = 9. The following properties of logrithmic functions re nlogs of corresponding properties of eponentil functions, consequence of how logrithms re defined. PROPERTIES Properties of Logrithms Let (the logrithmic bse) be positive rel number not equl to, let nd be positive rel numbers, nd let r be n rel number.. log = log + log ( the log of product is the sum of the logs ). log = log log ( the log of quotient is the difference of the logs ) = ( the log of something rised to power is the power times r. log rlog the log ) We illustrte the link between these properties nd the relted properties of eponents b proving the first one. Tr proving the second nd third s further prctice. Proof: Let m = log nd n = log. The equivlent eponentil forms of these two equtions re = m nd = n. Since we re interested in the product, note tht m n = m n =. The sttement m + = n cn then be converted to logrithmic form, giving us log = m + n. Referring bck to the definition of m nd n, we hve log = log + log. If the properties of logrithms pper strnge t first, remember tht the re just the properties of eponents restted in logrithmic form.

38 0 Chpter 5 CAUTION! Errors in working with logrithms often rise from incorrect recll of the logrithmic properties. The tble below highlights some common mistkes. Incorrect Sttements Correct Sttements log ( + )= log + log log = log + log log log log log log = log = log + log = log log log log log = log log log log = log log log log = z log ( z) = log log log z log logz ( z) = + log + log z In some situtions, we will find it useful to use properties of logrithms to decompose complicted epression into sum or difference of simpler epressions, while in other situtions we will do the reverse, combining sum or difference of logrithms into one logrithm. Emples nd illustrte these processes. EXAMPLE Epnding Logrithmic Epressions Use the properties of logrithms to epnd the following epressions s much s possible (tht is, decompose the epressions into sums or differences of the simplest possible terms) log ( 6 ) b. log c. log z Note: As long s the bse is the sme for ech term, its vlue does not ffect the use of the properties. Solutions: = + +. log 6 log 6 log log = log + log + log = + log + log Use the first propert to rewrite the epression s three terms. We cn evlute the first term nd rewrite the second nd third terms using the third propert.

39 Properties nd Applictions of Logrithms Section 5. b. log = log z z = log z = ( log + log log z ) = log + log log z Rewrite the rdicl s n eponent. Bring the eponent in front of the logrithm using the third propert. Epnd the epression using the first two properties. Appl the third propert to the terms tht result. c. Recll tht if bse is not eplicitl written, it is ssumed to be 0. This bse is convenient when working with numbers in scientific nottion log log 7. log 0 log = + log 7. log = log Epnd using the first nd second properties. Evlute the first two terms nd use the third propert on the lst term. or leve it in ect form. Use the It is pproprite to either evlute log 7. contet of the problem to decide which form is more convenient. EXAMPLE Condensing Logrithmic Epressions Use the properties of logrithms to condense the following epressions s much s possible (tht is, rewrite the epressions s sum or difference of s few logrithms s possible).. log log c. log 5 + log b b b. ln ln + ln Note: Often, there will be multiple orders in which we cn ppl the properties to find the finl result. Solutions:. log log log log = + = log = log + log 9 9 Use the third propert to mke the coefficients pper s eponents. Evlute the eponents. Combine terms using the first propert.

40 Chpter 5 b. ln ln ln ln ln + = + ln = ln + ln = ln or ln c. log 5+ log = log 5 + log b b b b = log b 5 or log b 5 Rewrite ech term to hve coefficient of or - using the third propert. We cn then combine the terms using the second propert. The finl nswer cn be written in severl different ws, two of which re shown. Rewrite the coefficient s n eponent, then combine terms. TOPIC The Chnge of Bse Formul The properties we just derived cn be used to provide n nswer to question bout logrithms tht hs been left unnswered thus fr. A specific illustrtion of the question rose in Emple : how do we determine the deciml form of number like log 9? Surprisingl, to nswer this question we will undo our work in Emple. We ssign vrible to the result log 9, convert the resulting logrithmic eqution into eponentil form, tke the nturl logrithm of both sides, nd then solve for the vrible. = log 9 Let equl the result from before, log 9. = 9 Convert the eqution to eponentil form. ln( )= ln 9 Tke the nturl logrithm of both sides. ln = ln 9 Move the vrible out of the eponent using the third propert of logrithms. ln 9 = Simplif. ln 7. Evlute with clcultor. While using logrithm with n bse will give the correct solution to this problem, if clcultor is to be used to pproimte the number log 9, there re (for most clcultors) onl two good choices: the nturl log nd the common log. If we hd done the work bove with the common logrithm, the finl nswer would hve been the sme. Tht is, log log And even though it would not be es to evlute, log 9 7. log for n llowble logrithmic bse.

41 Properties nd Applictions of Logrithms Section 5. More generll, logrithm with bse b cn be converted to logrithm with bse through the sme resoning. This llows us to evlute ll logrithmic epressions. THEOREM Chnge of Bse Formul Let nd b both be positive rel numbers, neither of them equl to, nd let be positive rel number. Then log log b =. log b EXAMPLE 5 Chnge of Bse Formul Evlute the following logrithmic epressions, using the bse of our choice.. log 7 5 b. log c. log π 5 Note: Both the common nd nturl logrithm work in solving these problems. Solutions: ln. log = ln7. 9 log b. log = log c. log π.585 log 5 5 = log π. 06 Appl the chnge of bse formul. Evlute using clcultor. Appl the chnge of bse formul. This time we use the common logrithm. Since the bse of the logrithm is frction, we should epect negtive nswer. Once gin, we ppl the chnge of bse formul, then evlute using clcultor. TOPIC Applictions of Logrithmic Functions Logrithms pper in mn different contets nd hve wide vriet of uses. This is due prtl to the fct tht logrithmic functions re the inverses of eponentil functions, nd prtl to the logrithmic properties we hve discussed. In fct, the mthemticin who cn be most credited for inventing logrithms, John Npier (550 67) of Scotlnd, ws inspired in his work b the convenience of wht we now cll logrithmic properties. Computtionll, logrithms re useful becuse the relocte eponents s coefficients, thus mking them esier to work with. Consider, for emple, ver lrge number such s 0 7 or ver smll number such s The common logrithm (used becuse it hs bse of 0) epresses these numbers on more comfortble scle:

42 Chpter 5 = + = log 0 log log 0 7 log log( 6 0 )= log 6+ log( 0 )= 9+ log 6 8. Npier, working long before the dvent of electronic clculting devices, devised logrithms in order to tke dvntge of this propert. In chemistr, the concentrtion of hdronium ions in solution determines its cidit. Since concentrtions re smll numbers tht vr over mn orders of mgnitude, it is convenient to epress cidit in terms of the ph scle, s follows. DEFINITION The ph Scle + + The ph of solution is defined to be log HO, where HO is the concentrtion of hdronium ions in units of moles / liter. Solutions with ph less thn 7 re sid to be cidic, while those with ph greter thn 7 re bsic. ph = 0 ph = ph = ph = ph = ph = 5 ph = 6 ph = 7 ph = 8 ph = 9 ph = 0 ph = ph = ph = ph = Btter Acid Hdrochloric Acid Secreted b Stomch Lining Lemon Juice, Gstric Acid, Vinegr Tomto Juice Soft Drinking Wter Urine, Sliv Pure Wter Se Wter Bking Sod Gret Slt Lke Ammoni Solution Sop Wter Bleches Liquid Drin Clener Grpefruit, Ornge Juice Figure : ph of Common Substnces

43 Properties nd Applictions of Logrithms Section 5. 5 EXAMPLE 6 The ph Scle If smple of ornge juice is determined + to hve HO concentrtion of moles / liter, wht is its ph? Solution: Appling the bove formul (nd using clcultor), the ph is equl to ph = log =.8. After doing this clcultion, the reson for the minus sign in the formul is more pprent. B multipling the log of the concentrtion b, the ph of solution is positive, which is convenient for comprtive purposes. The energ relesed during erthqukes cn vr gretl, but logrithms provide convenient w to nlze nd compre the intensit of erthqukes. DEFINITION The Richter Scle Erthquke intensit is mesured on the Richter scle (nmed for the Americn seismologist Chrles Richter, ). In the formul tht follows, I 0 is the intensit of just-discernible erthquke, I is the intensit of n erthquke being nlzed, nd R is its rnking on the Richter scle. I R = log I B this mesure, erthqukes rnge from clssifiction of smll ( R < 5. ), to moderte ( 5. R < 55.), to lrge ( 55. R < 65.), to mjor ( 65. R < 75.), nd finll to gretest ( 75. R). 0 The bse 0 logrithm mens tht ever increse of unit on the Richter scle corresponds to n increse b fctor of 0 in the intensit. This is chrcteristic of ll logrithmic scles. Also, note tht brel discernible erthquke hs rnk of 0, since log = 0.

44 6 Chpter 5 EXAMPLE 7 The Richter Scle. + log 5 The Jnur 00 erthquke in the stte of Gujrt in Indi ws 80, 000, 000 times s intense s 0-level erthquke. Wht ws the Richter rnking of this devstting event?. ln+ ln ln. + ln p lnq. + log 5. + log 9 log 9 6. log 6 p log 6 q 7. ln ln + p + 5lnq log b 5 5 log c 9. log + log 0. + log +. log +. + ln Solution: If we let I denote the intensit of the Gujrt erthquke, then I 80, 000, 000I 0 R = log I 7 = log( 8 0 ) 7 = log( 8) + log( 0 ) = log( 8) = 80, 000, 000I 0, so The Gujrt erthquke thus fell in the ctegor of gretest on the Richter scle. Sound intensit is nother quntit tht vries gretl, nd the mesure of how the humn er perceives intensit, in units clled decibels, is ver similr to the mesure of erthquke intensit. DEFINITION The Decibel Scle In the decibel scle, I 0 is the intensit of just-discernible sound, I is the intensit of the sound being nlzed, nd D is its decibel level: D = 0 log I I Decibel levels rnge from 0 for brel discernible sound, to 0 for the level of norml converstion, to 80 for hev trffic, to 0 for loud rock concert, nd finll (s fr s humns re concerned) to round 60, t which point the erdrum is likel to rupture. 0

45 Properties nd Applictions of Logrithms Section 5. 7 EXAMPLE 8 The Decibel Scle. log ( + z) log. + log 5. logb + log b log z b 6. ln7+ ln 7. + log + 8. blog 9. log b b c 0. log 5. log 5 ( + 5). ln. log + 5. log p q 5. ln p q 6. log 5 7. log 0 8. log b z 9. ln 0. log =. log 5. log Given tht I 0 = 0 wtts / meter, wht is the decibel level of jet irliner s engines t distnce of 5 meters, for which the sound intensit is 50 wtts / meter? Solution: 50 D = 0 log 0 = 0 log 5 0 = 0 log 5+ 7 In other words, the sound level would probbl not be literll er-splitting, but it would be ver pinful. Eercises Use the properties of logrithms to epnd the following epressions s much s possible. Simplif n numericl epressions tht cn be evluted without clcultor. See Emple.. log 5 5. ln. ln e p q. log ( 00) 5. log log p 6 q 7. ln pq 5 e 7. log 0. log ( ) b 8. log 5 9. log log 00 c z. log. log log 00, b 7. log b b c 6. ln ln( ln( e e )) ( ) 5. log b 8. ln ln e e z ( )

46 8 Chpter 5. ln 5. ln 6 5. log = 0 6. log 7 7. ln 8. log6 9. log 0. log. log 8. log.. e p b e Use the properties of logrithms to condense the following epressions s much s possible, writing ech nswer s single term with coefficient of. See Emple. 9. log log 0. log log 5 5. log 5 log 5. log + log ln ln ln. 5 ( log7 ( ) log 7 ( pq) ) 5. ln + ln p lnq 6. log5 log 5 7. log( 0 ) log 8. logb log + log b 9. ln z ln 0. log log. log 0 log 5. log0 log log ln5 + ln. ln 8 ln + ln log 6 log 6. log log ln8+ ln 8. log log+ log log log9 log 0. log log. log 8 ( ) 05. log 8 6. log + log 8 log 6 Use the properties of logrithms to write ech of the following s single term tht does not contin logrithm.. 5 log log log. 0 log log log b + log log 6 log 5. e ln + + ln 5. e ln + ln p 5 5( 6. e ln + ln ) 9. 0 log log log 5. log log ( 6 ) 5. log 6 log 9 Evlute the following logrithmic epressions. See Emple log log log log log log ln 6. log 6 6. log log π 6. log ln( log) 66. log

47 Properties nd Applictions of Logrithms Section , log6 68. log log log log log log log log log log log.. Without using clcultor, evlute the following epressions log log ln e + ln e 8. log ln e 5 log 8. log ( log log ) 8 Find the vlue of in ech of the following equtions. Epress our nswer s ect s possible, or s deciml rounded to the nerest hundredth. 85. log 0 = 86. log 6 79 = 87. log 59 = 88. log 65 = 89. log 79 = log = 8 9. log = 9. log 6, 807 = 7 9. log = 0 Solve the following ppliction problems. See Emples 6, 7, nd A certin brnd of tomto juice + hs HO concentrtion of moles / liter. Wht is the ph of this brnd? 95. One tpe of detergent, when dded to neutrl wter with ph of 7, + results in solution with HO concentrtion tht is times weker thn tht of the wter. Wht is the ph of the solution? Wht is the concentrtion of HO in lemon juice with ph of.? 97. The 99 Northridge, Cliforni erthquke mesured 6.7 on the Richter scle. Wht ws the intensit, reltive to 0-level erthquke, of this event? 98. How much stronger ws the 00 Gujrt erthquke (7.9 on the Richter scle) thn the 99 Northridge erthquke described in Eercise 97? 99. A construction worker operting jckhmmer would eperience noise with n intensit of 0 wtts / meter if it weren t for er protection. Given tht I 0 = 0 wtts / meter, wht is the decibel level for such noise? 00. A microphone picks up the sound of thunderclp nd mesures its decibel level s 05. Given tht I 0 = 0 wtts / meter, with wht sound intensit did the thunderclp rech the microphone? 9.,08,576

48 0 Chpter moles / liter 97. 5,0,87 times stronger 98. fctor of decibels wtts / meter , es 0.. decibels minutes b. 7:00 p.m. c. F; no 0. Mtt, lifegurd, hs to mke sure tht the ph of the swimming pool sts between 7. nd 7.6. If the ph is out of this rnge, he hs to dd chemicls tht lter the ph level of the pool. If Mtt mesures + the HO concentrtion in the swimming pool to be moles / liter, wht is the ph? Does he need to chnge the ph b dding chemicls to the wter? 0. The intensit of ct s soft purring is mesured to be Given tht I 0 = 0 wtts / meter, wht is the decibel level of this noise? 0. Newton s Lw of Cooling sttes tht the rte t which n object cools is proportionl to the difference between the temperture of the object nd the surrounding temperture. If C denotes the surrounding temperture nd T 0 denotes the temperture t time t = 0, the temperture of n object t time t is given b T()= t C+ ( T C) e kt 0, where k is constnt tht depends on the prticulr object under discussion.. You re hving friends over for te nd wnt to know how long fter boiling the wter it will be drinkble. If the temperture of our kitchen sts round 7F nd ou found online tht the constnt k for te is pproimtel 0.09, how mn minutes fter boiling the wter will the te be drinkble (ou prefer our te no wrmer thn 0F)? Recll tht wter boils t F. b. As ou intern for our locl crime scene investigtion deprtment, ou re sked to determine t wht time victim died. If ou re told k is pproimtel 0.97 for humn bod nd the bod s temperture ws 7F t :00.m., nd the bod hs been in storge building t constnt 60F, pproimtel wht time did the victim die? Recll the verge temperture for humn bod is 98.6F. Note in this sitution, t is mesured in hours. c. When helping our fther cook turke, ou were told to remove the turke when the thickest prt hd reched 80F. If ou remove the turke nd plce it on the tble in room tht is 7F, nd it cools to 55F in 0 minutes, wht will the temperture of the turke be t lunch time (n hour nd 5 minutes fter the turke is removed from the oven)? Should ou wrm the turke before eting?

49 Eponentil nd Logrithmic Equtions Section 5.5 Eponentil nd Logrithmic Equtions TOPICS Converting between eponentil nd logrithmic forms Further pplictions of eponentil nd logrithmic equtions TOPIC Converting Between Eponentil nd Logrithmic Forms At this point, we hve ll the tools we need to solve the most common sorts of eponentil nd logrithmic equtions. All tht is left is to develop our skill in using the tools. We hve lred solved mn eponentil nd logrithmic equtions, using elementr fcts bout eponentil nd logrithmic functions to obtin solutions. However, mn equtions require bit more work to solve. While there is no lgorithm to follow in deling with more complicted equtions, if given eqution doesn t ield solution esil, tr converting it from eponentil form to logrithmic form or vice vers. All of the properties of eponents nd their logrithmic counterprts re of gret use s well. For reference, the logrithmic properties tht we hve noted throughout Sections 5. nd 5. re restted here. PROPERTIES Summr of Logrithmic Properties. The equtions = nd = log re equivlent, nd re, respectivel, the eponentil form nd the logrithmic form of the sme sttement.. The inverse of the function f = is f = log, nd vice vers.. A consequence of the lst point is tht log ( ) = nd =. In prticulr, log = 0 nd log.. log = log + log ( the log of product is the sum of the logs ) 5. log log log = ( the log of quotient is the difference of the logs ) = ( the log of something rised to power is the power times r 6. log rlog the log ) The net severl emples illustrte tpicl uses of the properties, nd how converting between the eponentil nd logrithmic forms of n eqution cn led to solution.

50 Chpter 5 EXAMPLE Solving Eponentil Equtions Note: This eqution is not esil solved in eponentil form, since the two sides of the eqution do not hve the sme bse. Solve the eqution 5 =. Epress the nswer ectl nd s deciml pproimtion. Solution: There re two ws to convert the eqution into logrithmic form. We will eplore both, nd see tht the led to the sme nswer. The first method is to tke the nturl (or common) logrithm of both sides. 5 = 5 ln( )= ln ( 5) ln = ln ln 5 = ln ln 5 = ln ln = + 5ln 5 Tke the nturl logrithm of both sides. Use properties of logrithms to bring the vrible out of the eponent. Divide both sides b ln. Simplif. An ect form of the nswer. The second method is to rewrite the eqution using the definition of logrithms, then ppl the chnge of bse formul to work with nturl (or common) logrithms. 5 = 5 = log ln 5 = ln ln = + 5ln 5 The ke step to finding the ect nswer is to remove the vrible from the eponent, which is chieved b converting to logrithmic form. The ke step to finding deciml pproimtion is to chnge the bse (of the eponent nd logrithm) to either e or 0, llowing the use of clcultor. ln = ln 5 Rewrite the eqution using the definition of logrithms. Rewrite the logrithmic term using the chnge of bse formul. Appling the sme lgebr s bove leds to the sme ect nswer. An pproimte form of the nswer. We cn lso use clcultor to verif this solution in the originl eqution.

51 Eponentil nd Logrithmic Equtions Section 5.5 EXAMPLE Solving Eponentil Equtions + Solve the eqution 5 =. Epress the nswer ectl nd s deciml pproimtion. Solution: As in the first emple, tking logrithm of both sides is the ke. We will use the common logrithm this time, but the nturl logrithm would work just s well. + 5 = + log( 5 )= log( ) ( ) log5= ( + ) log log5 log 5= log+ log log5 log = log+ log 5 ( log5 log )= log + log 5 The ect nswer could pper in mn different forms, depending on the bse of the logrithm chosen nd the order of logrithmic properties used in simplifing the nswer. We could simplif it further s follows: log+ log 5 log8+ log 5 log 0 = = =. log5 log log5 log 5 log Tke the logrithm of both sides. Bring the eponents down using propert of logrithms, then multipl the terms out. Collect the terms with on one side, then fctor out. log+ log 5 = log5 log Simplif nd evlute with clcultor. EXAMPLE Solving Logrithmic Equtions Note: Since clcultors cn evlute eponents of n bse, it often does not mtter wht the bse is when we convert logrithmic equtions into their eponentil forms. = Solve the eqution log 7. Solution: Note tht rewriting this eqution using the chnge of bse formul does not help, since the vrible would still be trpped inside the logrithm. Insted, we use the definition of logrithms to rewrite the eqution in eponentil form. log 7 ( )= = 7 = 5 = 7 Rewrite the eqution in eponentil form. Simplif nd solve for.

52 Chpter 5 EXAMPLE Solving Logrithmic Equtions Note: We need to check for etrneous solutions when solving logrithmic equtions. In prticulr, remember tht logrithms of negtive numbers re undefined. ( ) Solve the eqution log = log + log. Solution: This is n emple of logrithmic eqution tht is not esil solved in logrithmic form. Once few properties of logrithms hve been utilized, the eqution cn be rewritten in ver fmilir form. log5 = log5( + ) log5 + log5 = log5 + = ( )= + = + 5 = 0 + ( )= 0 =, Combine terms using propert of logrithms. Equte the rguments since ech term hs the sme bse. Multipl both sides b. The result is qudrtic eqution. Rewrite the eqution with 0 on one side. Fctor. Solve using the Zero-Fctor Propert. A crucil step remins! While these two solutions definitel solve the qudrtic eqution, we must check tht the solve the initil logrithmic eqution, s the process of solving logrithmic equtions cn introduce etrneous solutions. If we check our two potentil solutions in the originl eqution, we quickl discover tht onl one of them is vlid. log5 log5 log5 = + We cn lred see tht is not solution to the eqution, becuse logrithms of negtive numbers re undefined. We move on to the second solution. ( ) ( ) log = log + log log5 = log59 log5 9 log5 = log5 log = log 5 5 Thus, the solution to the eqution is the single vlue =. Substitute into the eqution. Simplif. Combine the terms using propert of logrithms. A true sttement.

53 Eponentil nd Logrithmic Equtions Section TOPIC Further Applictions of Eponentil nd Logrithmic Equtions We will conclude our discussion of eponentil nd logrithmic equtions b revisiting some importnt pplictions. EXAMPLE 5 Compounding Interest Rit is sving up mone for down pment on new cr. She currentl hs $5500 but she knows she cn get lon t lower interest rte if she cn put down $6000. If she invests her $5500 in mone mrket ccount tht erns n nnul interest rte of.8% compounded monthl, how long will it tke her to ccumulte the $6000? Solution: We need to solve the compound interest formul for t, the mount of time Rit invests her mone. Given tht P = 5500, At ()= 6000, r = 0. 08, nd n =, we hve 6000 = t. Our solution needs to be deciml pproimtion to be of prcticl use, so we need to use the nturl or common logrithm to rewrite the eqution in logrithmic form = t = ln = +. ln 6000 ln 5500 = t ln ln 5500 t ln (. 00) = t 8. t t Divide both sides b Tke the nturl logrithm of both sides. Bring the vrible out of the eponent using propert of logrithms. Solve for t. Since t is mesured in ers, the solution tells us tht it will tke bit less thn er nd 0 months for the $5500 to grow to $6000.

54 6 Chpter 5 EXAMPLE 6 Rdiocrbon Dting We hve lred discussed how the rdioctive dec of crbon- cn be used to rrive t ge estimtes of crbon-bsed fossils, nd we constructed n eponentil function describing the rte of dec. A much more common form of the function is At ()= Ae, t 0 where At () is the mss of crbon- remining fter t ers, nd A 0 is the mss of crbon- initill. Use this formul to determine:. The hlf-life of crbon-. b. The ge of fossilized orgnism contining.5 grms of crbon-, given tht living orgnism of the sme size contins. grms of crbon-. Solutions:. We re looking for the vlue of t for which At () is hlf of A 0, or A 0 A 0 = Ae 0.000t t = e ln 0.000t = ln( e ) ln = 0.000t ln = t t 578 ers Thus, s we sw before, the hlf-life of crbon- is pproimtel 578 ers. b. We simpl plug the given informtion into the formul nd solve for t..5 =.e 0.000t t = e. ln.5 = 0.000t. ln.5. = t t 5 ers Divide both sides b A 0. According to the rdiocrbon dting, the fossil is bout 5 ers old.. Since the bse involved is e, using the nturl logrithm mkes clcultion much simpler. Simplif using propert of logrithms. Solve for t. Evlute. Divide both sides b.. Tke the nturl logrithm of both sides nd simplif using propert of logrithms. Solve for t. Evlute.

55 Eponentil nd Logrithmic Equtions Section ± = 0. = 8... = = 5 6. = 6 7. = 5 8. = = = 5. = 6. =± 7. = 6 Eercises Solve the following eponentil nd logrithmic equtions (round deciml pproimtions to the nerest hundredth). See Emples through. 5. e =. = = 7 5. e 5. = 0 = = = e 8. e 8 + e 8 e + = = 7 0. e 6 = 5. = 0. e = = = = 6. e = e + = 5. e. e = = = 8. = 5. e = = 0. 0 = log 5 = 6. log = 7. log+ log= 8. log ( ) log = 9. ln ln = 0. ln( 5) ln = 6. log ( )+ log =. log log = 5. log ( 8) log =. e ln 5 5 = log log = 6 6. log 8 ( 5 )= 6 7. log( 6) log ( 6)= e = e 8. ln( e)= log 9. ln ln 6 0. log( )+ log( + )=. log( )+ log( + )=. log ( 7 )= log 6. log + log 5. log ( 5)+ log ( + )= log ( ) π π π + ( ) = 5. log + log( ) =

56 8 Chpter = =. No solution. = 6 5. = 5 6. = 7. = 8 8. = 5 9. = = 5. = 5 5. = 5. No solution 5. = = = = 58. = ln =, 60. = ln 6. =, 6. f () = ln 6. f () = log 6. f () = ln f () = ln f () = 67. f () = f () = ln 69. f () = ln = f () = ln( + ) 7. f () = ln 5 7. f () = log e 7. f () = ln 5 = 7. log log = 6. log + log log log log ( ) = 9. log ( + ) log ( ) = + ( ) = ( + ) 50. ln + ln ln 6 + ( ) = ( + ) 5. log log log = 5. ln + ln + ( + ) = ( ) 5. log log log log + log ( + ) = 55. log ( ) + log ( + ) = 56. log + log ( ) = 57. ln( + ) + ln = 58. e e 0 = 0 (Hint: First solve for e.) = (Hint: First solve for.) e + e 8= = 0 Using the properties of logrithmic functions, simplif the following functions s much s possible. Write ech function s single term with coefficient of, if possible. = = ( 5 ) = = 8ln( ) 6. f 05. ln 6. f 05. log f ln 65. f = 66. f e ln 67. f = 0 log6 = + 6 = f ln ln 69. f ln ln 70. f = ln( + ) ln 7. ln( log f = e ) 7. f ( )= ln 5 7. f log0 5 = ln( 5 log ) Solve the following ppliction problems. See Emples 5 nd Assuming tht there re currentl 6 billion people on Erth nd growth rte of.9% per er, how long will it tke for the Erth s popultion to rech 0 billion? 75. How long does it tke for n investment to double in vlue if:. the investment is in monthl compounded svings ccount erning % er? b. the investment is in svings ccount erning 7% er tht is continuousl compounded?

57 Eponentil nd Logrithmic Equtions Section About 6 ers ers b ers 76. About 9079 ers hours 78. k , so the coffee reches 00 F when t 9.6 minutes ers ers b..90 ers c. $8, ers (bout 99 ds) ds 76. Assuming hlf-life of 578 ers, how long would it tke for grms of crbon- to dec to grm? 77. Suppose popultion of bcteri in Petri dish hs doubling time of one nd hlf hours. How long will it tke for n initil popultion of 0,000 bcteri to rech 00,000? 78. According to Newton s Lw of Cooling, the temperture T() t of hot object, t time t fter being plced in n environment with constnt temperture C, is given b T()= t C + ( T C) e kt 0, where T 0 is the temperture of the object t time t = 0 nd k is constnt tht depends on the object. If hot cup of coffee, initill t 90F, cools to 5F in 5 minutes when plced in room with constnt temperture of 75F, how long will it tke for the coffee to rech 00F? 79. Wne hs $,500 in high interest svings ccount t.66% nnul interest compounded monthl. Assum ing he mkes no deposits or withdrwls, how long will it tke for his investment to grow to $5,000? 80. Ben nd Cse both open mone mrket ccounts with.9% nnul interest compounded continuousl. Ben opens his ccount with $8700 while Cse opens her ccount with $00.. How long will it tke Ben s ccount to rech $0,000? b. How long will it tke Cse s ccount to rech $0,000? c. How much mone will be in Ben s ccount fter the time found in prt b? 8. Cesium-7 hs hlf-life of pproimtel 0 ers. How long would it tke for 60 grms of cesium-7 to dec to 59 grms? 8. A chemist, running tests on n unknown smple from n illegl wste dump, isoltes 50 grms of wht he suspects is rdioctive element. In order to help identif the element, he would like to know its hlf-life. He determines tht fter 0 ds onl grms of the originl element remins. Wht is the hlf-life of this mster element?

58 50 Chpter 5 Chpter 5 Project Eponentil Functions Computer viruses hve cost U.S. compnies billions of dollrs in dmges nd lost revenues over the lst few ers. One fctor tht mkes computer viruses so devstting is the rte t which the spred. A virus cn potentill spred cross the world in mtter of hours depending on its chrcteristics nd whom it ttcks. Consider the growth of the following virus. A new virus hs been creted nd is distributed to 00 computers in compn vi corporte e-mil. From these worksttions the virus continues to spred. Let t = 0 be the time of the first 00 infections, nd t t = 7 minutes the popultion of infected computers grows to 00. Assume the ntivirus compnies re not ble to identif the virus or slow its progress for hours, llowing the virus to grow eponentill.. Wht will the popultion of the infected computers be fter hour?. Wht will the popultion be fter hour 0 minutes?. Wht will the popultion be fter full hours? Suppose nother virus is developed nd relesed on the sme 00 computers. This t virus grows ccording to Pt () = ( 00), where t represents the number of hours from the time of introduction.. Wht is the doubling time for this virus? 5. How long will it tke for the virus to infect 000 computers, ccording to this model?