Chapter 4: Exponential and Logarithmic Functions

Transcription

1 Chaper 4: Eponenial and Logarihmic Funcions Secion 4.1 Eponenial Funcions Secion 4. Graphs of Eponenial Funcions... 3 Secion 4.3 Logarihmic Funcions... 4 Secion 4.4 Logarihmic Properies Secion 4.5 Graphs of Logarihmic Funcions... 6 Secion 4.6 Eponenial and Logarihmic Models Secion 4.7 Fiing Eponenials o Daa... Error! Bookmark no defined. Secion 4.1 Eponenial Funcions India is he second mos populous counry in he world, wih a populaion in 008 of aou 1.14 illion people. The populaion is growing y aou 1.34% each year 1. We migh ask if we can find a formula o model he populaion, P, as a funcion of ime,, in years afer 008, if he populaion coninues o grow a his rae. In linear growh, we had a consan rae of change a consan numer ha he oupu increased for each increase in inpu. For eample, in he equaion f ( ) 3 4, he slope ells us he oupu increases y hree each ime he inpu increases y one. This populaion scenario is differen we have a percen rae of change raher han a consan numer of people as our rae of change. To see he significance of his difference consider hese wo companies: Company A has 100 sores, and epands y opening 50 new sores a year Company B has 100 sores, and epands y increasing he numer of sores y 50% of heir oal each year. Looking a a few years of growh for hese companies: Year Sores, company A Sores, company B Saring wih 100 each = 150 They oh grow y 50 sores in he firs year = 00 Sore A grows y 50, Sore B grows y = 50 Sore A grows y 50, Sore B grows y % of (100) = % of (150) = % of (5) = World Bank, World Developmen Indicaors, as repored on hp:// rerieved Augus 0, 010 This chaper is par of Precalculus: An Invesigaion of Funcions Lippman & Rasmussen 011. This maerial is licensed under a Creaive Commons CC-BY-SA license.

2 16 Chaper 4 Noice ha wih he percen growh, each year he company is grows y 50% of he curren year s oal, so as he company grows larger, he numer of sores added in a year grows as well. To ry o simplify he calculaions, noice ha afer 1 year he numer of sores for company B was: (100) or equivalenly y facoring 100(1 0.50) 150 We can hink of his as he new numer of sores is he original 100% plus anoher 50%. Afer years, he numer of sores was: (150) or equivalenly y facoring 150(1 0.50) now recall he 150 came from 100(1+0.50). Susiuing ha, 100(1 0.50)(1 0.50) 100(1 0.50) 5 Afer 3 years, he numer of sores was: (5) or equivalenly y facoring 5(1 0.50) now recall he 5 came from 100(1 0.50). Susiuing ha, 100(1 0.50) (1 0.50) 100(1 0.50) From his, we can generalize, noicing ha o show a 50% increase, each year we muliply y a facor of (1+0.50), so afer n years, our equaion would e n B ( n) 100(1 0.50) In his equaion, he 100 represened he iniial quaniy, and he 0.50 was he percen growh rae. Generalizing furher, we arrive a he general form of eponenial funcions. Eponenial Funcion An eponenial growh or decay funcion is a funcion ha grows or shrinks a a consan percen growh rae. The equaion can e wrien in he form f ( ) a(1 r) or f ( ) a where = 1+r Where a is he iniial or saring value of he funcion r is he percen growh or decay rae, wrien as a decimal is he growh facor or growh muliplier. Since powers of negaive numers ehave srangely, we limi o posiive values. To see more clearly he difference eween eponenial and linear growh, compare he wo ales and graphs elow, which illusrae he growh of company A and B descried aove over a longer ime frame if he growh paerns were o coninue

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

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

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

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

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

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

9 Secion 4.1 Eponenial Funcions 3 Eample 9 A 59 plan is a college savings plan in which a relaive can inves money o pay for a child s laer college uiion, and he accoun grows a free. If Lily wans o se up a 59 accoun for her new granddaugher, wans he accoun o grow o $40,000 over 18 years, and she elieves he accoun will earn 6% compounded semi-annually (wice a year), how much will Lily need o inves in he accoun now? Since he accoun is earning 6%, r = 0.06 Since ineres is compounded wice a year, k = In his prolem, we don know how much we are saring wih, so we will e solving for a, he iniial amoun needed. We do know we wan he end amoun o e $40,000, so we will e looking for he value of a so ha A(18) = 40, ,000 A(18) a1 40,000 a(.8983) a 40,000 $13, (18) Lily will need o inves $13,801 o have $40,000 in 18 years. Try i now 5. Recalculae eample from aove wih quarerly compounding. Because of compounding hroughou he year, wih compound ineres he acual increase in a year is more han he annual percenage rae. If $1,000 were invesed a 10%, he ale elow shows he value afer 1 year a differen compounding frequencies: Frequency Value afer 1 year Annually $1100 Semiannually $ Quarerly $ Monhly $ Daily $ If we were o compue he acual percenage increase for he daily compounding, here was an increase of $ from an original amoun of $1,000, for a percenage increase of = % increase. This quaniy is called he annual percenage 1000 yield (APY).

10 4 Chaper 4 Noice ha given any saring amoun, he amoun afer 1 year would e k r A ( 1) a 1. To find he oal change, we would surac he original amoun, hen k o find he percenage change we would divide ha y he original amoun: k r a1 k a a 1 r k k 1 Annual Percenage Yield The annual percenage yield is he acual percen a quaniy increases in one year. I can e calculaed as r APY 1 1 k k Noice his is equivalen o finding he value of $1 afer 1 year, and suracing he original dollar. Eample 10 Bank A offers an accoun paying 1.% compounded quarerly. Bank B offers an accoun paying 1.1% compounded monhly. Which is offering a eer rae? We can compare hese raes using he annual percenage yield he acual percen increase in a year Bank A: APY = 1.054% Bank B: APY = % 1 4 Bank B s monhly compounding is no enough o cach up wih Bank A s eer APR. Bank A offers a eer rae. A Limi o Compounding As we saw earlier, he amoun we earn increases as we increase he compounding frequency. The ale, hough, shows ha he increase from annual o semi-annual compounding is larger han he increase from monhly o daily compounding. This migh lead us o elieve ha alhough increasing he frequency of compounding will increase our resul, here is an upper limi o his process.

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

12 6 Chaper 4 Eample 11 Radon- decays a a coninuous rae of 17.3% per day. How much will 100mg of Radon- decay o in 3 days? Since we are given a coninuous decay rae, we use he coninuous growh formula. Since he susance is decaying, we know he growh rae will e negaive: r = (3) f (3) 100e 59.51mg of Radon- will remain. Try i Now 6. Inerpre he following: S( ) 0e grams, and ime is measured in days. 0.1 if S() represens he growh of a susance in Coninuous growh is also ofen applied o compound ineres, allowing us o alk aou coninuous compounding. Eample 1 If $1000 is invesed in an accoun earning 10% compounded coninuously, find he value afer 1 year. Here, he coninuous growh rae is 10%, so r = We sar wih $1000, so a = To find he value afer 1 year, 0.10(1) f (1) 1000e $ Noice his is a $ increase for he year. As a percen increase, his is % increase over he original $ Noice ha his value is slighly larger han he amoun generaed y daily compounding in he ale compued earlier. The coninuous growh rae is like he nominal growh rae (or APR) i reflecs he growh rae efore compounding akes effec. This is differen han he annual growh rae used in he formula f ( ) a(1 r), which is like he annual percenage yield i reflecs he acual amoun he oupu grows in a year. While he coninuous growh rae in he eample aove was 10%, he acual annual yield was %. This means we could wrie wo differen looking u equivalen formulas for his accoun s growh: 0.10 f ( ) 1000e using he 10% coninuous growh rae f( ) 1000( ) using he % acual annual yield rae.

13 Secion 4.1 Eponenial Funcions 7 Imporan Topics of his Secion Percen growh Eponenial funcions Finding formulas Inerpreing equaions Graphs Eponenial Growh & Decay Compound ineres Annual Percen Yield Coninuous Growh Try i Now Answers 1. A & C are eponenial funcions, hey grow y a % no a consan numer.. B() is growing faser, u afer 3 years A() sill has a higher accoun alance (0.97) f ( ) $ An iniial susance weighing 0g is growing a a coninuous rae of 1% per day.

14 8 Chaper 4 Secion 4.1 Eercises For each ale elow, could he ale represen a funcion ha is linear, eponenial, or neiher? f() h() m() g() k() n() A populaion numers 11,000 organisms iniially and grows y 8.5% each year. Wrie an eponenial model for he populaion. 8. A populaion is currenly 6,000 and has een increasing y 1.% each day. Wrie an eponenial model for he populaion. 9. The fo populaion in a cerain region has an annual growh rae of 9 percen per year. I is esimaed ha he populaion in he year 010 was 3,900. Esimae he fo populaion in he year The amoun of area covered y lackerry ushes in a park has een growing y 1% each year. I is esimaed ha he area covered in 009 was 4,500 square fee. Esimae he area ha will e covered in A vehicle purchased for $3,500 depreciaes a a consan rae of 5% each year. Deermine he approimae value of he vehicle 1 years afer purchase. 1. A usiness purchases $15,000 of office furniure which depreciaes a a consan rae of 1% each year. Find he residual value of he furniure 6 years afer purchase.

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

16 30 Chaper If $6,000 is invesed in a ank accoun a an ineres rae of 9 per cen per year, find he amoun in he ank afer 5 years if ineres is compounded annually, quarerly, monhly, and coninuously. 31. Find he annual percenage yield (APY) for a savings accoun wih annual percenage rae of 3% compounded quarerly. 3. Find he annual percenage yield (APY) for a savings accoun wih annual percenage rae of 5% compounded monhly. 33. A populaion of aceria is growing according o he equaion measured in years. Esimae when he populaion will eceed A populaion of aceria is growing according o he equaion measured in years. Esimae when he populaion will eceed P( ) 1600e P( ) 100e , wih, wih 35. In 1968, he U.S. minimum wage was $1.60 per hour. In 1976, he minimum wage was $.30 per hour. Assume he minimum wage grows according o an eponenial model w (), where represens he ime in years afer [UW] a. Find a formula for w ().. Wha does he model predic for he minimum wage in 1960? c. If he minimum wage was $5.15 in 1996, is his aove, elow or equal o wha he model predics? 36. In 1989, research scieniss pulished a model for predicing he cumulaive numer 1980 of AIDS cases (in housands) repored in he Unied Saes: a , where is he year. This paper was considered a relief, since here was a fear he correc model would e of eponenial ype. Pick wo daa poins prediced y he research model a () o consruc a new eponenial model () for he numer of cumulaive AIDS cases. Discuss how he wo models differ and eplain he use of he word relief. [UW] 3

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

18 3 Chaper 4 Secion 4. Graphs of Eponenial Funcions Like wih linear funcions, he graph of an eponenial funcion is deermined y he values for he parameers in he funcion s formula. To ge a sense for he ehavior of eponenials, le us egin y looking more closely a he funcion f ( ). Lising a ale of values for his funcion: f() Noice ha: 1) This funcion is posiive for all values of. ) As increases, he funcion grows faser and faser (he rae of change increases). 3) As decreases, he funcion values grow smaller, approaching zero. 4) This is an eample of eponenial growh. 1 Looking a he funcion g( ) g() Noe his funcion is also posiive for all values of, u in his case grows as decreases, and decreases owards zero as increases. This is an eample of eponenial decay. You may noice from he ale ha his funcion appears o e he horizonal reflecion of he f ( ) ale. This is in fac he case: f ( ) ( ) 1 1 g( ) Looking a he graphs also confirms his relaionship:

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

20 34 Chaper 4 To ge a eer feeling for he effec of a and on he graph, eamine he ses of graphs elow. The firs se shows various graphs, where a remains he same and we only change he value for Noice ha he closer he value of is o 1, he less seep he graph will e. In he ne se of graphs, a is alered and our value for remains he same Noice ha changing he value for a changes he verical inercep. Since a is muliplying he erm, a acs as a verical srech facor, no as a shif. Noice also ha he long run ehavior for all of hese funcions is he same ecause he growh facor did no change and none of hese a values inroduced a verical flip.

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

22 36 Chaper 4 Ne consider he effec of a horizonal shif on an eponenial funcion. Shifing he 4 funcion f ( ) 3() four unis o he lef would give f ( 4) 3(). Employing eponen rules, we could rewrie his: 4 4 f ( 4) 3() 3() ( ) 48() Ineresingly, i urns ou ha a horizonal shif of an eponenial funcion corresponds wih a change in iniial value of he funcion. Lasly, consider he effec of a verical shif on an eponenial funcion. Shifing f ( ) 3() down 4 unis would give he equaion f ( ) 3() 4, yielding he graph Noice ha his graph is susanially differen han he asic eponenial graph. Unlike a asic eponenial, his graph does no have a horizonal asympoe a y = 0; due o he verical shif, he horizonal asympoe has also shifed o y = -4. We can see ha as, f( ) and as, f( ) 4. We have deermined ha a verical shif is he only ransformaion of an eponenial funcion ha changes he graph in a way ha canno e achieved y alering he parameers a and in he asic eponenial funcion f ( ) a. Transformaions of Eponenials Any ransformed eponenial can e wrien in he form f ( ) a c where y = c is he horizonal asympoe. Noe ha, due o he shif, he verical inercep is shifed o (0, a+c). Try i Now. Wrie he equaion and graph he eponenial funcion descried as follows: f ( ) e is verically sreched y a facor of, flipped across he y ais and shifed up 4 unis.

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

24 38 Chaper 4 Eample 4 Find an equaion for he graph skeched elow. Looking a his graph, i appears o have a horizonal asympoe a y = 5, suggesing an equaion of he form f ( ) a 5. To find values for a and, we can idenify wo oher poins on he graph. I appears he graph passes hrough (0,) and (-1,3), so we can use hose poins. Susiuing in (0,) allows us o solve for a 0 a 5 a 5 a 3 Susiuing in (-1,3) allows us o solve for The final formula for our funcion is f ( ) 3(1.5) 5. Try i Now 3. Given he graph of he ransformed eponenial funcion, find a formula and descrie he long run ehavior.

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

26 40 Chaper 4 Secion 4. Eercises Mach each funcion wih one of he graphs elow. 1. f f f f f f A B C D E F If all he graphs o he righ have equaions wih form f a, 7. Which graph has he larges value for? A B C D E 8. Which graph has he smalles value for? F 9. Which graph has he larges value for a? 10. Which graph has he smalles value for a? Skech a graph of each of he following ransformaions of f 11. f 1. g 13. h f f k Saring wih he graph of f 4, find a formula for he funcion ha resuls from 17. Shifing f( ) 4 unis upwards 18. Shifing f( ) 3 unis downwards 19. Shifing f( ) unis lef 0. Shifing f( ) 5 unis righ 1. Reflecing f( ) aou he -ais. Reflecing f( ) aou he y-ais

27 Secion 4. Graphs of Eponenial Funcions 41 Descrie he long run ehavior, as and of each funcion f 3 3. f f f f f 3 1 Find a formula for each funcion graphed as a ransformaion of f Find an equaion for he eponenial funcion graphed

28 4 Chaper 4 Secion 4.3 Logarihmic Funcions A populaion of 50 flies is epeced o doule every week, leading o a funcion of he form f ( ) 50(), where represens he numer of weeks ha have passed. When will his populaion reach 500? Trying o solve his prolem leads o: () Dividing oh sides y 50 o isolae he eponenial 10 While we have se up eponenial models and used hem o make predicions, you may have noiced ha solving eponenial equaions has no ye een menioned. The reason is simple: none of he algeraic ools discussed so far are sufficien o solve eponenial equaions. Consider he equaion 10 aove. We know ha 3 8 and 4 16, so i is clear ha mus e some value eween 3 and 4 since g ( ) is increasing. We could use echnoy o creae a ale of values or graph o eer esimae he soluion. From he graph, we could eer esimae he soluion o e around 3.3. This resul is sill fairly unsaisfacory, and since he eponenial funcion is one-o-one, i would e grea o have an inverse funcion. None of he funcions we have already discussed would serve as an inverse funcion and so we mus inroduce a new funcion, named as he inverse of an eponenial funcion. Since eponenial funcions have differen ases, we will define corresponding arihms of differen ases as well. Logarihm The arihm (ase ) funcion, wrien funcion (ase ),., is he inverse of he eponenial Since he arihm and eponenial are inverses, i follows ha: Properies of Logs: Inverse Properies

29 Secion 4.3 Logarihmic Funcions 43 Recall also from he definiion of an inverse funcion ha if Applying his o he eponenial and arihmic funcions: 1 f ( a) c, hen f ( c) a. Logarihm Equivalen o an Eponenial The saemen a c is equivalen o he saemen ( c) a. a Alernaively, we could show his y saring wih he eponenial funcion c, hen a aking he ase of oh sides, giving ( c). Using he inverse propery of s we see ha ( c ) a. Since is a funcion, i is mos correcly wrien as ( c), using parenheses o denoe funcion evaluaion, jus as we would wih f(c). However, when he inpu is a single variale or numer, i is common o see he parenheses dropped and he epression wrien as c. Eample 1 Wrie hese eponenial equaions as arihmic equaions: is equivalen o (8) is equivalen o 5 (5) is equivalen o Eample Wrie hese arihmic equaions as eponenial equaions: is equivalen o 6 1/ 6 3 is equivalen o Try i Now Wrie he eponenial equaion 4 16 as a arihmic equaion.

30 44 Chaper 4 By esalishing he relaionship eween eponenial and arihmic funcions, we can now solve asic arihmic and eponenial equaions y rewriing. Eample 3 Solve 4 for. By rewriing his epression as an eponenial, 4, so = 16 Eample 4 Solve 10 for. By rewriing his epression as a arihm, we ge (10) While his does define a soluion, and an eac soluion a ha, you may find i somewha unsaisfying since i is difficul o compare his epression o he decimal esimae we made earlier. Also, giving an eac epression for a soluion is no always useful ofen we really need a decimal approimaion o he soluion. Luckily, his is a ask calculaors and compuers are quie adep a. Unluckily for us, mos calculaors and compuers will only evaluae arihms of wo ases. Happily, his ends up no eing a prolem, as we ll see riefly. Common and Naural Logarihms The common is he arihm wih ase 10, and is ypically wrien ( ). The naural is he arihm wih ase e, and is ypically wrien ln( ). Eample 5 Evaluae ( 1000) using he definiion of he common. To evaluae ( 1000), we can say (1000), hen rewrie ino eponenial form using he common ase of From his, we migh recognize ha 1000 is he cue of 10, so = 3. We also can use he inverse propery of s o 3 wrie Values of he common numer numer as (numer) eponenial

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

32 46 Chaper 4 Eample 9 Rewrie 4ln( ) using he eponen propery for s. 4 Using he propery in reverse, 4ln( ) ln Try i Now 1 3. Rewrie using he eponen propery for s: ln. The eponen propery allows us o find a mehod for changing he ase of a arihmic epression. Properies of Logs: Change of Base c ( A) A ( ) c Proof: Le A. Rewriing as an eponenial gives A. Taking he ase c of oh sides of his equaion gives c c A Now uilizing he eponen propery for s on he lef side, c c A Dividing, we oain c A c A or replacing our epression for, A c Wih his change of ase formula, we can finally find a good decimal approimaion o our quesion from he eginning of he secion. c Eample 10 Evaluae (10) using he change of ase formula. According o he change of ase formula, we can rewrie he ase as a arihm of any oher ase. Since our calculaors can evaluae he naural, we migh choose o use he naural arihm, which is he ase e: e 10 ln10 10 e ln Using our calculaors o evaluae his,

33 Secion 4.3 Logarihmic Funcions 47 ln10 ln This finally allows us o answer our original quesion he populaion of flies we discussed a he eginning of he secion will ake 3.3 weeks o grow o 500. Eample 11 Evaluae 5 (100) using he change of ase formula. We can rewrie his epression using any oher ase. If our calculaors are ale o evaluae he common arihm, we could rewrie using he common, ase (100) While we were ale o solve he asic eponenial equaion 10 y rewriing in arihmic form and hen using he change of ase formula o evaluae he arihm, he proof of he change of ase formula illuminaes an alernaive approach o solving eponenial equaions. Solving eponenial equaions: 1. Isolae he eponenial epressions when possile. Take he arihm of oh sides 3. Uilize he eponen propery for arihms o pull he variale ou of he eponen 4. Use algera o solve for he variale. Eample 1 Solve 10 for. Using his alernaive approach, raher han rewrie his eponenial ino arihmic form, we will ake he arihm of oh sides of he equaion. Since we ofen wish o evaluae he resul o a decimal answer, we will usually uilize eiher he common or naural. For his eample, we ll use he naural : ln ln(10) Uilizing he eponen propery for s, ln ln(10) Now dividing y ln(), ln(10).861 ln Noice ha his resul maches he resul we found using he change of ase formula.

34 48 Chaper 4 Eample 13 In he firs secion, we prediced he populaion (in illions) of India years afer 008 y using he funcion f ( ) 1.14( ). If he populaion coninues following his rend, when will he populaion reach illion? We need o solve for he so ha f() = 1.14(1.0134) Divide y 1.14 o isolae he eponenial epression Take he arihm of oh sides of he equaion 1.14 ln ln Apply he eponen propery on he righ side 1.14 ln ln Divide oh sides y ln(1.0134) 1.14 ln years ln If his growh rae coninues, he model predics he populaion of India will reach illion aou 4 years afer 008, or approimaely in he year 050. Try i Now 4. Solve 5(0.93) 10. In addiion o solving eponenial equaions, arihmic epressions are common in many physical siuaions. Eample 14 In chemisry, ph is a measure of he acidiy or asiciy of a liquid. The ph is relaed o he concenraion of hydrogen ions, [H + ], measured in moles per lier, y he equaion H ph. If a liquid has concenraion of moles per lier, deermine he ph. Deermine he hydrogen ion concenraion of a liquid wih ph of 7. To answer he firs quesion, we evaluae he epression While we could use our calculaors for his, we do no really need hem here, since we can use he inverse propery of s: ( 4) 4

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

36 50 Chaper 4 Eample 16 A company s sales can e modeled y he funcion years. Find he annual growh rae. S 0.1 ( ) 5000e, wih measured in k 0.1 Noing ha 1 r e, hen r e , so he annual growh rae is 1.75%. The sales funcion could also e wrien in he form S ( ) 5000( ). Imporan Topics of his Secion The Logarihmic funcion as he inverse of he eponenial funcion Wriing arihmic & eponenial epressions Properies of s Inverse properies Eponenial properies Change of ase Common Naural Solving eponenial equaions Try i Now Answers ln( ) ln() ln(0.93) 1. 4

37 Secion 4.3 Logarihmic Funcions 51 Secion 4.3 Eercises Rewrie each equaion in eponenial form 1. 4( q) m. 3( ) k 3. ( a ) c 4. ( z p ) u 5. v 6. r s 7. ln w n 8. ln y Rewrie each equaion in arihmic form y y a p v 15. d c k e k 1. h 16. z n y e L Solve for ( ) ( ) ( ) ln 4. ln Simplify each epression using arihm properies , ln e ln e Evaluae using your calculaor ln ln 0.0 Solve each equaion for he variale e e e e

38 5 Chaper 4 f ae. Conver he equaion ino coninuous growh form, k 57. f f f f f a. Conver he equaion ino annual growh form, f 150e 6. f 100e f 50e f 80e 65. The populaion of Kenya was 39.8 million in 009 and has een growing y aou.6% each year. If his rend coninues, when will he populaion eceed 45 million? 66. The populaion of Algeria was 34.9 million in 009 and has een growing y aou 1.5% each year. If his rend coninues, when will he populaion eceed 45 million? 67. The populaion of Seale grew from 563,374 in 000 o 608,660 in 010. If he populaion coninues o grow eponenially a he same rae, when will he populaion eceed 1 million people? 68. The median household income (adjused for inflaion) in Seale grew from $4,948 in 1990 o $45,736 in 000. If i coninues o grow eponenially a he same rae, when will median income eceed $50,000? 69. A scienis egins wih 100 mg of a radioacive susance. Afer 4 hours, i has decayed o 80 mg. How long afer he process egan will i ake o decay o 15 mg? 70. A scienis egins wih 100 mg of a radioacive susance. Afer 6 days, i has decayed o 60 mg. How long afer he process egan will i ake o decay o 10 mg? 71. If $1000 is invesed in an accoun earning 3% compounded monhly, how long will i ake he accoun o grow in value o $1500? 7. If $1000 is invesed in an accoun earning % compounded quarerly, how long will i ake he accoun o grow in value o $1300?

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

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

41 Secion 4.4 Logarihmic Properies 55 Eample 3 4 y Rewrie ln 7 as a sum or difference of s Firs, noicing we have a quoien of wo epressions, we can uilize he difference propery of s o wrie 4 y 4 ln ln ln(7) 7 y Then seeing he produc in he firs erm, we use he sum propery 4 4 ln y ln(7) ln ln( y) ln(7 ) Finally, we could use he eponen propery on he firs erm ln 4 ln( y) ln(7) 4ln( ) ln( y) ln(7 ) Ineresingly, solving eponenial equaions was no he reason arihms were originally developed. Hisorically, up unil he adven of calculaors and compuers, he power of arihms was ha hese properies reduced muliplicaion, division, roos, or powers o e evaluaed using addiion, suracion, division and muliplicaion, respecively, which are much easier o compue wihou a calculaor. Large ooks were pulished lising he arihms of numers, such as in he ale o he righ. To find he produc of wo numers, he sum of propery was used. Suppose for eample we didn know he value of imes 3. Using he sum propery of s: ( 3) () (3) value (value) Using he ale, ( 3) () (3) We can hen use he ale again in reverse, looking for as an oupu of he arihm. From ha we can deermine: ( 3) (6). By doing addiion and he ale of s, we were ale o deermine 3 6. Likewise, o compue a cue roo like 3 8 1/ ( 8) 8 (8) ( ) So 8. () 3

42 56 Chaper 4 Alhough hese calculaions are simple and insignifican hey illusrae he same idea ha was used for hundreds of years as an efficien way o calculae he produc, quoien, roos, and powers of large and complicaed numers, eiher using ales of arihms or mechanical ools called slide rules. These properies sill have oher pracical applicaions for inerpreing changes in eponenial and arihmic relaionships. Eample 4 Recall ha in chemisry, ph H. If he concenraion of hydrogen ions in a liquid is douled, wha is he affec on ph? Suppose C is he original concenraion of hydrogen ions, and P is he original ph of he liquid, so P C. If he concenraion is douled, he new concenraion is C. Then he ph of he new liquid is ph C Using he sum propery of s, ph C () ( C) () ( C) Since P C, he new ph is ph P ( ) P When he concenraion of hydrogen ions is douled, he ph decreases y Log properies in solving equaions The arihm properies ofen arise when solving prolems involving arihms. Eample 5 Solve ( 50 5) ( ). In order o rewrie in eponenial form, we need a single arihmic epression on he lef side of he equaion. Using he difference propery of s, we can rewrie he lef side: 50 5 Rewriing in eponenial form reduces his o an algeraic equaion:

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