SECTION 5-9 CHAPTER 5 Sction 5-. An ponntial function is a function whr th variabl appars in an ponnt.. If b >, th function is an incrasing function. If < b <, th function is a dcrasing function. 5. A positiv numbr raisd to any ral powr will giv a positiv rsult. 7. (A) Th graph of y = (.) is dcrasing and passs through th point (,. ) = (, 5). This corrsponds to graph g. (B) Th graph of y = is incrasing and passs through th point (, ). This corrsponds to graph n. (C) Th graph of y = is dcrasing and passs through th point (, ). This corrsponds to graph f. (D) Th graph of y = 4 is incrasing and passs through th point (, 4). This corrsponds to graph m. 9. 6.4. 7.54..649 5. 4.469 7. 4 = + 4 = + 9. = ( ) = + =. 4 y 5 z = 4 5 z yz. 5 = 5 ( + ) = 5 = 5. Th graph of g is th sam as th graph of f strtchd vrtically by a factor of. Thrfor g is incrasing and th graph has horizontal asymptot y =. 7. Th graph of g is th sam as th graph of f rflctd through th y ais and shrunk vrtically by a factor of. Thrfor g is dcrasing and th graph has horizontal asymptot y =. 9. Th graph of g is th sam as th graph of f shiftd upward units. Thrfor g is incrasing and th graph has horizontal asymptot y =.. Th graph of g is th sam as th graph of f shiftd units to th lft. Thrfor g is incrasing and th graph has horizontal asymptot y =.
CHAPTER 5 EXPONENTIAL AND LOGARITHMIC FUNCTIONS. 5 = 5 4 if and only if = 4 = = 5. 7 = 7 + if and only if = + = ( )( + ) = =, 7. 4 5 6 6 = 5 4 4 = 4 5 5 if 6 + = 6 = = if and only 9. ( ) 5 = ( ) 5 if and only if = = = 4. = if = or =. Sinc is nvr, th only solution is =. 4. 5 = ( 5) = = or = or 5 = nvr = 5 =, 5 45. 9 = ( ) = = if and only if = + = ()( ) = =, 47. 5 + = 5 (5 ) + = (5 ) 5 +6 = 5 if and only if + 6 = = 6 49. 4 +7 = 8 + ( ) +7 = ( ) + 4+4 = +6 if and only if 4 + 4 = + 6 = 8 5. a = a a = a a 4 = (a ) a 4 = (a )(a + )(a + ) = a = or a = This dos not violat th ponntial proprty mntiond bcaus a = and a ngativ ar cludd from considration in th statmnt of th proprty. 55. Th graph of g is th sam as th graph of f rflctd through th ais; g is incrasing; horizontal asymptot: y =. 5.,,, =, =, =. = for all ral ; th function f() = is nithr incrasing nor dcrasing and is qual to f() =, thus th variabl is ffctivly not in th ponnt at all. 57. Th graph of g is th sam as th graph of f strtchd horizontally by a factor of and shiftd upward units; g is dcrasing; horizontal asymptot: y =.
SECTION 5-59. Th graph of g is th sam as th graph of f strtchd vrtically by a factor of 5; g is incrasing; horizontal asymptot: y =. 6. Th graph of g is th sam as th graph of f shiftd units to th right, strtchd vrtically by a factor of, and shiftd upward unit; g is incrasing; horizontal asymptot: y =. 6. Th graph of g is th sam as th graph of f shiftd units to th right, rflctd in th origin, strtchd vrtically by a factor of 4, and shiftd upward units; g is incrasing; horizontal asymptot: y =. ( ) ( ) 65. = = 6 6 4 67. ( + ) + ( ) = ( ) + ( )( ) + ( ) + ( ) ( )( ) + ( ) Common Errors: 4 = + + + + = + 69. Eamining th graph of y = f(), w obtain Thr ar no local trma and no intrcpts. Th y intrcpt is.4. As, y, so th lin y = is a horizontal asymptot. 7. Eamining th graph of y = s(), w obtain 5 5 Thr is a local maimum at s() =, and is th y intrcpt. Thr is no intrcpt. As or, y, so th lin y = (th ais) is a horizontal asymptot
CHAPTER 5 EXPONENTIAL AND LOGARITHMIC FUNCTIONS 7. Eamining th graph of y = F(), w obtain 5 Thr ar no local trma and no intrcpts. Whn =, F() = = 5 is th y intrcpt. As, y, so th lin y = (th ais) is a horizontal asymptot. As, y, so th lin y = is also a horizontal asymptot. 75. Th local minimum is f() =, so zro is th y intrcpt. Thr ar no intrcpts or horizontal asymptots; f() as and. 77. Eamining th graph of y = f(), w obtain As, f() = ( + ) / sms to approach a valu nar. A tabl of valus nar = yilds Although f() is not dfind, as, f() sms to approach a numbr nar.78. In fact, it approachs, sinc as, u =, and f() = u u must approach as u. 79. Mak a tabl of valus, substituting in ach rqustd valu:.4.4.44.44.44.444.696.6577.66475.6659.6658.66545 Th approimat valu of w gt.66544. is.66545 to si dcimal placs. Using a calculator to comput dirctly, 8. 8.
SECTION 5-85. Hr ar graphs of f () =, f () =, and f () =. In ach cas as, f n (). Th lin y = is a horizontal asymptot. As, f () and f (), whil f (). It appars that as, f n () if n is vn and f n () if n is odd. f : f : f :.4.5.4. As confirmation of ths obsrvations, w show th graph of f 4 = (not rquird). 4 5 87. W us th compound intrst formula n A = P r to find P: P = A n m r m m = 65 r =.65 A =, n = 65 7, P = = $4,56. to th narst dollar 657.65 65 n 89. W us th Continuous Compound Intrst Formula A = P rt P = 5,5 r =.68 (A) t = 6.5 A = 5,5 (.68)(6.5) = $78. (B) t = 7 A = 5,5 (.68)(7) = $5,5.85 9. W us th compound intrst formula A = P r m For th first account, P =, r =.8, m = 65. Lt y = A, thn y = ( +.8/65) whr is th numbr of compounding priods (days). For th scond account, P = 5, r =.5, m = 65. Lt y = A, thn y = 5( +.5/65) Eamining th graphs of y and y, w obtain th graphs at th right. Th graphs intrsct at = 66.5 days. Comparing th amounts in th accounts, w s that th first account is worth mor than th scond for 67 days.. 9. W us th compound intrst formula A = P r m For th first account, P =,, r =.49, m = 65. Lt y = A, thn y = ( +.49/65) whr is th numbr of compounding priods (days). For th scond account, P =,, r =.5, m = 4. Lt y = A, thn y = ( +.5/4) 4/65 whr is th numbr of days. Eamining th graphs of y and y, w obtain th graph at th right. Th two graphs ar just about indistinguishabl from on anothr. Eamining a tabl of valus, w obtain: n
4 CHAPTER 5 EXPONENTIAL AND LOGARITHMIC FUNCTIONS Th two accounts ar trmly clos in valu, but th scond account is always largr than th first. Th first will nvr b largr than th scond. 95. W us th Continuous Compound Intrst Formula A = P rt P = A or P = A -rt rt A =, r =.6 t = P =, (.6)() P = $6,464.5 99. W us th compound intrst formula Sction 5- A = P r m = 5 m P = 4, r =.6 A = 4,.6 5 (A) n = (5)(.5), hnc n n (5)(.5) 97. W us th compound intrst formula A = P r m Flagstar Bank: P = 5, r =. m = 4 n = (4)() (4)() A = 5,. = $5,488.6 4 UmbrllaBank.com: P = 5, r =. m = 65 n = (65)() (65)() A = 5,. = $5,47.85 65 Allid First Bank: P = 5, r =.96 m = n = ()() A = 5,.96 ()() = $5,46.7 [Not: If m = 65/7 is usd th answrs will diffr vry slightly.] (B) n = (5)() = 5, hnc A = 4,.6 A = 4,.6 5 5 = $4,.75 = $7,85.95. Doubling tim is th tim it taks a population to doubl. Half-lif is th tim it taks for half of an initial quantity of a radioactiv substanc to dcay.. Eponntial growth is th simpl modl A = A kt, i.. unlimitd growth. Limitd growth modls mor ralistically incorporat th fact that thr is a rasonabl maimum valu for A. 5 n 5. Us th doubling tim modl A = A () t/d with A, d 5. A = () t/5 9. Us th half-lif modl A, h 6. A t A t h A 6 with rt 7. Us th continuous growth modl A A with.t A,, r.. A,. Us th ponntial dcay modl A A kt with.4t A 4, k.4. A 4
SECTION 5-5. n L 5. Us th doubling tim modl: P = P t/d 4 8 4 6 5 6 64 7 8 8 56 9 5, 4 7. Us th doubling tim modl A A with A,, d. A, t whr t is yars aftr 97. (A) For t = : A,, 5,8 (B) For t = 5: A, 5 47,8, 6 t h Substituting P = and d =.4, w hav P = ( t/.4 ) (A) t = 7, hnc P = ( 7/.4 ) = 75.5 76 flis (B) t = 4, hnc P = ( 4/.4 ) = 57. 57 flis t/h 9. Us th half-lif modl A = A = A t/h Substituting A = 5 and h =, w hav A = 5( t/ ) (A) t = 5, hnc A = 5( 5/ ) = 9 pounds (B) t =, hnc A = 5( / ) = 7.9 pounds rt. Us th continuous growth modl A A with A = 6.8, r =.88, t = 8 = A 6.8.88() = 7.8 billion rt. Us th continuous growth modl A A. Blow is a graph of A and A. Lt A = th population of Russia and A = th population of Nigria. For Russia, A =.4 8, r =.7 8.7t A.4 For Nigria, A =.9 8, r =.56 8.56t A.9 From th graph, assuming t = in 5, it appars that th two populations bcam qual whn t was approimatly.5, in 8. Aftr that th population of Nigria will b gratr than that of Russia.
6 CHAPTER 5 EXPONENTIAL AND LOGARITHMIC FUNCTIONS 5. A tabl of valus can b gnratd by a graphing calculator and yilds t P 75 7 7 68 4 65 5 6 6 6 7 59 8 57 9 55 5 rt 9. Us th continuous growth modl A A with A =. million, r =.7 (A) In 4, assuming t = in 7, substitut t = 7. A =..7(7) = 9. million (B) In, substitut t =. A =..7() = 45. million 7. I = I.94d (A) d = 5 I = I.94(5) =.6I 6% (B) d = I = I.94() =.9I 9%. T = T m + (T T m ) kt T m = 4 T = 7 k =.4 t = T = 4 + (7 4).4() T = 5. As t incrass without bound,.t approachs, hnc q =.9(.t ) approachs.9. Hnc.9 coulomb is th maimum charg on th capacitor. 5. (A) Eamining th graph of N(t), w obtain th graphs blow. Aftr yars, 5 dr will b prsnt. Aftr 6 yars, 7 dr will b prsnt. 5 (B) Applying a built-in routin, w obtain th graph at th right. It will tak yars for th hrd to grow to 5 dr. (C) As t incrass without bound,.4t approachs, hnc N =.4t 4 approachs. Hnc is th numbr of dr th island can support. 7. Entr th data. Comput th rgrssion quation.
SECTION 5-7 Th modl givs y = 49.(.869477). Clarly, whn =, y = $4,9 is th stimatd purchas pric. Applying a built-in routin, w obtain th graph at th right. Whn =, th stimatd valu of th van is $,959. 5 9. (A) Th indpndnt variabl is yars sinc 98, so ntr, 5,, 5,, and 5 as L. Th dpndnt variabl is powr gnration in North Amrica, so ntr th North Amrica column as L. Thn us th logistic rgrssion command from th STAT CALC mnu. Th modl is 96 y =.69.7 (B) Sinc = corrsponds to 98, us = to prdict powr gnration in. 96 y = = 89. billion kilowatt hours.69().7 Us = 4 to prdict powr gnration in. 96 y = = 9.6 billion kilowatt hours.69(4).7 Sction 5-. Th ponntial function f() = b for b >, b and th logarithmic function g() = log b ar invrs functions for ach othr.. Th rang of th ponntial function is th positiv ral numbrs, hnc th domain of th logarithmic function must also b th positiv ral numbrs. 5. log 5 = log / log 5 or log /log 5. 7. 8 = 4 9.. =. =6. log 4 8 = 6 9. Mak a tabl of valus for ach function: f ( ) f ( ) log / 7 / 7 / 9 / 9 / / 9 9 7 7 5. log = 5 7. log / 8 7 =
8 CHAPTER 5 EXPONENTIAL AND LOGARITHMIC FUNCTIONS. Mak a tabl of valus for ach function: f ( ) / f ( ) log / 7 / 8 7 / 8 9 / 4 9 / 4 / / / / 4 / 9 4 / 9 8 / 7 8 / 7. 5. 7. 4 9. log. = log =. log 7 = log =. log / = log / = 5. 5 7. log 5 5 = log 5 5 / = 9. 4.69 4..995 4. log 7 = ln ln 7 =.8 45. log 5.4 = ln.4 =.9759 ln 5 using th chang of bas formula using th chang of bas formula 5.7.77 47. = =,8 49. = = 6.648 4 =.6648.8655.96 5. = = 47.7 5. =.676 55. Writ log = in quivalnt ponntial form. = = 4 59. Writ log b 6 = in quivalnt ponntial form. 6 = b b = 6 b = 4 sinc bass ar rquird to b positiv 6. Writ log 4 = in quivalnt ponntial form. = 4 / = 57. log 4 6 = log 4 4 = y = 6. Writ log b = in quivalnt ponntial form. = b This statmnt is tru if b is any ral numbr cpt. Howvr, bass ar rquird to b positiv and is not allowd, so th original statmnt is tru if b is any positiv ral numbr cpt. 65. log / 9 = log / = log / = log / = 67. Writ log b = in quivalnt ponntial 69. Writ log 8 = 4 in quivalnt ponntial form. form = b / = b / ( ) / = (b / ) / (If two numbrs ar qual th rsults ar qual if thy ar raisd to th sam ponnt.) (/) = b /(/) = b b = 8 4/ = = (8 / ) 4 = 4 = 6 7. Writ log 6 8 = y in quivalnt ponntial form. 6 y = 8 ( 4 ) y = 4y = if and only if 4y = y = 4 7. 4.959 75. 7.86 77..8 79. log logy
SECTION 5-9 8. log( 4 y ) = log 4 + log y = 4 log + log y 8. ln y 85. ln + 5 ln y ln z = ln + ln y 5 ln z= ln( y 5 ) ln z 87. log (y) = log + log y = + = 5 = ln y z 89. log log log y log log y ( ) y 9. Th graph of g is th sam as th graph of f shiftd upward units; g is incrasing. Domain: (, ) Vrtical asymptot: = 9. Th graph of g is th sam as th graph of f shiftd units to th right; g is dcrasing. Domain: (, ) Vrtical asymptot: = 95. Th graph of g is th sam as th graph of f rflctd through th ais and shiftd downward unit; g is dcrasing. Domain: (, ) Vrtical asymptot: = 97. Th graph of g is th sam as th graph of f rflctd through th ais, strtchd vrtically by a factor of, and shiftd upward 5 units. g is dcrasing. Domain: (, ) Vrtical asymptot: = 99. Writ y = log 5 In ponntial form: 5 y = Intrchang and y: 5 = y Thrfor f - () = 5.. (A)Writ y = log ( ) In ponntial form: y = = y Intrchang and y: y = Thrfor, f - () =. Writ y = 4 log ( + ) y = log ( + ) 4 In ponntial form: y/4 = + = y/4 Intrchang and y: y = /4 Thrfor f - () = /4 (B) Th graph is th sam as th graph of y = rflctd through th ais and shiftd units upward.
CHAPTER 5 EXPONENTIAL AND LOGARITHMIC FUNCTIONS (C) 5. Th inquality sign in th last stp rvrss bcaus log is ngativ. 7. 9. - - -. Lt u log b M and v log b N -. Changing ach quation to ponntial form, u b M and M b Thn w can writ M/N as N b M quation to logarithmic form: log b u v N M Finally, rcall th way w dfind u and v in th first lin of our proof: log log M log N Sction 5-4 v b N. u uv b using a familiar proprty of ponnts. Now chang this v b b b. Answrs will vary.. Th intnsity of a sound and th nrgy rlasd by an arthquak can vary from trmly small to trmly larg. A logarithmic scal can condns this variation into a rang that can b asily comprhndd. 5. W us th dcibl formula D = log (A) I = I D = log I I D = log D = dcibls (B) I =. I =. D = log.. D = dcibls I I 7. W us th dcibl formula D = log I I I = I D = log I D = log I I I D D = log I log I I I = log I I = log I I I I = log I = log = dcibls I 9. W us th magnitud formula M = log E with E =.99 7, E = 4.4 M = log 7.99 = 8.6 4.4 E N
SECTION 5-4. W us th magnitud formula M = log E E For th Long Bach arthquak, For th Anchorag arthquak, 6. = log E 8. = log E E E 9.45 = log E.45 = log E E E (Chang to ponntial form) (Chang to ponntial form) E = 9.45 E =.45 E E E = E 9.45 E = E.45 Now w can compar th nrgy lvls by dividing th mor powrful (Anchorag) by th lss (Long Bach):.45 E = E = 9.45 E E E = E, or tims as powrful. Us th magnitud formula E 4 4.4 M log with E.4, E : E 4.4 M log 6.5 4.4 5. Us th magnitud formula E 4.4 M log with E.8, E :.8 M log. 4.4 E 7. W us th rockt quation. 9. (A) ph = log[h + ] = log(4.6 9 ) = 8.. v = c ln Wt Sinc this is gratr than 7, th substanc is basic. Wb (B) ph = log[h + ] = log(9. 4 ) =. v =.57 ln (9.8) Sinc this is lss than 7, th substanc is acidic. v = 7.67 km/s. Sinc ph = log[h + ], w hav 5. = log[h + ], or [H + ] = 5. = 6. 6 mols pr litr. m = 6.5 log L (B) W compar L for m = with L for m = 6 L = 6.5 log L 6 = 6.5 log L (A) W find m whn L = L L L m = 6.5 log L 5 =.5 log L =.5 log L L L L m = 6.5 log = log L = log L m = 6 L L L = L = L L L = L L = L Hnc L = L =. Th star of magnitud is tims brightr. L L 5. (A) Entr th yars sinc 995 as L. Entr th valus shown in th column hadd % with hom accss as L. Us th logarithmic rgrssion modl from th STAT CALC mnu.
CHAPTER 5 EXPONENTIAL AND LOGARITHMIC FUNCTIONS Th modl is y =.9 + 4. ln. Evaluating this for = (yar 8) yilds 7.7%. Evaluating for = (yar 5) yilds 84.%. (B) No; th prdictd prcntag gos ovr somtim around 4. Sction 5-5. Th logarithm function is th invrs of th ponntial function, morovr, log b M p = plog b M. This proprty of logarithms can oftn b usd to gt a variabl out of an ponnt in solving an quation.. If log b u = log b v, thn u = v bcaus th logarithm is a on-to-on function. 5. ln mans to tak th logarithm of, thn squar th rsult. ln mans to squar, thn tak th logarithm of th rsult. 7. =.47 = log.47 = log.47 =.46 5. =.46 =.46 ln = ln.46 8 ln = ln.46 8.46 ln = 8 ln = 4.. log + log( ) = log[( )] = ( ) = = = ( 5)( + ) = = 5 or 5. log( + ) log( ) = log = = = + = ( ) + = = 9 = 9 9. + = 9 + = log 9 = log 9 7. log 5 = 5 = = 5 log = 9 =. 9. log ( t 4) = = t 4 t = 4 + t = 4 + t = 4. =.65 = ln.65 =.9 Common Error: log log log Chck: log 5 + log(5 ) = log( ) + log( ) is not dfind. = 5 Common Error: log Chck: log log 9 9 log 9 log 9 log. + 68 = 7 = 9 = ln 9 = ln9 =.97. log 5 + log = log(5) = 5 = 5 = =? =? = =
SECTION 5-5 7. =.5 ln = ln.5 ln ln.5 = = 4. 9..4 + 5 = No solution. Both trms on th lft sid ar always positiv, so thy can nvr add to.. = 5. 5 =. ln 5 =. ln 5 =. =.7. =. = ln. = ln. = ± ln. = ±. 5. log(5 ) = log( + ) 5 = + 4 = 5 7. log log 5 = log log( ) log = log 5 = 5 Ecludd valu: 5( ) = 5( ) 5 ( ) = = = 4 5 Chck: = log 5 log 5 = log log ( 5)( + ) = log( ) is not dfind = 5, Solution: 5 9. ln = ln( ) ln( ) ln = ln = Ecludd valu: ( ) = ( ) ( ) = = 4 + = = = b b 4ac a a =, b = 4, c = ( 4) ( 4) 4()() = 4 () = ± Chck: ln( + )? = ln[( + ) ] ln[( + ) ] ln( + )? = ln( + ) ln ln( + ) =? ln ln( + ) = ln( + ) ln( ) is not dfind if = Solution: +
4 CHAPTER 5 EXPONENTIAL AND LOGARITHMIC FUNCTIONS 4. log( + ) = log( ) log( + ) + log( ) = Chck: log 89? = log 89 log[( + )( )] = 4 4 ( + )( ) = = log 89? = log 89 4 = 4 = b b 4ac log 89? = log 89 a 4 a =, b =, c = = ( ) ( ) 4()( ) log 89? = log log 89 () 4 = 89? = log 4 4 89? = log 4 89 89 9 = log 89 log( ) is not dfind if = 89. Solution: = 89 4 4 4. ln( + ) = ln( + ) + = + = = Chck: ln( + ) is not dfind No solution. 45. (ln ) = ln 4 (ln ) = 4 ln (ln ) 4 ln = ln [(ln ) 4] = ln (ln )(ln + ) = ln = ln = ln + = = ln = ln = = = Chck: 47. ln(ln ) = ln = ln = = (ln )? ln 4 (ln )? ln( ) 4 (ln )? ln( ) 4 8 8 8 8 Solution:,, 49. A = P rt A = rt P ln A = rt P ln A = r t P r = ln A t P 5. D = log I I D = log I I I = D/ I I = I ( D/ ) 5. M = 6.5 log I I 6 M =.5 log I I 6 M = log I.5 I I = (6-M)/.5 I I = I [ (6-M)/.5 ]
SECTION 5-5 5 55. I = E R ( Rt/L ) L R 59. y = y = y = RI = E( Rt/L ) RI = Rt/L E RI = Rt/L E RI = Rt/L E RI + = Rt/L E RI = Rt/L E ln RI = Rt E L RI ln = t E t = L R ln y = y( + ) = y + y = + y = y + y = ( y) = y y = ln y y = ln y y RI E 57. y = y = + y = + y = ( ) + = ( ) y + This quation is quadratic in = = = = a b b 4ac a, b y, c ( y) ( y) 4()() () y 4y 4 ( y y ) = y ± y = ln(y ± y ) 6. Graphing y = and applying a built-in routin, w obtain 5 5 Th rquird solution of =,, is.8. 6. Graphing y = and applying a built-in routin, w obtain 5 5 Th rquird solution of =,, is.57.
6 CHAPTER 5 EXPONENTIAL AND LOGARITHMIC FUNCTIONS 65. Graphing y = ln + and applying a built-in routin, w obtain 5 67. Graphing y = ln + and applying a built-in routin, w obtain 5 5 Th rquird solution of ln + =,, is.4. 5 Th rquird solution of ln + =,, is.7. 69. To find th doubling tim w rplac A in A = P( +.7) n with P and solv for n. P = P(.7) n = (.7) n ln = n ln.7 n = ln ln.7 n = yars to th narst yar 7. W solv A = P rt for r, with A =,5, P =,, t =,5 =, r().5 = r r = ln (.5) r = ln.5 =.96 or 9.6% 7. (A) W r givn P.5 (w could us.5 million, but if you look carfully at th calculations blow, you ll s that th millions will cancl out anyhow),. for P, and for t (sinc May 7 is two yars aftr May 5)...5 r. r.5. ln r.5. ln.5 r.67 Th annual growth rat is.67%. 75. W solv P = P rt for t with P = P, r =.4. P = P.4t =.4t ln =.4t t = ln.4 t = 6 yars to th narst yar.67t (B) P.5 ; plug in for P and solv for t..67t.5.67t.5 ln.67t.5 ln.5 t 7.6.67 Th illgal immigrant population is prdictd to rach million nar th nd of, which is 7.6 yars aftr May 5. 77. W r givn A 5, A, t 6 : A t/ h A 6/ h 5 5 6/ h 6/ h ln ln 5 6 ln ln 5 h hln 6ln 5 6ln(/ ) h.58 ln(/ 5) Th half-lif is about.58 hours.
CHAPTER 5-5 7 79. Lt A rprsnt th amount of Carbon-4 originally prsnt. Thn th amount lft in was.89a. Plug this in for A, and solv for t:.89a A.4t.4t.4t ln.89 ln.89 ln.89. 4t ln.89 t,. 4 Th sampl was about, yars old. 8. W solv q =.9(.t ) for t with q =.7.7 =.9(.t ).7 =.t.9 7 =.t 9 =.t 9 =.t 9 ln =.t 9 t = ln 9. t = 7.5 sconds 87. (A) Plug in M = 7. and solv for E: E 7 log 4.4 E log 4.4 / E 4.4 / 4.4 4 E 7.94 jouls 89. First, find th nrgy rlasd by on magnitud 7.5 arthquak: E 7.5 log 4.4.5 log E4.4 8. Lt A rprsnt th amount of Carbon-4 originally prsnt. Thn th amount lft in 4 was.88a. Plug this in for A, and solv for t:.4t.88a A.4t.4t ln.88 ln.88 ln.88. 4t ln.88 t,. 4 It was, yars old in 4, so it was mad in. 85. First, w solv T = T m + (T T m ) kt for k, with T = 6.5, T m = 4, T = 7, t = 6.5 = 4 + (7 4) k().5 = k = k (B).5 ln.5 = k k = ln.5 k =.4 Now w solv T = T m + (T T m ).4t for t, with T = 5, T m = 4, T = 7 5 = 4 + (7 4).4t =.4t =.4t ln =.4t t = ln.4 t =.9 hours 4 7.94 jouls 4.88 jouls/day.76 days Finally, divid by th nrgy consumption pr yar: 6 5.64 jouls.5 7.5 jouls/yar So this nrgy could powr th U.S. for.5 yars, or about 86 days.
8 CHAPTER 5 EXPONENTIAL AND LOGARITHMIC FUNCTIONS.5 E 4.4.5 4.4 5 E 4.47 jouls Now multiply by twlv to gt th nrgy rlasd by twlv such arthquaks: 5 6 4.47 5.64 CHAPTER 5 REVIEW. (A) Th graph of y = log passs through (, ) and (, ). This corrsponds to graph m. (B) Th graph of y =.5 passs through (, ) and (,.5). This corrsponds to graph f. (C) Th graph of y = log.5 passs through (, ) and (.5, ). This corrsponds to graph n. (D) Th graph of y = passs through (, ) and (, ). This corrsponds to graph g. (5-, 5-). log m = n (5-). ln = y (5-) 4. = y (5-) 5. y = (5-) 6. (A) Mak a tabl of valus: (B) Th function in part (B) is th invrs of th on graphd in part (A), so its graph is a rflction about th lin y = of th graph to th lft. To plot points, just switch th and y coordinats of th points from 4 9 4 6 64 th tabl in part (A). 6 4 9 7 (5-) 7. 7 = 7 (+) ( ) 7 = 7 + + = 7 (5-) 8. = [ ( ) ] = ( ) = = (5-) 9. log = = = 8 (5-) (5-). log 5 = 5 = = 5 sinc bass ar rstrictd positiv (5-). = 4, = ln 4, =.9. log 7 = log = = (5-) 4. ln =.57 =.57 =.984 (5-). = 7.5 = log 7.5 =.4 5. log =. =. = (5-) 6..45 (5-) 7. Not dfind. ( is not in th domain of th 8.. (5-) 9..59 (5-) logarithm function.) (5-) 5. log a log b log c log a log c log b. a 5 ln ln a ln b b. log log log log a c log b log a c b (5-). 5 log log 5 log 5 5ln a ln b 5ln a ln b (5-) 4. log (45) 5 5 4 5 4 5
CHAPTER 5 REVIEW 9 or, using th chang-of-bas formula ln ln 5. ln( 5) 5 5 6 8. = = = ( )( + ) = =, 6. ln( ) = ln( + ) = + = 4 Chck: ln( 4 )? ln(4 + ) ln 7 ln 7 9. 4 = ( ) = ( ) = ( ) = = = = log 5 7. log( ) = log( ) log( ) = log( ) = ( ) = + = + 4 = = 7 4 7 4 Chck: log( )? log( ) log? log. = 8 8 = ( 9) = ( )( + ) = = = + = nvr = = Solution:,. log /4 6 = log /4 4 = log /4 = 4 = 5. log = log = log = 9. = log(.56-7 ) = 6.67 (5-) 4.. =.5.5 = ln. = ln..5 = 9.. log 9 = = 9 = 9 = 9 = 9 = ± 9 = sinc bass ar rstrictd positiv 6. = (. ) = 4.8 (5-) 4. = ln 4 ln. =.66 (5-) 7. = log 5 log = or ln log5 ln 5 =.95 (5-) 4. 5 = 5() 5 = 5 5 = ln 5 = ln ln 5 ln = =. 44. 5 = 7.8 ( )log 5 = log 7.8 = log 7.8 log5 =. log 6 = 6 / = 64 = = 64 4. log 5 = 5 5 = 5 = 8. ln =.8 =.8 =.4 (5-) 4. 4, =,5. 4,,5 =.. = ln 4,,5 = ln 4,.,5 =.9 log 7.8 log 5 =.
4 CHAPTER 5 EXPONENTIAL AND LOGARITHMIC FUNCTIONS 45. = = = = ( ) = ( ) = 46. log log 9 = log = 9 = 9 = = 47. log log = log 4 log( + 4) log = log 4 4 = 4 cludd valu: 4 4 ( + 4) = ( + 4) 4 4 ( + 4) = + 4 = + 4 = ( + 6)( ) = = 6 = This quation is quadratic in : = = 4 b b ac a, a b, c ( ) ( ) 4()( ) = 8 = ± = ln( ± ) is ngativ, hnc not in th domain of th logarithm function. = ln( + ) =.88 Chck: log( ) log(9 )? = log(7,) log(,7) =? log 7,? =,7 log = Chck: log( 6) is not dfind? log log = log 4 log( + 4) log? = log 4 6 log = log Solution: 48. ln( + ) ln = ln ln = ln = = 4 + = 4 = = 49. ln( + ) ln( ) = ln ln = ln = Ecludd valu: ( ) = ( ) + = = Chck: ln( + ) ln? = ln ln 4? = ln ln 4 = ln 4 Chck: ln is not dfind ln ln? = ln ln( + + ) ln? = ln
CHAPTER 5 REVIEW 4 = = b b 4ac a a =, b =, c = ( ) ( ) 4()( ) = () 5. (log ) = log 9 (log ) = 9 log (log ) 9 log = log [(log ) 9] = log (log )(log + ) = log = log = log + = = log = log = = = ln(4 + ) ln? = ln ln 4? = ln ln 4? = ln ln 4 4 ln? = ln? = ln ln 8 6? = ln ln = ln Solution: Chck: (log )? 5. ln(log ) = = log 9 = log = = (log )? = log( ) 9 7 = 7 (log )? = log( ) 9 7 = 7 Solution:,, 5. ( + )( ) ( ) = + + = + + = (5-) 5. ( + )( ) ( ) = ( ) ( ) [( ) + ( ) ] = [ + ] = + = (5-) 54. Th graph of g is th sam as th graph of f rflctd through th ais, shrunk vrtically by a factor of, and shiftd upward units; g is dcrasing. Domain: all ral numbrs Horizontal asymptot: y = 55. Th graph of g is th sam as th graph of f strtchd vrtically by a factor of and shiftd downward 4 units; g is incrasing. Domain: all ral numbrs Horizontal asymptot: y = 4 (5-) (5-)
4 CHAPTER 5 EXPONENTIAL AND LOGARITHMIC FUNCTIONS 56. Th graph of g is th sam as th graph of f shiftd downward units; g is incrasing. Domain: (, ) Vrtical asymptot: = 57. Th graph of g is th sam as th graph of f strtchd vrtically by a factor of and shiftd upward unit; g is dcrasing. Domain: (, ) Vrtical asymptot: = (5-) 58. If th graph of y = is rflctd in th ais, y is rplacd by y and th graph bcoms th graph of y = or y =. If th graph of y = is rflctd in th y ais, is rplacd by and th graph bcoms th graph of y = or y = or y =. (5-) 59. (A) For >, y = / dcrass from / to whil ln( + ) incrass from to. Consquntly, th graphs can intrsct at actly on point. (B) Graphing y = / and y = 4 ln( + ) w obtain (5-) Th solution of / = 4 ln( + ) is =.58. 6. Eamining th graph of f() = 4 + ln, w obtain 5 Th zros ar at.8 and.87. 6. Graphing y = and y = 8 log, w obtain Th graphs intrsct at (.,.) and (.65, 4.5).
CHAPTER 5 REVIEW 4 6. D = log I I D = log I I D/ = I I I D/ = I I = I ( D/ ) 65. r = P i ( ) n i r P = i ( ) n i P = ( ) i n r i Pi = ( + i) n r Pi = ( + i) n r Pi = ( + i) n r ln Pi = n ln ( + i) r Pi ln r = n ln( i) Pi n = ln r ln( i) 6. y = y = = ln( y) = ln( y) = ± ln( y) 64. = ln I k I k = ln I I I = k I I = I ( k ) 66. ln y = 5t + ln c ln y ln c = 5t y ln c = 5t y c = 5t y = c 5t 67. y = log = log y y 4 4 8 8 Domain f = (, ) = Rang f - Rang f = (, ) = Domain f - (5-) 68. If log = y, thn w would hav to hav y = ; that is, = for arbitrary positiv, which is impossibl. (5-) 69. Lt u = log b M and v = log b N; thn M = b u and N = b v. Thus, log(mn) = log b (b u b v ) = log b b u+v = u + v = log b M + log b N. (5-) 7. W solv P = P (.) t for t, using P = P. P = P (.) t = (.) t ln = t ln. ln = t ln. t =.4 yars (5-) 7. W solv P = P.t for t using P = P. P = P.t =.t ln =.t ln = t. t =. yars (5-)
44 CHAPTER 5 EXPONENTIAL AND LOGARITHMIC FUNCTIONS 7. A = original amount.a = prcnt of original amount W solv A = A -.4t for t, using A =.A..A = A -.4t. = -.4t ln. = -.4t ln. = t.4 t = 7, yars (5-) 7. (A) Whn t =, N =. As t incrass by /, N doubls. Hnc N = () t / N = t (or N = 4 t ) (B) W solv N = 4 t for t, using N = 9 9 = 4 t 9 = t log 4 t = 9 log 4 t = 5 days (5-) 74. W us A = P rt with P =, r =., and t = =. A =.() A =.5 6 dollars (5-) 75.(A) t p, 5 67 449 5 5 5 9 (B)As t tnds to infinity, P appars to tnd to. (5-) 76. M = log E E = 4.4 E W us E =.99 4 M = log 4.99 4.4 M = log(.99 9.6 ) M = (log.99 + 9.6) M = (.99 + 9.6) M = 6.6 (5-4) 78. W us th givn formula twic, with I =,I D = log I D = log I I I 77. W solv M = log E for E, using E E = 4.4, M = 8. 8. = log E 4.4 (8.) = log E 4.4.45 = log E D D = log I log I = log I I = log I I I I I I = log, I = log, = 5 dcibls I Th lvl of th loudr sound is 5 dcibls mor. (5-) 4.4 E =.45 4.4 E = 4.4.45 E = 6.85 or 7.8 6 jouls (5-4)
CHAPTER 5 REVIEW 45 79. I = I kd To find k, w solv for k using I = I and d = 7.6 W now find th dpth at which % of th surfac light rmains. W solv I = I.94d for d with =.I I = I k(7.6).i = I.94d. =.94d = 7.6k.94d = ln. 7.6k = ln d = ln..94 d = 489 ft (5-) k = ln 7.6 k =.94 8. W solv N = for t with N =..5t 9 =.5t 9.5t = 9.5 = + 9.5t.5 = 9.5t.5 =.5t 9.5t = ln.5 9.5 t = ln 9.5 t = yars (5-) 8. (A) Th indpndnt variabl is yars sinc 98, so ntr, 5,, 5,, and 5 as L. Th dpndnt variabl is Mdicar pnditurs, so ntr that column as L. Thn us th ponntial rgrssion command on th STAT CALC mnu. I Th ponntial modl is y 4..9. To find total pnditurs in and, w plug in and 4 for : y 4 y 4..9 574; 4 4..9,6 Epnditurs ar prdictd to b $574 billion in and $,6 billion in. (B) Graph y 4..9 and y 9 and us th INTERSECT command: Epnditurs ar prdictd to rach $9 billion in 5. (5-) 8. (A) Th indpndnt variabl is yars sinc 99, so ntr 4, 7,,, 6 as L. Th dpndnt variabl is th numbr of subscribrs, so ntr th subscribrs column as L. Thn us th logarithmic rgrssion
46 CHAPTER 5 EXPONENTIAL AND LOGARITHMIC FUNCTIONS command from th STAT CALC mnu. Th modl is y = 99. + 4.7 ln. Evaluating this at = 5 (yar 5) givs 6.5 million subscribrs. (B) With th sam data as in part (A), us th logistic rgrssion command from th STAT CALC mnu. 54.9 Th modl is y =. Evaluating this at = 5 (yar 5) givs 6.8 million.556.94 subscribrs. (C) Plot both modls, togthr with th givn data points, on th sam scrn. Clarly, th logistic modl fits th data bttr. Morovr, th logarithmic modl prdicts that th numbr of subscribrs bcoms infinit, vntually, which is absurd. Th logistic modl prdicts vntual lvling off nar 54.9 million, which still sms high compard with th US population, but is mor rasonabl. Th logistic modl wins on both critria spcifid in th problm. (5-, 5-4)