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Unformatted text preview: a... Section 4.5 2. y = lnx5/2 4. y = In x”2
5 l
—21nx —21nx
[—3 ,/_i
3 2x ) 2x
I __§ / _l
y(l)—2 )(Uz 3
10. y = ln(1  x)”2 = Eln(l  x) v,_§(—l)_ 3
y 21x 2(xl) 14. y: 7
x v,_ .rzil/x! — 2xlnx : x— 2xlnx : l  21nx 4 x4 18. y = in X3 22. y = ln(x\/4 + x2) = lnx +1n(4 + x2)”1 / = l + x _ £2.53.
y x 4+x2 x(4+x2)
+ .T
26. f(x) = 1n]1 _ :x = 1n(l + e") — ln(l — e")
’ _ ex _ ._e,\’
f(x) 1 + e‘ l  e'“
= e~"(l — e“) + e‘(l + e") z 28'"
(l + e‘)(l — e") l  el“
1
30. log3x = “111.: Derivatives of Logarithmic Functions Section 4.5 Derivatives of Logarithmic Functions 149
6. f(x) = In 2): 8. y = ln(l ~ x3)
, . ~ 3 _ l V, _ _ 2x f('x)_2x_x “  lx2
: 2x
x2  1
12. y = (In x32 = (2 In x)2 = 4(1n x)2
v’ = 4[2(lnx)(l)] = 81""
x x
x '7
. = : —  +
16 y lnx2 +1 lnx ln(x 1)
,_l_ l _i_ 2x __ l‘X2
‘—x 12+l(2x)_ x2+l_x(x2+l)
20 i=ln erl=l[ln( +l)—1n( —1)]
. \ x _ l 2 x x
yl[1_1]=1=1
‘ 2 x+1x—1 xZ—I 1—x2
24. f(x) = xlne‘2 = x062) = x3
f'(x) = 3x2
28. 3x = e"“"”
32. log512 = ﬁlm 12 =3 1.544 (calculator)
In 32
. l 7 . = z  . 8
36 ogh/3 32 Ina/3) 854
42. y = 65'"
1 y’ = ln6  65‘(5) = 51n6  65" 150 Chapter 4 Exponential and Logarithmic Functions
W 44. f(x) = 10‘:
f’(x) = (ln lO)lO“‘(2x) = 2x(ln IONO": 48. v,:x(]/x)7—lnx= l —jnx
x‘ x~
v’(e) = 0
1
. —  = 0(x  e)
6
l .
y = ; Tangent lme
52. In xy + 5x = 30 lnx+lny+5x=30 l Idy
—+——+ =
x ydx 5 0
dv l
_= ___s
dx (x )V
z_x_5y=_y(l+5x)
x (I ~ _ _2
f(x)_ x2 60. T = 87.97 + 34.961np + 7.91 J; d_T = 34.96 + 7.91
dP p 2J5
dT 34.96 7.91 At =60,—=—+
p @ a) zjﬁ 62. f(x) = Zlnx3 = 6lnx I _ 9
f (x) — x
At (e, 6). the slope of the tangent line is f’(e) = 6/e.
. 6
Tangent lme: y — 6 = :(x — e)
6
)7 : —x e 46. y = 23““
y’ = 3m + x3~+l In 3 = 3WD + xln 3] 50. g(x) = log2(3x — I) = In 2 t ___3__
gm‘ax l)ln2 3
321n2 g’(11)= y~5= 54. 4x)! + ln(x2y) = 7 ll
\1 4xy+ 2lnx+ lny 2 y,
4g,'+4y+_+._
x y II
0 <4x + l)y’ = 4y — 2
y x y : ~ 1.093 degrees per pounds per square inch. ln(3x — 1) I M=Lm ’\ 4x +(1/y) 4x2y + x
58. f(x) = logmx
ﬂ") = 111—110”2 2 x21_n]10
64. f(x) = ln(x\/m) = lnx + %1n(x + 3) At (1.20, 0.90), the slope of the tangent line is f’(l.20) = Tangent line: y — 0.90 = 52—?(x ~ 1.20) 21y — 18.90 = 20x — 24 0 = 20x — 21y — 5.10
0 = 200x — 210y — 51 20 1. Section 4.5 Derivatives of Logarithmic Functions 151 7
.l" , ,lnx 68. v = — — in r
66. . =.‘l .'=.'—  '
f(r) l og3t In 3 2
,._l' _‘ \,/=x_1_x2l
f (.x) — E0 + 21 lnx) . x A.
l M = =
‘f,(l):___ ) 0whenx 1.
ln 3 I .2 + ]
0 l y”=l+;:lxz
‘ ’ ln 3 U — I)
x I [Notez x = ~ 1 is not in the domain of the function]
y = _ Since y”(l) = 2 > 0, there is a relative minimum at
’ (l, Moreover, since y” > 0 on (0. 00), it follows that
the graph is concave upward in its domain and there are
no inflection points.
5 :I.
4
3
1 2 I A
1 .
70. y = xlnx 72. y = (In x)2
l 1 2 l
y’ = x() + (ln.r)(l) = 1 + lnx y’ = 2(ln x)<—) = nx
x x x
y’=0whenx=e". y’=0whenx= l.
',,_ l 7”: x(2/x) — 21nx)(l)_ 2(1 — lnx)
.\ x r " x3 X:
Since y”(e‘ ’) > 0, there is a relative minimum at Since y”(1) > 0. there is a relative minimum at (l , 0).
(e“, —e“). Moreover. since y” > 0 on (0. 00), it fol Since y” = 0 when x = e, it follows that there is an
lows that the graph is concave upward in its domain and inﬂection point at (e. I). there are no inﬂection points. ‘ \ > »\
'1': imaammm
74. x = 1000 — plnp 76. x = 300 — 501n(lnp)
(ix 1 dx [/11 50
—. = _ _ + = _ _. : _ __ = _
dp 0 Mp) (In/0(1)] (l + lnp) dp 0 solnp plnp
dx dx —50
: — = — + z — _ _ = ' —— = z — _ _
When p 10,dp (1 In 10) 3 3 When [1 10 dp 101nm 217l 152 Chapter 4 Exponential and Logarithmic Functions W 500
78. x—m 80. C—100+25x—1201nx,)t21
— C 100 lnx
500 : '—— = — —— —__
ln(p3+I)=T (a)C x X +25 120x
[,2 + I = eSOO/x (b) 6,, : —x1200 _ I2O[x(l/x)x2— In x]
P = e500x _ 1
= 100 £0 _ I...
.311 = %(e500/x _ l)—i/2(esoo/_t~)< _ : x
S tt’ C’ = 0,
250 esno/x 6 mg
2 x2 ' e500} _ I 100 = 120“ — In x)
Whenp=10. ~2=1nx*1
500 dp
= z .3 — = — . .
x mum) 108 4anddx 0215 16_l=lnx Note that dp/dx = 1 /[dx/dp] from Exercise 75. Minimum of 5.81 at x = 6.25. 5.315 82. I=m, x> 1000
(a) 5?. 2000 0 (b) Ifx = 1167.41. 1‘ = 20 and the total amount is (1 167.41)(20)(12) z $280,178.
(c) Ifx = 1068.45,t = 30 and the total amount is (1068.45)(30)(12) z $384,642. (d) [I — ‘ (lnx — 6.7968)2 ~5.315/x
(In x — 6.7968)2 2 —5.315
x(lnx — 6.7968)2 When x = 1167.41, the instantaneous rate of change is —5.315
ll67.41(ln 1167.41 — 6.7968)2 == — 0.064. When x = 1068.45, the instantaneous rate of change is 
‘5'315 a: —0.158.
1068.45(1n 1068.45 — 6.7968)2 (e) For a higher monthly payment, the term is shorter and the overall payment total smaller. Section 4.6 Exponential Growth and Decay 153 Section 4.6 Exponential Growth and Decay 2. Since 3 = %whent = 0. it follows that C = Moreover. 4. Since y = 2 when t = 0, it follows that C = 2. Moreover.
since y = 5 when! = 5. we have 5 = §e5k and since y = 1 when! = 5, we have 1 = 225* and
k=%ln]0’~*0.4605. Clintz .0386.
Thus, v = £10460“. Thus, v = ze—onxhi 6. Using the fact that y = éwhen t = 3 and y = 5 when t = 4. we have% = Ce“ and 5 = Ce“. From these two equations we have 5
% = $6“ = 5e“. and e" = 10 Thus, k = In 10 and we have y = 02”" 1". Since 5 = Ce‘“" '0, it follows that S l
C = e41n 10 : Therefore,
y 2 1 e:ln 10 Q, ] €2.3026!
” 2000 2000 ‘
dv 2 ' d3,
. —e = —— . + = 5. i
8 dt 3y 10 dt 2"
y = 20 when t = 0: y = 2050/” _v = 18 when! = 0: y = 18e52’
dv 2 —1°/3)i . 2 42/1): 2 d." s, q2
—'=   =—  =—v =l . ~’=5.21~'=5.2
dt 20( 3)e 3[20e ] 3) dt 8(5 2)e ( 8e ) y
Exponential decay Exponential growth
12. From Example 1 we have 14. Since y = Ce"! = 3eDW/2V57'51', we have
V = Celt! = Ce[ln(l/2)/1599]l t = ‘000: y = 3e[ln(1/2)/5715)1000 z 166 grams
1.5 = Ce['“(‘/2V‘599]'°°° = C ~ 2.314. t = 10.000: y = 3e[‘““/2V‘7‘5]“"°°° *2 0.89 grams.
The initial quantity is 2.314 grams.
Whent = 10,000,
y = 2.314e['“‘“21/599119000 z 0.03 grams.
16. Since y 2 Ce’" = Ce['"“/2)/24“‘“’]’, we have 18. 0.9957C = Ce‘““/2V"
0.4 = Ce['""/2V24“0°]'0~0m => C a 0.533. when t = 1 and h is the halflife
The initial quantity is 0.533 grams. 0.9957 = emH/ZW
= 1 1
When t 1000, “09957) : n(2)
V = 0‘533eﬂnl/2)/24.l()0]10()0 m 0518 h ln(i) h : ln(0.9957)
h = 160.85 years 20. 0.30C = Ce[ln1I/2)/5715]i ln(1/2)
’5715 = 5715 In 0.30
ln(l/2) In 0.30 = 1‘ z 9927 years 154 Chapter 4 Exponential and Logarithmic Functions 1
22. (0. 8). (20,) 24. (a) Let t = 0 represent 1960. 2
VI : 8eky’ 'V = C6,“ = 2.3,“
12
_ = 1140) = __ g
%: 86:01:”l =21_01n(%)= 151016z 411386 12 2.3e => 40k 111(2'3) => k 0,0413
y = 2.3e"1“‘“3’ Exponential model VI ._. 86—().1386t
1970: y(10) = 2.3e004'3W z 3.5 million .Vz : 80)”
l : 8(2)20k1 => k = .Llo = ‘ln 16 z _02 ‘9803 YOU} ’~' 5.3 m1llion
2 2 20 g2 16 20 In 2 . 1900: y(30) z 7.9 million
."2 = 8(2)“"2’ (b) 24 = 2_3e().04131
kl=(ln2)k2 _ 1 (3£>~
’7 0.0413 1" 2.3 56'8 years (c) Increasing by 0.0413, or 4.13% each year. 26. Since A = 20.000e"~“’5’. the time to double is given by
the following. 40,000 = 20000120105! 28. Since A = 10,000e" and A = 20.000 when t = 5, we
have the following. 20,000 = 10,000e5’ 1“ 2 = 0105' r = 1—"3—2 ~ 0.1386 = 13.86%
In 2
I — 0.105 ~ 6601 years Amount after 10 years: Amount after 10 years: A = 10.0009”ln 2” 51"") = $40,000 A = 20.000e0"°5“°) ~ $57,153.02 Amount after 25 years: Amount after 25 years: A = 10.000eﬂIn W512” = $320,000 A = 20,000e0105‘25’ z $276,091.48 30. SinceA = 2000e" and A = 6008.33 whent = 25, we 32. (a) By equating A = P0 + i)’ and A = Pe" we have the
have the following. following. 6008.33 = 2000e25’
1n 3.004165 = 25r P(l + i)’ = Fe”
(1 + i1r= (er)! r z 0.044 = 4.4%
The time to double is given by
4000 = 2000:3004“
1n 2 = 0.044t
t z 15.753 years. Amount after 10 years: A = 2000e°'°44('°’ z $3105.41 1 + i = e’
Therefore,i = e’ — 1. (b) lfr = 0.06. theni = e006  1 z 0.0618 or 6.18%. 36. 38. 42. 4
n=4: i=(l — l z0.077l4=7.7l4%
.5”
n = l2: 1': (I + — I % 0.07763 g 7.763%
0.075
= : '= +————
n 365 1 (l 365 Continuous: i = elm” — l 2 0.07788 *~v 7.788% 7.7l4% 7.763% 7.788% 365
>  I z 0.07788 a 7.788% Section 4.6 Exponential Growth and Decay 7.788% (a) For r = 10%, the approximate time necessary for the investment to double is % = 7 years. (b) For r = 7%, the approximate time necessary for the investment to double is Q = 10 years. (a) Let t = 0 correspond to 1990.
Date points: (0, 150), (10. 1074) y = 150(1.21756)’ = lSOe""%35’ Exponential models ;:: 924t+—150 (c) 2006: y = 92,406) + 150 ~ $16284 million ((1) 2000 0 20
0 Answers will vary. ‘ (a) Since N = 20 when t = 30, we have
20 = 30(1 — e30") _ ln(l/3)
“ m N = 300 _ 90.03661).
(b) 25 = 30(1 _ e—(unsm) t = ln( 1 /6)
~0.0366 k a —0.0366 ~ 49 days Linear model (b) 2006: y = 150e““’°"5"“’ 3 $3499 million 40. S = 30(1 — 3’”) (a) Since S = 5 when t = l. we have 5=30(l—e")
e"=l—%
k=mg [Notez S is in thousands of units]
Therefore, S = 30(1 — el'"‘5/")3’). (mummU—wmmo=m I630 The saturation point for the market is 30,000 units. (c) When I = 5. we have S=3m1—AW%WW~11%4 Thus. S ~ 17,944 units.
((0 1 5 5 156 Chapter 4 Exponential and Logarithmic Functions 44. (21) Since p = Ce“ where p = 45 when x = 1000 and p = 40 when x = 1200. we have the following.
45 : CeIOOOk and : CeIZUOk
In 45 = In C + lOOOk
ln40 = lnC +1200k
In 45 — In 40 = —200k _ ln(45/40) ~ _
— _ 200 0.0005889 Therefore, we have 45 = Ce”’°°"°°"05"89) which implies that C m 81.0915 and p = 8l.0915e“""°°5“9*". k (b) Since R = xp = 81.0915xe‘0'0"05839", we have the following.
R’ = 8l.09]5[__0.0005889xe—0.0005889); + e‘0.0005889.t]
= 8l.0915e“)"m°5389‘[1 — 0.0005889x]
= 0
Since R’ = 0 when x = I/0.0005889 z l698 units. we have 17 = 81.0915e“0‘0°"5“39“698’ ~ $29.83. 46. A = Ve‘O‘l‘l' 48. (a) If 10 = 1, then we have
: 075$ —0.04z
100.000e e R: In] andgj: lnl'
: 100,000e(0.75./}—0.041) I 10 1" 10
0 75 Therefore, I = e331“ ‘0 ~ 199.526.2315.
A [m = 100,0“)(7 _ 0.04)e(o.75/1—0.04n : 0 In I
2 ’  (b) 2R = In 10 implies that 1 = eZR'n '0 = (eR’“ '0)?
3—7/5. = .04 The intensity is squared if R is doubled.
t
dR l 1 l
0.75 (C) — = (—) =
=—__=.5 d] 1101 1110
x/i (0.040(2) 937 n n
t = 87.89 « 88
The timber should be harvested in 2078 to maximize the
present value.
Review Exercises for Chapter 4
, 3 _ ~i/3 V3 3
2. 10V : (JR) = 43 = 64 4. Q) = (%) = 3
6. (9V3)(3'/3) = (9  3W3 = 271/3 = 3 8. 56)” = 41(2)3 = % = 2
10. 32“ = 23"+1 12. x5/2 = 243 = 35
(25).\‘ = 23x+l J; =
~5x = 3x + l x :
8x = l
l
x = —§
14. e‘5 2 el‘“ 16. 4x2 = e5
—5 = 2x + l , l i
x = 26*
x = 3
e.5/2
x — :t 2 ...
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 Winter '08
 chuchel
 Logarithmic Functions, Tangent lme

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