Thomas Calculus 13th [Solutions]

Section 1.6 Inverse Functions and Logarithms 39
43. (a)
ln 7.2
e 7.2
(b)
2
ln x
e 1 1 (c)
ln x
2 2
e x
ln x ln y ln( x/ y)
e e x
y
44. (a)
2 2
ln( x y ) 2 2
e x y (b) e ln 0.3 1 1
ln 0.3
e 0.3
(c)
ln ln 2 ln( /2)
e
x e
x x
2
45. (a)
(c)
1/2 1
e
2ln e 2ln e (2) ln e 1
(b) ln(ln e ) ln( eln e) ln e 1
2
2 2
( x y ) 2 2 2 2
ln e ( x y )ln e x y
46. (a)
(c)
sec
ln e (sec )(ln e ) sec
(b)
2ln x ln x 2
ln( e ) ln e ln x 2ln x
2
x
( e ) x x
ln e ( e )(ln e)
e
47. ln y = 2t + 4
ln y 2t 4 2t
4
e e y e 48. ln y = t + 5
ln y t 5 t 5
e e y e
49. ln(y 40) = 5t
ln( y 40) 5t 5t 5t
e e y 40 e y e 40
50. ln(1 2y) = t
ln(1 2 y) t t t
e e 1 2y e 2y e 1 y e t 1
2
51. ln(y 1) ln 2 = x + ln x ln(y 1) ln 2 ln x = x
x
x
y 1 2xe y 2xe
1
ln
y 1
2x
y 1 ln
x y 1 x
x e e e
2x
2x
52.
2
y
ln( y 1) ln( y 1) ln(sin x) ln ln(sin x ) ln(y 1) = ln(sin x)
y 1
y 1 = sin x y = sin x + 1
2 1
ln( y 1) ln(sin x)
e e
53. (a)
(b)
(c)
54. (a)
(b)
(c)
2k
2k
2
e 4 ln e ln 4 2k ln e ln 2 2k = 2 ln 2 k = ln 2
10k 10k 10k
100e 200 e 2 ln e ln 2 10k ln e = ln 2 10k = ln 2 k ln 2
10
k /1000 k /1000
e a ln e ln a k ln e ln a k ln a k 1000ln a
1000 1000
5k
1 5k
1
e ln e ln 4 5k ln e = ln 4 5k = ln 4 k ln 4
4
5
k k 1 k 1
80e 1 e 80 ln e ln80 k ln e = ln 80 k = ln 80
(ln 0.8) k ln 0.8 k k
e 0.8 ( e ) 0.8 (0.8) 0.8 k = 1
55. (a)
(b)
(c)
0.3t
0.3t
3
e 27 ln e ln 3 ( 0.3t)ln e = 3 ln 3 0.3t = 3 ln 3 t = 10 ln 3
kt 1 kt 1
e ln e ln 2 kt ln e ln 2 t ln 2
2
k
(ln 0.2) t ln 0.2 t t t
e 0.4 ( e ) 0.4 0.2 0.4 ln 0.2 ln 0.4 t ln 0.2 = ln 0.4 t ln 0.4
ln 0.2
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