Thomas Calculus 13th [Solutions]

514 Chapter 7 Integrals and Transcendental Functions
65. From zooming in on the graph at the right, we
estimate the third root to be x 0.76666
ln 2
ln
66. The functions f ( x)
x and g( x ) 2 x appear
to have identical graphs for x > 0. This is no
ln 2 ln 2 ln x ln 2 ln x ln x
accident, because x e ( e ) 2 .
67. (a) The point of tangency is (p, ln p) and m 1
tangent p
since dy 1
dx x
. The tangent line passes through (0, 0)
the equation of the tangent line is y 1 x . The tangent line also passes through
p
(p, ln p) ln p 1 p 1 p e , and the tangent line equation is y 1 x.
p
e
d 2 y
(b) 1 for x 0 y = ln x is concave downward over its domain. Therefore, y = ln x lies below the
2 2
dx x
graph of y 1 x for all x > 0, x e, and ln x x for x > 0, x e.
e
e
e
(c) Multiplying by e, e ln x < x or ln x x.
e
ln x x
(d) Exponentiating both sides of ln x x , we have e e , or
e
(e) Let x = to see that e . Therefore, e is bigger.
e
e x
x e for all positive x e.
68. Using Newtons Method: f(x) = ln(x) 1 1
ln( xn
) 1
f ( x) xn 1 xn x
1 n 1 xn[2 ln( xn
)].
x
Then x1 2, x2 2.61370564, x 3 2.71624393, and x 5 2.71828183. Many other methods may be used.
For example, graph y = ln x 1 and determine the zero of y.
69. (a) log ln 8
3 8 1.89279 (b) log ln 0.5
ln 3
7 0.5 0.35621
ln 7
(c)
ln17
ln 7
log20 17 0.94575 (d) log
ln 20
0.5 7 2.80735
ln 0.5
(e) ln x (log 10 x )(ln10) 2.3ln10 5.29595 (f) ln x (log 2 x)(ln 2) 1.4ln 2 0.97041
(g) ln x (log 2 x )(ln 2) 1.5ln 2 1.03972 (h) ln x (log 10 x)(ln10) 0.7 ln10
x n
70. (a) ln10 log ln10 ln ln
ln 2 10 x x x log
ln 2 ln10 ln 2 2 x (b) ln a log ln ln ln
ln 4 x a x x
b ln b ln a ln b
logb
x
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