Answer
a) $x=\frac{1}{4}(7-\ln6)$
b) $x=\frac{1}{3}(e^2+10)$
Work Step by Step
a) $e^{7-4x}=6$
$7-4x=\ln6$
$4x=7-\ln6$
$x=\frac{1}{4}(7-\ln6)$
b) $\ln(3x-10)=2$
$3x-10=e^2$
$3x=e^2+10$
$x=\frac{1}{3}(e^2+10)$
Calculus: Early Transcendentals 9th Edition
by Stewart, James
Published by
Cengage Learning
ISBN 10:
1337613924
ISBN 13:
978-1-33761-392-7
Chapter 1 - Section 1.5 - Inverse Functions and Logarithms - 1.5 Exercises - Page 65: 51
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