Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 1 - Section 1.5 - Inverse Functions and Logarithms - 1.5 Exercises - Page 67: 51


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) $\ln(3x-10)=2$ $3x-10=e^2$ $3x=e^2+10$ $x=\frac{1}{3}(e^2+10)$
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