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: 56


a) $\frac{1}{3}\lt x\lt\frac{1+\ln2}{3}$ b) $x\gt\frac{1}{e}$

Work Step by Step

a) $1\lt e^{3x-1}\lt2$ $0=\ln1\lt3x-1\lt\ln2$ $1\lt3x\lt1+\ln2$ $\frac{1}{3}\lt x\lt\frac{1+\ln2}{3}$ b)$1-2\ln x\lt3$ $2\ln x\gt-2$ $\ln x\gt-1$ $x\gt e^{-1}=\frac{1}{e}$
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