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: 23

Answer

$y=\frac{1}{2}\ln x+\frac{1}{2}$

Work Step by Step

$y=e^{2x-1}$ $2x-1=\ln y$ $2x=\ln y +1$ $x=\frac{1}{2}\ln y+\frac{1}{2}$ $y\leftrightharpoons x$ $y=\frac{1}{2}\ln x+\frac{1}{2}$
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