Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 6 - Inverse Functions - Review - Exercises - Page 505: 17


$$x= \sqrt{e+1}$$

Work Step by Step

Given $$\ln (x+1)+\ln (x-1)=1$$ Then \begin{align*} \ln (x+1)+\ln (x-1)&=1\\ \ln (x+1)(x-1)&=1\\ \ln (x^2-1)&=\ln e\\ x^2-1&=e \end{align*} Hence $$x= \sqrt{e+1}$$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.