Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 10: Infinite Sequences and Series - Section 10.2 - Infinite Series - Exercises 10.2 - Page 580: 90


$b=\ln (\dfrac{8}{9})$

Work Step by Step

Given: $1+e^b+e^{2b}=\dfrac{1}{1-e^b}=9$ This implies that $\dfrac{1}{9}=1-e^b$ or, $e^b=1 -\dfrac{1}{9}=\dfrac{8}{9}$ Hence, $b=\ln (\dfrac{8}{9})$
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.