University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 9 - Section 9.2 - Infinite Series - Exercises - Page 498: 90


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

Work Step by Step

We have $1+e^b+e^{2b}=\dfrac{1}{1-e^b}=9$ and $\dfrac{1}{9}=1-e^b$ or, $e^b=\dfrac{8}{9}$ Thus, $b=\ln (\dfrac{8}{9})$
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