Answer
$b=\ln (\dfrac{8}{9})$
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, $b=\ln (\dfrac{8}{9})$
Thomas' Calculus 13th Edition
by Thomas Jr., George B.
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
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