Answer
$\ln (\dfrac{8}{9})$
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, $b=\ln (\dfrac{8}{9})$
University Calculus: Early Transcendentals (3rd Edition)
by Hass, Joel R.; Weir, Maurice D.; Thomas Jr., George B.
Published by
Pearson
ISBN 10:
0321999584
ISBN 13:
978-0-32199-958-0
Chapter 9 - Section 9.2 - Infinite Series - Exercises - Page 498: 90
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