Precalculus (6th Edition)

Published by Pearson
ISBN 10: 013421742X
ISBN 13: 978-0-13421-742-0

Chapter 4 - Inverse, Exponential, and Logarithmic Functions - Summary Exercises on Inverse, Exponential, and Logarithmic Functions - Exercises - Page 449: 43

Answer

$-\frac{1}{3}$

Work Step by Step

As $\frac{1}{3}=3^{-1}$ and $9=3^2$, we have $3^{-x-1}=3^{2x}$, thus $2x=-x-1$, $3x=-1$ and $x=-\frac{1}{3}$
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