Intermediate Algebra (6th Edition)

Published by Pearson
ISBN 10: 0321785045
ISBN 13: 978-0-32178-504-6

Chapter 9 - Review: 32



Work Step by Step

We are given that $3^{x}=\frac{1}{9}$. Both of these numbers are powers of 3, so we can rewrite the equation as $3^{x}=3^{-2}$. We know that $3^{-2}=\frac{1}{3^{2}}=\frac{1}{9}$. From the uniqueness of $b^{x}$, we know that $b^{x}=b^{y}$ is equivalent to $x=y$ (when $b\gt0$ and $b\ne1$). Therefore, $x=-2$.
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