Intermediate Algebra (6th Edition)

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

Chapter 9 - Review: 33



Work Step by Step

We are given that $2^{3x}=\frac{1}{16}$. Both of these numbers are powers of 2, so we can rewrite the equation as $2^{3x}=2^{-4}$. We know that $2^{-4}=\frac{1}{2^{4}}=\frac{1}{16}$. 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, $3x=-4$. To solve for x, divide both sides by 3. $x=-\frac{4}{3}$
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