Intermediate Algebra (6th Edition)

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

Chapter 9 - Section 9.3 - Exponential Functions - Exercise Set - Page 559: 28



Work Step by Step

We are given that $\frac{1}{27}=3^{2x}$. Both of these numbers are powers of 3, so we can rewrite the equation as $3^{-3}=3^{2x}$. We know that $3^{-3}=\frac{1}{27}=\frac{1}{3^{3}}$. 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, $-3=2x$. Divide both sides by 2. $x=-\frac{3}{2}$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.