Intermediate Algebra (6th Edition)

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

Chapter 9 - Review: 35



Work Step by Step

We are given that $9^{x+1}=243$. Both of these numbers are powers of 3, so we can rewrite the equation as $(3^{2})^{x+1}=3^{2x+2}=3^{5}$. 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, $2x+2=5$. To solve for x, subtract 2 from both sides. $2x=3$ Divide both sides by 2. $x=\frac{3}{2}$
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