Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

Published by Pearson
ISBN 10: 0-32184-874-8
ISBN 13: 978-0-32184-874-1

Chapter 12 - Exponential Functions and Logarithmic Functions - Connecting: The Concepts - Exercises - Page 824: 8


$x=-\displaystyle \frac{2}{5}$

Work Step by Step

We want to use the principle of exponential equality, $b^{m}=b^{n}$ is equivalent to $m=n$, but the bases are not equal. So, we rewrite $9$ as $3^{2}$ and $27$ as $3^{3}$ $(3^{2})^{x+1}=(3^{3})^{-x}\qquad$ ... apply $(a^{m})^{n}=a^{mn}$ $3^{2(x+1)}=3^{-3x}$ ... and now, we apply the principle of exponential equality, $ 2x+2=-3x\qquad$... add $3x-2$ $5x=-2$ $x=-\displaystyle \frac{2}{5}$
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