College Algebra (11th Edition)

Published by Pearson
ISBN 10: 0321671791
ISBN 13: 978-0-32167-179-0

Chapter 4 - Section 4.2 - Exponential Functions - 4.2 Exercises - Page 410: 70



Work Step by Step

We rewrite both sides, so that the base is $2$ on both sides. $32^{2x}=16^{x-1}$ $(2^5)^{2x}=(2^4)^{x-1}$ By the law of exponents: $(a^x)^y=a^{xy}$ Left side: $(2^5)^{2x}=2^{10x}$ Right side: $(2^4)^{x-1}=2^{4x-4}$ The rewritten equation: $2^{10x}=2^{4x-4}$ Then, since both sides have the same base, we set the exponents equal, because the two sides are equal only if the powers are equal too: if $b^m=b^n$ and $b \ne0$ , $b \ne1$ then $x=y$ $10x=4x-4 \\10x-4x=-4$ $6x=-4 \\\frac{6x}{6}=\frac{-4}{6}$ $x=-\frac{2}{3}$
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