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: 25



Work Step by Step

We are given that $32^{2x-3}=2$. Both of these numbers are powers of 2, so we can rewrite the equation as $(2^{5})^{2x-3}=2^{10x-15}=2^{1}$. 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, $10x-15=1$. Add 15 to both sides.. $10x=16$ Divide both sides by 10. $x=\frac{16}{10}=\frac{8}{5}$
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