Intermediate Algebra (6th Edition)

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

Chapter 9 - Sections 9.1-9.6 - Integrated Review - Functions and Properties of Logarithms: 22



Work Step by Step

We are given that $4^{x-1}=8^{x+2}$. Both of these numbers are powers of 2, so we can rewrite the equation as $(2^{2})^{x-1}=(2^{3})^{x+2}$. This can be rewritten as $2^{2x-2}=2^{3x+6}$ 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, $2x-2=3x+6$. To solve for x, subtract 6 from both sides. $2x-8=3x$ Next, subtract 2x from both sides. $x=-8$
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