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: 35



Work Step by Step

We are given that $81^{x-1}=27^{2x}$. Both of these numbers are powers of 3, so we can rewrite the equation as $(3^{4})^{x-1}=(3^{3})^{2x}$. Therefore, $3^{4x-4}=3^{6x}$ 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, $4x-4=6x$. Subtract 4x from both sides. $2x=-4$ Divide both sides by 2. $x=-2$
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