Intermediate Algebra for College Students (7th Edition)

Published by Pearson
ISBN 10: 0-13417-894-7
ISBN 13: 978-0-13417-894-3

Chapter 9 - Section 9.5 - Exponential and Logarithmic Equations - Exercise Set: 7

Answer

$x=2$

Work Step by Step

Note that $64=4^3$. Write $64$ as $4^3$ to obtain: $4^{2x-1}=4^3$ RECALL: If $a^x = a^y$, then $x=y$. Use the rule above to obtain: $4^{2x-1}=4^3 \\2x-1=3 \\2x=3+1 \\2x=4 \\x=\frac{4}{2} \\x=2$
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