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: 17

Answer

$x = -\dfrac{1}{4}$

Work Step by Step

Note that $\dfrac{1}{\sqrt{2}} = \dfrac{1}{2^{\frac{1}{2}}}=2^{-\frac{1}{2}}$. Thus, the given expression is equivalent to: $4^x=2^{-\frac{1}{2}}$ Write $4$ as $2^2$ to obtain: $(2^2)^x=2^{-\frac{1}{2}}$ Use the rule $(a^m)^n=a^{mn}$ to obtain: $2^{2x} = 2^{-\frac{1}{2}}$ Use the rule $a^m=a^n \longrightarrow m=n$ to obtain: $2x = -\dfrac{1}{2}$ Divide 2 on both sides of the equation to obtain: $x = -\dfrac{1}{4}$
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