College Algebra (6th Edition)

Published by Pearson
ISBN 10: 0-32178-228-3
ISBN 13: 978-0-32178-228-1

Chapter 4 - Exponential and Logarithmic Functions - Exercise Set 4.4 - Page 490: 17



Work Step by Step

We are given the exponential equation $4^{x}=\frac{1}{\sqrt2}$. We can express each side using a common base and then solve for $x$. $4^{x}=(2^{2})^{x}=2^{2x}$ $\frac{1}{\sqrt2}=\frac{1}{2^{\frac{1}{2}}}=2^{-\frac{1}{2}}$ $2^{2x}=2^{-\frac{1}{2}}$ Take the natural log of both sides. $ln(2^{-\frac{1}{2}})=ln(2^{-\frac{1}{2}})$ $2xln(2)=-\frac{1}{2}ln(2)$ Divide both sides by $ln(2)$. $2x=-\frac{1}{2}$ Divide both sides by 2. $x=-\frac{1}{4}$
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