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: 18



Work Step by Step

We are given the exponential equation $9^{x}=\frac{1}{\sqrt[3]3}$. We can express each side using a common base and then solve for $x$. $9^{x}=(3^{2})^{x}=3^{2x}$ $\frac{1}{\sqrt[3]3}=\frac{1}{3^{\frac{1}{3}}}=3^{-\frac{1}{3}}$ $3^{2x}=3^{-\frac{1}{3}}$ Take the natural log of both sides. $ln(3^{2x})=ln(3^{-\frac{1}{3}})$ $2xln(3)=-\frac{1}{3}ln(3)$ Divide both sides by $ln(3)$. $2x=-\frac{1}{3}$ Divide both sides by 2. $x=-\frac{1}{6}$
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