Elementary Algebra

Published by Cengage Learning
ISBN 10: 1285194055
ISBN 13: 978-1-28519-405-9

Chapter 11 - Additional Topics - 11.3 - Fractional Exponents - Problem Set 11.3 - Page 489: 77



Work Step by Step

First, we will simplify the expression within the parenthesis using the rule $\frac{a^{m}}{a^{n}}=a^{m-n}$: $(\frac{3x^{\frac{1}{3}}}{2x^{\frac{1}{2}}})^{2}$ =$(\frac{3}{2}\times\frac{x^{\frac{1}{3}}}{x^{\frac{1}{2}}})^{2}$ =$(\frac{3}{2}\times x^{\frac{1}{3}-\frac{1}{2}})^{2}$ =$(\frac{3}{2}\times x^{\frac{2-3}{6}})^{2}$ =$(\frac{3}{2}\times x^{\frac{-1}{6}})^{2}$ Now, we raise each term in the expression to the power of $2$: $(\frac{3}{2}\times x^{\frac{-1}{6}})^{2}$ =$(\frac{3}{2})^{2}\times (x^{\frac{-1}{6}})^{2}$ =$(\frac{3^{2}}{2^{2}})\times (x^{\frac{-2}{6}})$ =$(\frac{9}{4})\times (x^{\frac{-1}{3}})$ =$(\frac{9}{4})\times \frac{1}{x^{\frac{1}{3}}}$ =$\frac{9}{4x^{\frac{1}{3}}}$
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