Elementary Algebra

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

Chapter 11 - Additional Topics - Chapter 11 Review Problem Set - Page 520: 34



Work Step by Step

Using the rule $a^{m}\times a^{n}=a^{m+n}$, we obtain: $9a^{\frac{1}{2}}\times 4a^{-\frac{1}{3}}$ $=(9\times4)\times(a^{\frac{1}{2}}\times a^{-\frac{1}{3}})$ $=36(a^{\frac{1}{2}+(-\frac{1}{3})})$ $=36a^{\frac{1(3)-1(2)}{6}}$ $=36a^{\frac{3-2}{6}}$ $=36a^{\frac{1}{6}}$
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