Intermediate Algebra (6th Edition)

Published by Pearson
ISBN 10: 0321785045
ISBN 13: 978-0-32178-504-6

Chapter 7 - Sections 7.1-7.5 - Integrated Review - Radicals and Rational Exponents: 14

Answer

$16x^{\frac{1}{2}}$

Work Step by Step

To simplify, we can use the power rule, which holds that $(a^{m})^{n}=a^{m\times n}$ (where a is a real number, and m and n are integers). $\frac{(2x^{\frac{1}{3}})^{4}}{x^{\frac{5}{6}}}=\frac{2^{4}x^{\frac{1}{3}\times4}}{x^{\frac{5}{6}}}=\frac{16x^{\frac{4}{3}}}{x^{\frac{5}{6}}}$ Next, we can use the quotient rule, which holds that $\frac{a^{m}}{a^{n}}=a^{m-n}$ (where a is a nonzero real number, and m and n are integers). $16(x^{\frac{4}{3}-\frac{5}{6}})=16(x^{\frac{8}{6}-\frac{5}{6}})=16(x^{\frac{3}{6}})=16(x^{\frac{1}{2}})=16x^{\frac{1}{2}}$
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