Intermediate Algebra (12th Edition)

Published by Pearson
ISBN 10: 0321969359
ISBN 13: 978-0-32196-935-4

Chapter 7 - Section 7.2 - Rational Exponents - 7.2 Exercises: 32

Answer

$\frac{1}{81}$

Work Step by Step

We know that $a^{-\frac{m}{n}}=\frac{1}{a^{\frac{m}{n}}}$, where $a^{\frac{m}{n}}$ is a real number. Therefore, $27^{-\frac{4}{3}}=\frac{1}{27^{\frac{4}{3}}}$. We know that $a^{\frac{m}{n}}=\sqrt[n] a^{m}=\sqrt[n] (a)^{m}$, where all indicated roots are real numbers. Therefore, $\frac{1}{27^{\frac{4}{3}}}=\frac{1}{\sqrt[3] 27^{4}}=\frac{1}{(\sqrt[3] 27)^{4}}=\frac{1}{3^{4}}=\frac{1}{81}$. $\sqrt[3] 27=3$, because $3^{3}=27$
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