Intermediate Algebra (6th Edition)

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

Chapter 7 - Section 7.2 - Rational Exponents - Exercise Set: 27

Answer

$\frac{64}{27}$

Work Step by Step

We know that $a^{\frac{m}{n}}=(\sqrt[n] a)^{m}$ (where m and n are positive integers greater than 1 with $\frac{m}{n}$ in simplest form and $\sqrt[n] a$ is a real number). Therefore, $(\frac{16}{9})^{\frac{3}{2}}=(\sqrt \frac{16}{9})^{3}=(\frac{4}{3})^{3}=\frac{4}{3}\times\frac{4}{3}\times\frac{4}{3}=\frac{4\times4\times4}{3\times3\times3}=\frac{64}{27}$. We know that $\sqrt \frac{16}{9}=\frac{4}{3}$, because $(\frac{4}{3})^{2}=\frac{16}{9}$
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