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: 30

Answer

$\frac{1}{16}$

Work Step by Step

We know that $(a)^{-\frac{m}{n}}=\frac{1}{(a)^{\frac{m}{n}}}$ (as long as $(a)^{\frac{m}{n}}$ is a real number). Therefore, $64^{-\frac{2}{3}}=\frac{1}{64^{\frac{2}{3}}}$ We also 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). $\frac{1}{64^{\frac{2}{3}}}=\frac{1}{(\sqrt[3] 64)^{2}}=\frac{1}{4^{2}}=\frac{1}{4\times4}=\frac{1}{16}$ We know that $\sqrt[3] 64=4$, because $4^{3}=64$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.