Intermediate Algebra for College Students (7th Edition)

Published by Pearson
ISBN 10: 0-13417-894-7
ISBN 13: 978-0-13417-894-3

Chapter 7 - Section 7.2 - Rational Exponents - Exercise Set - Page 524: 137

Answer

See the explanation below.

Work Step by Step

Since, $p^{\frac{m}{n}}=\sqrt[n] {p^m}$ Here, $n$ signifies the index of the radical $a$ and $m$ signifies the power of the exponent and when we take the reciprocal of the exponent it becomes positive such that :$p^{\frac{-m}{n}}=\frac{1}{p^{\frac{m}{n}}}$ Thus, $9^{\frac{-3}{2}}=\frac{1}{9^{\frac{3}{2}}}$ or, $=\dfrac{1}{3^3}$ or, $=\dfrac{1}{27}$
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