Intermediate Algebra (6th Edition)

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

Chapter 5 - Section 5.2 - More Work with Exponents and Scientific Notation - Exercise Set - Page 268: 1

Answer

$\frac{1}{9}$

Work Step by Step

We are given the expression $(3^{-1})^{2}$. 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). $(3^{-1})^{2}=3^{-1\times2}=3^{-2}$ To simplify this into a positive exponent, we know that $a^{-n}=\frac{1}{a^{n}}$ (where a is a nonzero real number and n is a positive integer). Therefore, $3^{-2}=\frac{1}{3^{2}}=\frac{1}{3\times3}=\frac{1}{9}$
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