Algebra: A Combined Approach (4th Edition)

Published by Pearson
ISBN 10: 0321726391
ISBN 13: 978-0-32172-639-1

Chapter 5 - Section 5.2 - Negative Exponents and Scientific Notation - Exercise Set: 17

Answer

$\frac{4}{9}$

Work Step by Step

We know that $a^{−n}=\frac{1}{a^{n}}$ and $\frac{1}{a^{-n}}=a^{n}$ (as long as a is a nonzero real number and n is an integer). Therefore, $3^{-2}+3^{-1}=\frac{1}{3^{2}}+\frac{1}{3}=\frac{1}{9}+\frac{1}{3}=\frac{1}{9}+\frac{1\times3}{3\times3}=\frac{1}{9}+\frac{3}{9}=\frac{1+3}{9}=\frac{4}{9}$.
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