Intermediate Algebra (6th Edition)

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

Chapter 5 - Section 5.1 - Exponents and Scientific Notation - Exercise Set: 85

Answer

$\frac{a^{5}}{81}$

Work Step by Step

We are given the expression $\frac{9^{-5}a^{4}}{9^{-3}a^{-1}}$. To simplify, we can use the quotient rule, which holds that $\frac{a^{m}}{a^{n}}=a^{m-n}$ (where a is a nonzero real number, and m and n are integers). $(9^{-5-(-3)})\times(a^{4-(-1)})=9^{-2}\times a^{5}$ To simplify this into only positive exponents, we know that $a^{-n}=\frac{1}{a^{n}}$ (where a is a nonzero real number and n is a positive integer). $\frac{a^{5}}{9^{2}}=\frac{a^{5}}{9\times9}=\frac{a^{5}}{81}$
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