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: 86

Answer

$\frac{b^{7}}{121}$

Work Step by Step

We are given the expression $\frac{11^{-9}b^{3}}{11^{-7}b^{-4}}$. 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). $(11^{-9-(-7)})\times(b^{3-(-4)})=11^{-2}\times b^{7}$ 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). $\frac{b^{7}}{11^{2}}=\frac{b^{7}}{11\times11}=\frac{b^{7}}{121}$
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