Intermediate Algebra for College Students (7th Edition)

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

Chapter 1 - Section 1.6 - Properties of Integral Exponents - Exercise Set - Page 80: 83

Answer

$a^{8}b^{12}$

Work Step by Step

To solve $(\frac{a^{-2}}{b^{3}})^{-4}$, RECALL: (i) The quotients-to-powers rule states that: $(\frac{a}{b})^n=\frac{a^n}{b^n}$ (ii) The power-rule states that $(a^m)^n=a^{mn}$ (iii) The negative-exponent rule states that: $a^{−m} =\frac{1}{a^m}$ and $\frac{1}{a^{-m}} = a^{m}$ Hence, using quotients-to-powers rule and power rule: $(\frac{a^{-2}}{b^{3}})^{-4}$ = $\frac{(a^{-2})^{-4}}{(b^{3})^{-4}}$ Using the power rule: $=\frac{a^{-2(-4)}}{b^{3(-4)}}$ $=\frac{a^{8}}{b^{-12}}$ Using negative-exponent rule, $=\frac{a^{8}}{b^{-12}}= a^{8}b^{12}$
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