Precalculus (6th Edition)

Published by Pearson
ISBN 10: 013421742X
ISBN 13: 978-0-13421-742-0

Chapter R - Review of Basic Concepts - R.6 Rational Exponents - R.6 Exercises - Page 63: 34

Answer

$\color{blue}{\dfrac{3}{5a^3b^5}}$

Work Step by Step

RECALL: (1) $a^{-m} = \dfrac{1}{a^m}$ (2) $\dfrac{a^m}{a^n} = a^{m-n}$ (3) $\left(\dfrac{a}{b}\right)^m=\dfrac{a^m}{b^m}$ (4) $(ab)^m = a^mb^m$ Divide the coefficients together by canceling common factors, then use rule (2) above to simplify the variables to obtain: $\require{cancel}=\dfrac{\cancel{15}3}{\cancel{25}5}a^{-5-(-2)}b^{-1-4} \\=\dfrac{3}{5}a^{-5+2}b^{-5} \\=\dfrac{3}{5}a^{-3}b^{-5}$ Use rule (1) above to obtain: $=\dfrac{3}{5} \cdot \dfrac{1}{a^3} \cdot \dfrac{1}{b^5} \\=\color{blue}{\dfrac{3}{5a^3b^5}}$
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