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: 41

Answer

$\color{blue}{\dfrac{1}{2pq}}$

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) $(ab)^m = a^mb^m$ (5) $(a^m)^n=a^{mn}$ (6) $a^m \cdot a^n = a^{m+n}$ (7) $a^0=1, a\ne0$ Use rule (6) above to obtain: $=\dfrac{3pq^{1+2}}{6p^2q^4} \\=\dfrac{3pq^3}{6p^2q^4}$ Divide the coefficients by canceling out the common factors to obtain: $\require{cancel} =\dfrac{\cancel{3}pq^3}{\cancel{6}2p^2q^4} \\=\dfrac{pq^3}{2p^2q^4}$ Use rule (2) above to obtain: $=\dfrac{1}{2} \cdot p^{1-2} \cdot q^{3-4} \\=\dfrac{1}{2} \cdot p^{-1} \cdot q^{-1}$ Use rule (1) above to obtain: $=\dfrac{1}{2} \cdot \dfrac{1}{p^1} \cdot \dfrac{1}{q^1} \\=\dfrac{1}{2} \cdot \dfrac{1}{p} \cdot \dfrac{1}{q} \\=\color{blue}{\dfrac{1}{2pq}}$
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