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: 43

Answer

$\color{blue}{\dfrac{4}{a^2}}$

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 (5) above to obtain: $=\dfrac{4a^5(a^{-1\cdot3})}{a^{-2\cdot(-2)}} \\=\dfrac{4a^5\cdot a^{-3}}{a^4}$ Use rule (6) above to obtain: $=\dfrac{4a^{5+(-3)}}{a^4} \\=\dfrac{4a^2}{a^4}$ Use rule (2) above to obtain: $=4a^{2-4} \\=4a^{-2}$ Use rule (1) above to obtain: $=4 \cdot \dfrac{1}{a^2} \\=\color{blue}{\dfrac{4}{a^2}}$
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