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: 45

Answer

$\color{blue}{\dfrac{5}{x^2}}$

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 (4) above to obtain: $=\dfrac{(5^{-2}x^{-2}) \cdot 5^{-3}(x^3)^{-3}}{(5^{-2})^{3}(x^{-3})^3}$ Use rule (5) above to obtain: $=\dfrac{5^{-2}x^{-2}\cdot 5^{-3}(x^{3(-3)})}{5^{-3(2)}x^{-3(3)}} \\=\dfrac{5^{-2}x^{-2} \cdot 5^{-3}x^{-9}}{5^{-6}x^{-9}}$ Use rule (6) above to obtain: $=\dfrac{5^{-2+(-3)}x^{-2+(-9)}}{5^{-6}x^{-9}} \\=\dfrac{5^{-5}x^{-11}}{5^{-6}x^{-9}}$ Use rule (2) above to obtain: $=5^{-5-(-6)}x^{-11-(-9)} \\=5^{-5+6}x^{-11+9} \\=5^1x^{-2} \\=5x^{-2}$ Use rule (1) above to obtain: $=5 \cdot \dfrac{1}{x^2} \\=\color{blue}{\dfrac{5}{x^2}}$
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