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 - Test - Page 102: 37

Answer

$$(\frac{-6x^{-5}y}{2x^{3}y^{-4}})^{-2} = \frac{x^{16}}{9y^{10}}$$

Work Step by Step

$$(\frac{-6x^{-5}y}{2x^{3}y^{-4}})^{-2}$$ Recall quotient to power rule: $(\frac{a}{b})^n = \frac{a^n}{b^n}$ Thus, $$(\frac{-6x^{-5}y}{2x^{3}y^{-4}})^{-2}$$ $$=\frac{(-6x^{-5}y)^{-2}}{(2x^{3}y^{-4})^{-2}}$$ Recall the power rule: $(a^{m})^{n}=a^{mn}$ Thus, $$=\frac{(-6)^{-2}x^{-5(-2)}y^{-2}}{2^{-2}x^{3(-2)}y^{-4(-2)}}$$ $$=\frac{(-6)^{-2}x^{10}y^{-2}}{2^{-2}x^{-6}y^{8}}$$ Recall the quotient rule: $\frac{a^{m}}{a^{n}}=a^{m-n}$ and $\frac{a^{n}}{a^{m}}=\frac{1}{a^{m-n}}$ Thus, $$=\frac{(-6)^{-2}}{2^{-2}} \cdot\frac{x^{10-(-6)}}{y^{8-(-2)}}$$ $$=\frac{(-6)^{-2}}{2^{-2}} \cdot\frac{x^{16}}{y^{10}}$$ Recall the negative exponent rule: $a^{−n}=\frac{1}{a^{n}}$ and $\frac{1}{a^{-n}} = a^{n}$ Thus, $$=\frac{2^{2}}{(-6)^{2}} \cdot\frac{x^{16}}{y^{10}}$$ $$=\frac{4x^{16}}{36y^{10}}$$ $$=\frac{x^{16}}{9y^{10}}$$
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