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 - Section 1.6 - Properties of Integral Exponents - Exercise Set: 69

Answer

$\dfrac{y^{8}}{25x^6}$

Work Step by Step

(i) The products-to-powers rule states that: $(ab)^n=a^nb^n$ (ii) The power rule states that: $(a^m)^n=a^{mn}$ (iii) The negative-exponent rule states that: $a^{-m} = \dfrac{1}{a^m}$ and $\dfrac{1}{a^{-m}} = a^m$ Use the products-to-powers rule to find: $=5^{-2}(x^{3})^{-2}(y^{-4})^{-2}$ Use the power rule to find: $=5^{-2}x^{3(-2)}y^{-4(-2)} \\=5^{-2}x^{-6}y^{8}$ Use the negative-exponent rule to find: $=\dfrac{1}{5^2}\cdot \dfrac{1}{x^{6}} \cdot y^{8} \\=\dfrac{1}{25} \cdot \dfrac{1}{x^6} \cdot y^{8} \\=\dfrac{y^{8}}{25x^6}$
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