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: 71

Answer

$\dfrac{x^{20}}{16y^{16}z^{8}}$

Work Step by Step

RECALL: (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: $=(-2)^{-4}(x^{-5})^{-4}(y^{4})^{-4}(z^2)^{-4}$ Use the power rule to find: $=(-2)^{-4}x^{-5(-4)}y^{4(-4)}z^{2(-4)} \\=(-2)^{-4}x^{20}y^{-16}z^{-8}$ Use the negative-exponent rule to find: $=\dfrac{1}{(-2)^4}\cdot x^{20} \dfrac{1}{y^{16}} \cdot \dfrac{1}{z^{8}} \\=\dfrac{1}{16} \cdot x^{20} \cdot \dfrac{1}{y^{16}} \cdot \dfrac{1}{z^{8}} \\=\dfrac{x^{20}}{16y^{16}z^{8}}$
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