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 - Page 80: 78

Answer

$-\dfrac{32x^5}{y^5}$

Work Step by Step

The given expression can be written as: $=\left(\dfrac{-2x}{y}\right)^5$ (i) The quotients-to-powers rule states that: $(\frac{a}{b})^n=\frac{a^n}{b^n}$ (ii) The products-to-powers rule states that: $(xy)^n = x^ny^n$ Use the quotients-to-powers rule to find: $=\dfrac{(-2x)^5}{y^5}$ Use the product-to-powers rule to find: $\\=\dfrac{(-2)^5x^5}{y^5} \\=\dfrac{-32x^5}{y^5} \\=-\dfrac{32x^5}{y^5}$
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