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: 122



Work Step by Step

$$(-4x^{-4}y^{5})^{-2}(-2x^{5}y^{-6})$$ Simplify the first term $(-4x^{-4}y^{5})^{-2}$ Recall the power rule: $(a^{m})^{n}=a^{mn}$ Thus, $$(-4x^{-4}y^{5})^{-2} = -4^{-2}x^{(-4)(-2)}y^{5(-2)}=-4^{-2}x^{8}y^{-10}$$ Rewrite the equation: $$(-4^{-2}x^{8}y^{-10})(-2x^{5}y^{-6})$$ Recall the product rule: $a^{m}⋅a^{n}=a^{m+n}$ Thus, $$(-4^{-2}x^{8}y^{-10})(-2x^{5}y^{-6}) = (-4^{-2})(-2)(x^{8+5}y^{-10-6}) = (-4^{-2})(-2)(x^{13}y^{-16})$$ Recall the negative exponent rule: $a^{−n}=\frac{1}{a^{n}}$ and $\frac{1}{a^{-n}} = a^{n}$ Thus, $$(-4^{-2})(-2)(x^{13}y^{-16}) = \frac{(-2)(x^{13})}{-4^{2}y^{16}}$$ $$= \frac{-2x^{13}}{16y^{16}}$$ $$= \frac{-x^{13}}{8y^{16}}$$ Using the fraction rule: $\frac{-a}{b} = -\frac{a}{b}$ $$\frac{-x^{13}}{8y^{16}} = -\frac{x^{13}}{8y^{16}}$$
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