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: 121


$$ \frac{y^{5}}{8x^{14}}$$

Work Step by Step

$$(-4x^{3}y^{-5})^{-2}(2x^{-8}y^{-5})$$ Simplify the first term $(-4x^{3}y^{-5})^{-2}$ Recall the power rule: $(a^{m})^{n}=a^{mn}$ Thus, $$(-4x^{3}y^{-5})^{-2} = -4^{-2}x^{3(-2)}y^{-5(-2)}=-4^{-2}x^{-6}y^{10}$$ Rewrite the equation: $$(-4^{-2}x^{-6}y^{10})(2x^{-8}y^{-5})$$ Recall the product rule: $a^{m}⋅a^{n}=a^{m+n}$ Thus, $$(-4^{-2}x^{-6}y^{10})(2x^{-8}y^{-5}) = (-4^{-2})(2)(x^{-6-8}y^{10-5}) = (-4^{-2})(2)(x^{-14}y^{5})$$ Recall the negative exponent rule: $a^{−n}=\frac{1}{a^{n}}$ and $\frac{1}{a^{-n}} = a^{n}$ Thus, $$(-4^{-2})(2)(x^{-14}y^{5}) = \frac{(2)(y^{5})}{-4^{2}x^{14}}$$ $$= \frac{(2)(y^{5})}{16x^{14}}$$ $$= \frac{y^{5}}{8x^{14}}$$
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