## Intermediate Algebra for College Students (7th Edition)

Published by Pearson

# Chapter 1 - Section 1.6 - Properties of Integral Exponents - Exercise Set - Page 80: 123

#### Answer

$$\frac{1}{128x^{7}y^{16}}$$

#### Work Step by Step

$$\frac{(2x^{2}y^{4})^{-1}(4xy^{3})^{-3}}{(x^{2}y)^{-5}(x^{3}y^{2})^{4}}$$ Simplify the numerator: $(2x^{2}y^{4})^{-1}(4xy^{3})^{-3}$ Recall the power rule: $(a^{m})^{n}=a^{mn}$ Thus, $$(2x^{2}y^{4})^{-1}(4xy^{3})^{-3} = (2^{-1}x^{2(-1)}y^{4(-1)})(4^{-3}x^{-3}y^{3(-3)}) = (2^{-1}x^{-2}y^{-4})(4^{-3}x^{-3}y^{-9})$$ Recall the product rule: $a^{m}⋅a^{n}=a^{m+n}$ Thus, $$(2^{-1})(4^{-3})(x^{-2-3})(y^{-4-9}) = (2^{-1})(4^{-3})(x^{-5})(y^{-13})$$ Recall the negative exponent rule: $a^{−n}=\frac{1}{a^{n}}$ and $\frac{1}{a^{-n}} = a^{n}$ Thus, $$(2^{-1})(4^{-3})(x^{-5})(y^{-13}) = \frac{1}{(2^{1})(4^{3})(x^{5})(y^{13})}$$ Simplify the denominator:$(x^{2}y)^{-5}(x^{3}y^{2})^{4}$ Using power rule: $(x^{2(-5)}y^{-5})(x^{3(4)}y^{2(4)}) = (x^{-10}y^{-5})(x^{12}y^{8})$ Using the product rule: $(x^{-10}y^{-5})(x^{12}y^{8}) = x^{-10+12}y^{-5+8} = x^{2}y^{3}$ Rewrite the equation using the simplified forms. $$(\frac{1}{(2^{1})(4^{3})(x^{5})(y^{13})})(x^{2}y^{3}) = \frac{x^{2}y^{3}}{(2^{1})(4^{3})(x^{5})(y^{13})}$$ $$=\frac{x^{2}y^{3}}{(2)(64)(x^{5})(y^{13})}$$ $$=\frac{x^{2}y^{3}}{128(x^{5})(y^{13})}$$ Recall the quotient rule: $\frac{a^{m}}{a^{n}}=a^{m-n}$ and $\frac{a^{n}}{a^{m}}=\frac{1}{a^{m+n}}$ if $m>n$ Thus, $$=\frac{x^{2}y^{3}}{128(x^{5})(y^{13})} = \frac{1}{128(x^{5+2})(y^{13+3})} = \frac{1}{128x^{7}y^{16}}$$

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