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: 124



Work Step by Step

$$\frac{(3x^{3}y^{2})^{-1}(2x^{3}y)^{-2}}{(xy^{2})^{-5}(x^{2}y^{3})^{3}}$$ Simplify the numerator: $(3x^{3}y^{2})^{-1}(2x^{3}y)^{-2}$ Recall the power rule: $(a^{m})^{n}=a^{mn}$ Thus, $$(3x^{3}y^{2})^{-1}(2x^{3}y)^{-2} = (3^{-1}x^{3(-1)}y^{2(-1)})(2^{-2}x^{3(-2)}y^{-2}) = (3^{-1}x^{-3}y^{-2})(2^{-2}x^{-6}y^{-2})$$ Recall the product rule: $a^{m}⋅a^{n}=a^{m+n}$ Thus, $$(3^{-1}x^{-3}y^{-2})(2^{-2}x^{-6}y^{-2}) = (3^{-1})(2^{-2})(x^{-3-6})(y^{-2-2}) = (3^{-1})(2^{-2})(x^{-9})(y^{-4})$$ Recall the negative exponent rule: $a^{−n}=\frac{1}{a^{n}}$ and $\frac{1}{a^{-n}} = a^{n}$ Thus, $$(3^{-1})(2^{-2})(x^{-9})(y^{-4}) = \frac{1}{(3^{1})(2^{2})(x^{9})(y^{4})}$$ Simplify the denominator:$(xy^{2})^{-5}(x^{2}y^{3})^{3}$ Using power rule: $(x^{-5}y^{2(-5)})(x^{2(3)}y^{3(3)})=(x^{-5}y^{-10})(x^{6}y^{9})$ Using the product rule: $(x^{-5}y^{-10})(x^{6}y^{9}) = x^{-5+6}y^{-10+9} = x^{1}y^{-1}$ Rewrite the equation using the simplified forms. $$(\frac{1}{(3^{1})(2^{2})(x^{9})(y^{4})})(x^{1}y^{-1}) = \frac{x^{1}y^{-1}}{(3^{1})(2^{2})(x^{9})(y^{4})} = \frac{x^{1}y^{-1}}{12(x^{9})(y^{4})}$$ 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, $$=\frac{x^{1}y^{-1}}{12(x^{9})(y^{4})} = \frac{1}{12(x^{9+1})(y^{4-1})} = \frac{1}{12x^{10}y^{3}}$$
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