College Algebra (6th Edition)

Published by Pearson
ISBN 10: 0-32178-228-3
ISBN 13: 978-0-32178-228-1

Chapter P - Prerequisites: Fundamental Concepts of Algebra - Exercise Set P.2: 114

Answer

$\frac{x^{26}y^{6}}{4}$

Work Step by Step

$$\frac{(2^{-1}x^{-3}y^{-1})^{-2}(2x^{-6}y^{4})^{-2}(9x^{3}y^{-3})^{0}}{(2x^{-4}y^{-6})^{2}}$$ When a product is raised to an exponent, raise each factor to that exponent. Remember, any non-zero base raised to the power of zero equals one. $$=\frac{(2^{(-1\times-2)}x^{(-3\times-2)}y^{(-1\times-2)})(2^{-2}x^{(-6\times-2)}y^{(4\times-2)})(1)}{2^{2}x^{(-4\times2)}y^{(-6\times2)}}$$ $$=\frac{(2^{2}x^{6}y^{2})(2^{-2}x^{12}y^{-8})}{2^{2}x^{-8}y^{-12}}$$ Group like terms and simplify. $$=(\frac{2^{2}\times2^{-2}}{2^{2}})(\frac{x^{6}\times x^{12}}{x^{-8}})(\frac{y^{2}\times y^{-8}}{y^{-12}})$$ $$=(\frac{2^{(2+(-2))}}{2^{2}})(\frac{x^{(6+12)}}{x^{-8}})(\frac{y^{(2+(-8))}}{y^{-12}})$$ $$=(\frac{2^{0}}{2^{2}})(\frac{x^{18}}{x^{-8}})(\frac{y^{-6}}{y^{-12}})$$ When dividing exponential expressions with the same non-zero base, subtract the exponent in the denominator from the exponent in the numerator. $$=2^{(0-2)}x^{(18-(-8))}y^{(-6-(-12))}$$ $$=2^{-2}x^{(18+8)}y^{(-6+12)}$$ $$=2^{-2}x^{26}y^{6}$$ When an exponent is negative, write the expression as a fraction and move the base from the numerator to the denominator (or vise-versa) and make the exponent positive. $$=\frac{x^{26}y^{6}}{2^{2}}$$ $$=\frac{x^{26}y^{6}}{4}$$
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