Intermediate Algebra (6th Edition)

Published by Pearson
ISBN 10: 0321785045
ISBN 13: 978-0-32178-504-6

Chapter 5 - Review - Page 330: 40

Answer

$\dfrac{x^{3}y^{10}}{3z^{12}}$

Work Step by Step

Using laws of exponents, then, \begin{array}{l} \dfrac{3x^{5}}{y^{-4}}\cdot\dfrac{(3xy^{-3})^{-2}}{(z^{-3})^{-4}} \\\\= \dfrac{3x^{5}}{y^{-4}}\cdot\dfrac{3^{-2}x^{-2}y^{-3(-2)}}{z^{-3(-4)}} \\\\= \dfrac{3x^{5}}{y^{-4}}\cdot\dfrac{3^{-2}x^{-2}y^{6}}{z^{12}} \\\\= \dfrac{3x^{5}(3^{-2}x^{-2}y^{6})}{y^{-4}(z^{12})} \\\\= \dfrac{3^{1+(-2)}x^{5+(-2)}y^{6-(-4)}}{z^{12}} \\\\= \dfrac{3^{-1}x^{3}y^{10}}{z^{12}} \\\\= \dfrac{x^{3}y^{10}}{3z^{12}} .\end{array}
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