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 - Review Exercises - Page 100: 103


$$(\frac{3xy^{3}}{5x^{-3}y^{-4}})^{2}=\frac{9}{25}x^{8}y^{14} $$

Work Step by Step

$$(\frac{3xy^{3}}{5x^{-3}y^{-4}})^{2}$$ Recall the quotient rule: $\frac{a^{m}}{a^{n}}=a^{m-n}$ and $\frac{a^{n}}{a^{m}}=\frac{1}{a^{m-n}}$ Thus, $$(\frac{3xy^{3}}{5x^{-3}y^{-4}})^{2}$$ $$(\frac{3}{5}x^{1-(-3)}y^{3-(-4)})^{2}$$ $$(\frac{3}{5}x^{4}y^{7})^{2}$$ Recall product to power rule: $(ab)^n$ =$a^{m}b^{n}$ and power rule: $(a^{m})^{n}=a^{mn}$ Thus, $$(\frac{3}{5}x^{4}y^{7})^{2}$$ $$(\frac{3}{5})^2x^{4(2)}y^{7(2)} $$ $$(\frac{3}{5})^2x^{8}y^{14} $$ $$\frac{9}{25}x^{8}y^{14} $$
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