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: 119


$$\frac{1}{3x^{4}y^{4}} + \frac{8x^3}{y^{6}}$$

Work Step by Step

$$(\frac{3x^4}{y^{-4}})^{-1}+ (\frac{2x}{y^2})^{3}$$ Simplify the first term: $(\frac{3x^4}{y^{-4}})^{-1}$ Recall the power rule: $(a^{m})^{n}=a^{mn}$ Thus, $$(\frac{3x^4}{y^{-4}})^{-1} = \frac{3^{-1}x^{4\cdot-1}}{y^{-4\cdot-1}} = \frac{3^{-1}x^{-4}}{y^{4}}$$ Recall the negative exponent rule: $a^{−n}=\frac{1}{a^{n}}$ and $\frac{1}{a^{-n}} = a^{n}$ Thus, $$\frac{3^{-1}x^{-4}}{y^{4}} = \frac{1}{3x^{4}y^{4}}$$ Simplify the second term: $(\frac{2x}{y^2})^{3}$ Recall the power rule: $(a^{m})^{n}=a^{mn}$ Thus, $$(\frac{2x}{y^2})^{3} = \frac{2^{3}x^3}{y^{2\cdot3}} = \frac{8x^3}{y^{6}}$$ Rewrite the equation: $$\frac{1}{3x^{4}y^{4}} + \frac{8x^3}{y^{6}}$$
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