Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 1305071751
ISBN 13: 978-1-30507-175-9

Chapter 1 - Section 1.2 - Exponents and Radicals - 1.2 Exercises - Page 25: 109

Answer

(a)$$(\frac{a}{b})^{-n} = \frac{a^{-n}}{b^{-n}} = \frac{\frac{1}{a^n}}{\frac{1}{b^n}} = \frac{1}{a^n}\div\frac{1}{b^n} = \frac{1}{a^n}\times\frac{b^n}{1} = \frac{b^n}{a^n}$$ (b) $$\frac{a^{-n}}{b^{-m}} = \frac{\frac{1}{a^n}}{\frac{1}{b^m}} = \frac{1}{a^n}\div\frac{1}{b^m} = \frac{1}{a^n}\times\frac{b^m}{1} = \frac{b^m}{a^n}$$

Work Step by Step

(a) According to the law 6: $(\frac{a}{b})^{-n} = \frac{b^n}{a^n}$ To prove this let's try to simplify left side of this equation $$(\frac{a}{b})^{-n} = \frac{a^{-n}}{b^{-n}} = \frac{\frac{1}{a^n}}{\frac{1}{b^n}} = \frac{1}{a^n}\div\frac{1}{b^n} = \frac{1}{a^n}\times\frac{b^n}{1} = \frac{b^n}{a^n}$$ (b) We will follow the same idea this time, simplify left-hand side of the equation and try to make it equal to right-hand side of the equation. According to the law 7, we have: $\frac{a^{-n}}{b^{-m}}=\frac{b^m}{a^n}$ $$\frac{a^{-n}}{b^{-m}} = \frac{\frac{1}{a^n}}{\frac{1}{b^m}} = \frac{1}{a^n}\div\frac{1}{b^m} = \frac{1}{a^n}\times\frac{b^m}{1} = \frac{b^m}{a^n}$$
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