Discrete Mathematics with Applications 4th Edition

Published by Cengage Learning
ISBN 10: 0-49539-132-8
ISBN 13: 978-0-49539-132-6

Chapter 7 - Functions - Exercise Set 7.1 - Page 394: 21

Answer

$proof\,of\,:\\ \log_{b}\frac{1}{u}=-\log_{b}u\\ let\,x=\log_{b}\frac{1}{u}\\ \Leftrightarrow b^x=\frac{1}{u}\\ \Leftrightarrow b^x=u^{-1}\\ \Leftrightarrow (b^{x})^{-1}=(u^{-1})^{-1}\\ \Leftrightarrow b^{-x}=u^{1}=u\\ \Leftrightarrow \log_{b}u=-x \\ \Leftrightarrow -\log_{b}u=x \\ \Leftrightarrow \log_{b}\frac{1}{u}=-\log_{b}u\\ $

Work Step by Step

$proof\,of\,:\\ \log_{b}\frac{1}{u}=-\log_{b}u\\ let\,x=\log_{b}\frac{1}{u}\\ \Leftrightarrow b^x=\frac{1}{u}\\ \Leftrightarrow b^x=u^{-1}\\ \Leftrightarrow (b^{x})^{-1}=(u^{-1})^{-1}\\ \Leftrightarrow b^{-x}=u^{1}=u\\ \Leftrightarrow \log_{b}u=-x \\ \Leftrightarrow -\log_{b}u=x \\ \Leftrightarrow \log_{b}\frac{1}{u}=-\log_{b}u\\ $
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