Precalculus (10th Edition)

Published by Pearson
ISBN 10: 0-32197-907-9
ISBN 13: 978-0-32197-907-0

Chapter 5 - Exponential and Logarithmic Functions - 5.5 Properties of Logarithms - 5.5 Assess Your Understanding - Page 306: 107


Use the properties the $\log_a {x}-\log_a {y}=\log_a {(\frac{x}{y})}$ and $\log_a{1}=0$ to obtain: $f(\frac{1}{x}) \\=\log_a{(\frac{1}{x})} \\=\log_a{1}-\log_a{x} \\=0-\log_a{x} \\=-\log_a{x} \\=-f(x)$

Work Step by Step

We know that $\log_a {x}-\log_a {y}=\log_a {(\frac{x}{y})}$. Hence, $f(\frac{1}{x})=\log_a{(\frac{1}{x})}=\log_a{1}-\log_a{x}.$ We also know that $\log_a{1}=0$. Thus, $\log_a{1}-\log_a{x}\\ =0-\log_a{x}\\ =-\log_a{x}\\ =-f(x)$
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