## Precalculus (10th Edition)

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