## Precalculus (10th Edition)

Since $\log_a {x}-\log_a {y}=\log_a {(\frac{x}{y})}$ and $\log_a{1}=0$, then \begin{align*} \log_a{\left(\frac{1}{N}\right)}&=\log_a{1}-\log_a{N}\\ &=0-\log_a{N}\\&=-\log_a{N}\end{align*}
We know that $\log_a {x}-\log_a {y}=\log_a {(\frac{x}{y})}$. Hence, $\log_a{\left(\frac{1}{N}\right)}=\log_a{1}-\log_a{N}.$ We also know that $\log_a{1}=0$. Thus, $\log_a{1}-\log_a{N}\\ =0-\log_a{N}\\ =-\log_a{N}$