## College Algebra (10th Edition)

$-2$
Since $9=3^2$, the given expression is equivalent to: $\log_3{(\frac{1}{3^2})}$ Using the rule $\frac{1}{a^m} = a^{-m}, a\ne0$ gives: $\log_3{(\frac{1}{3^2})}=\log_3{(3^{-2})}$ RECALL: $\log_a{(a^n)} = n, a \gt 0, a \ne1$ Using the property above gives: $\log_3{(3^{-2})} = -2$ Thus, $\log_3{(\frac{1}{9})} = -2$.