## Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

$-2$
$\because y=\log_a x \text{ is equivalent to } x= a^y$ $\therefore y = \log_3 \left(\dfrac{1}{9}\right) \text{ means } 3^y=\dfrac{1}{9}$ Note that: $\dfrac{1}{9}= \dfrac{1}{3^2} = 3^{-2}$ $\therefore 3^y=3^{-2}$ Use the rule $a^m=a^n \implies m=n$ to obtain: $y=-2$ Therefore, $\log_3 \left(\dfrac{1}{9}\right) = \boxed{-2}$