Answer
$\displaystyle \log_{3}\left(\frac{1}{9}\right)=-2$
Work Step by Step
From the definition of logarithms,
for $a>0$, $a\neq 1$, and $x > 0$,
$y=\log_{a}x$ means $a^{y}=x$.
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$ 3^{-2}=\displaystyle \frac{1}{9}\quad$ means $\displaystyle \quad \log_{3}\left(\frac{1}{9}\right)=-2$