Answer
$-2$
Work Step by Step
RECALL:
$\log_b{b^x} = x.$
Note that
$\dfrac{1}{9}=\dfrac{1}{3^3} = 3^{-2}.$
Thus, the given expression can be written as
$\log_3{3^{-2}}.$
Use the rule above to obtain
$-2.$
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