Answer
$\frac{1}{18}=x$
Work Step by Step
$\log_a x+\log_a y=\log_a (x\cdot y)$, hence $\log_3x+\log_32=\log_3{(2x)}$.
We know that if $a^x=y$, then $\log_a y=x$ and vice versa.
Hence here $3^{-2}=2x\\\frac{1}{9}=2x\\\frac{1}{18}=x$
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