Answer
$log_{3}\frac{12}{z}$
Work Step by Step
$\div$We know that $log_{b}\frac{x}{y}=log_{b}x-log_{b}y$ (where $x$, $y$, and $b$ are positive real numbers and $b\ne1$).
Therefore, $log_{3}12-log_{3}z=log_{3}\frac{12}{z}$.
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