Answer
$\dfrac {2}{3}\log _{3}m+\dfrac {4}{3}\log _{3}4-\dfrac {2}{3}\log _{3}t$
Work Step by Step
$\log _{3}\sqrt [3] {\dfrac {m^{2}n^{4}}{t^{2}}}=\log _{3}\left( m^{2/3}\times n^{4/3}\times t^{-2/3}\right) =\dfrac {2}{3}\log _{3}m+\dfrac {4}{3}\log _{3}4-\dfrac {2}{3}\log _{3}t$
