Answer
$\log \dfrac {1}{x^{2}\sqrt {x}}=-\log x^{2}\sqrt {x}$
Work Step by Step
$\log _{3}\sqrt {x}-\log _{3}x^{3}=\log _{3}\dfrac {\sqrt {x}}{x^{3}}=\log \dfrac {1}{x^{2}\sqrt {x}}=-\log_{3} (\frac{1}{x^{5/2}})$
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