Answer
$\dfrac{1}{3} \log_3 x-4$
Work Step by Step
Recall the log rules:
(a) $\log_b{\dfrac{p}{q}}=\log_b{p} - \log_b{q}$ (Quotient Rule).
(b) $\log{a^n}=n \log x $
(c) $\log_a{a^x}=x $
Use the rule (a) with $ p=\sqrt[3] x $ and $ q=81$ to obtain
$\log_3 \dfrac{\sqrt[3] x}{81}=\log_3 x^{1/3}-\log_3 81$
Simplify the second term by applying rules (b) and (c) to obtain:
$ \log_3 x^{1/3}-\log_3 3^4=\dfrac{1}{3} \log_3 x-4$
