Answer
$\ln{\left(\dfrac{x^{\frac{1}{3}}y^{\frac{1}{3}}}{z^4}\right)}$
Work Step by Step
Distribute $\frac{1}{3}$ to obtain:
\begin{align*}
\frac{1}{3}\ln{x}+\frac{1}{3}\ln{y}-4\ln{z}
\end{align*}
Recall:
(1) $n\cdot \ln{a}=\ln{a^n}$
(2) $\ln{a}+\ln{b}=\ln{(ab)}$
(3) $\ln{a}-\ln{b} = \ln{\left(\frac{a}{b}\right)}$
Use rule (1) above to obtain:
\begin{align*}
\frac{1}{3}\ln{x}+\frac{1}{3}\ln{y}-4\ln{z}&=\ln{\left(x^{\frac{1}{3}}\right)}+\ln{\left(y^{\frac{1}{3}}\right)}-\ln{\left(z^4\right)}
\end{align*}
Use rule (2) above to obtain:
\begin{align*}
\ln{\left(x^{\frac{1}{3}}\right)}+\ln{\left(y^{\frac{1}{3}}\right)}-\ln{\left(z^4\right)}&=\ln{\left(x^{\frac{1}{3}}y^{\frac{1}{3}}\right)}-\ln{(z^4)}
\end{align*}
Use rule (3) above to obtain:
\begin{align*}
\ln{\left(x^{\frac{1}{3}}y^{\frac{1}{3}}\right)}-\ln{(z^4)}&=\ln{\left(\frac{x^{\frac{1}{3}}y^{\frac{1}{3}}}{z^4}\right)}
\end{align*}