Answer
$2 \ln x+\frac{1}{2} \ln (1-x)$
Work Step by Step
Given: $\quad \quad \ln \left(x^{2} \sqrt{1-x}\right)$
Use the rule $\ln A B=\ln A+\ln B$ to obtain:
\begin{align*}
\ln\left(x^{2} \sqrt{1-x}\right)&=\ln x^{2}+\ln \sqrt{1-x}\\
&=\ln x^{2}+\ln (1-x)^{1 / 2}
\end{align*}
Using the rule $\quad\ln a^{m}=m \ln a$ gives:
$$\ln\left(x^{2} \sqrt{1-x}\right)=2 \ln x+\frac{1}{2} \ln (1-x)$$
