Answer
$2\ln{(x)}+\frac{1}{2}\ln{(1-x)}$
Work Step by Step
Recall:
(1) $\sqrt[m]{a}=a^{\frac{1}{m}}$
(2) $\log_a {x^n}=n\cdot \log_a {x}$.
(3) $\log_a{xy}=\log_a{x} +\log_a{y}$
(4) $\log_a{\frac{x}{y}}=\log_a{x} -\log_a{y}$
Use Rule (3) above to obtain:
$\ln{(x^2\cdot \sqrt{1-x})}\\
=\ln {(x^2)}+\ln{\sqrt{1-x}}\\
=\ln {(x^2)}+\ln{(1-x)^{\frac{1}{2}}}$.
Use Rule (2) above to obtain:
$\ln {(x^2)}+\ln{(1-x)^{\frac{1}{2}}}\\
=2\ln{(x)}+\frac{1}{2}\ln{(1-x)}$
