Answer
$2\ln{(2x+3)}-2\ln{(x-2)}-2\ln{(x-1)}$
Work Step by Step
The given expression is equivalent to:
$$\ln{\left(\frac{2x+3}{(x-2)(x-1)}\right)^2}$$
Recall:
(1) $\log_a{(mn)}=\log_a{m} + \log_a{n}$
(2) $\log_a{\left(\frac{m}{n}\right)}=\log_a{m} - \log_a{n}$
(3) $\log_a{a^m}=m\log_a{m}$
Use rule (3) above to obtain:
\begin{align*}
&=2\ln{\left(\frac{2x+3}{(x-2)(x-1)}\right)}
\end{align*}
Use rule (2) above to obtain:
\begin{align*}
&=2\left[\ln{(2x+3)}-\ln{(x-2)(x-1)}\right]\\
\end{align*}
Use rule (1) above to obtain:
\begin{align*}
&=2\left[\ln{(2x+3)}-\left(\ln{(x-2)}+\ln{(x-1)}\right)\right]\\
&=2\left[\ln{(2x+3)}-\ln{(x-2)}-\ln{(x-1)}\right]\\
&=2\ln{(2x+3)}-2\ln{(x-2)}-2\ln{(x-1)}\\
\end{align*}
