Answer
(a) $\ln 5$
(b) $ \ln (x-3) $
(c) $ 2\ln t $
Work Step by Step
(a) Since
\begin{align*}
\ln \sin\theta - \ln \frac{\sin \theta}{5}
&= \ln \sin\theta - \ln \sin \theta +\ln 5 \\
&=\ln 5
\end{align*}
(b) Since
\begin{align*}
\ln \left(3 x^{2}-9 x\right)+\ln \left(\frac{1}{3 x}\right)&= \ln \left(3 x^{2}-9 x\right) \left(\frac{1}{3 x}\right)\\
&= \ln (x-3)
\end{align*}
(c) \begin{align*}
\frac{1}{2} \ln \left(4 t^{4}\right)-\ln 2&=\frac{1}{2} \ln \left(2^2 t^{4}\right)-\ln 2\\
&=\frac{1}{2}\left( \ln (2^2 )+\ln (t^{4})\right)-\ln 2 \\
&=\left( \ln (2 )+2\ln t \right)-\ln 2\\
&=2\ln t
\end{align*}
