Simple use of log I am struggling to see how we can go from the first expression to the second:
$$\begin{align}
2\log_3 12 - 4\log_3 6 &= \log_3 \left ( \frac{4^2 \cdot 3^2}{2^4 \cdot 3^4} \right )\\
&= \log_3 (3^{-2}) = -2
\end{align}$$

First note that
\[ \log_{b}(x^{n})=n\log_{b}(x) \]
\[ \log_{b}(x)-\log_{b}(y)=\log_{b}\left(\frac{x}{y}\right) \]
Here are the steps
\[
2\log_{3} 12 - 4\log_{3} 6 = 2\log_{3} (4\cdot 3) - 4\log_{3} (2\cdot 3) 
\]
\[
=\log_{3} (4\cdot 3)^{2} - \log_{3} (2\cdot 3)^{4} = \log_{3} (4^{2}\cdot 3^{2}) - \log_{3} (2^{4}\cdot 3^{4})
\]
\[
= \log_{3} \frac{4^{2}\cdot 3^{2}}{2^{4}\cdot 3^{4}} = \log_{3} \frac{16}{16\cdot 3^{2}}=\log_{3} \frac{1}{3^{2}}=\log_{3} 3^{-2}=-2\log_{3} 3 = -2
\]
