<s>[INST] $\log_{12} 2=m$ what's $\log_6 16$ in function of $m$? Given $\log_{12} 2=m$ what's $\log_6 16$ in function of $m$?
$\log_6 16 = \dfrac{\log_{12} 16}{\log_{12} 6}$
$\dfrac{\log_{12} 2^4}{\log_{12} 6}$
$\dfrac{4\log_{12} 2}{\log_{12} 6}$
$\dfrac{4\log_{12} 2}{\log_{12} 2+\log_{12} 3}$
$\dfrac{4m}{m+\log_{12} 3}$
And this looks like a dead end for me.
 [/INST] Writing everything without logarithms:
$$
\begin{array}{ccc}
12^m=2&\therefore&3=2^{1/m-2}\\
6^x=16&\therefore&3=2^{4/x-1}
\end{array}
$$
Thus we get
$$
\begin{array}{ccc}
1/m-2=4/x-1&\therefore&x=\frac{4m}{1-m}
\end{array}
$$
</s>