$\log_{12} 2=m$ 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.

Here is a zero-cleverness solution: write everything in terms of the natural logarithm $\log$ (or any other logarithm you like). Recall that $\log_ab=\log b/\log a$.
Hence your hypothesis is that $m=\log2/\log12$, or $\log2=m(\log3+2\log2)$, and you look for $k=\log16/\log6=4\log2/(\log2+\log3)$. 
Both $m$ and $k$ are functions of the ratio $r=\log3/\log2$, hence let us try this. One gets $1=m(r+2)$ and one wants $k=4/(1+r)$. Well, $r=m^{-1}-2$ hence $k=4/(m^{-1}-1)=4m/(1-m)$.
An epsilon-cleverness solution is to use at the onset logarithms of base $2$ and to mimick the proof above (the algebraic manipulations become a tad simpler).
