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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.

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    I will post the answer later because I have not enough reputation to answer my own question for 8 hours2011-09-21

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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} $

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    This is at least square-root-of-epsilon-clever! And possibly more... I like the way you avoid the logs.2011-09-21
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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).