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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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HINT Simplify $\log_{12} 3 + 2\log_{12} 2$. Now can you express the denominator in terms of $m$? – Srivatsan Sep 21 '11 at 15:46
You mean $\log_{12} 3 +2\log_{12} 2$ where? – Kaeser Sep 21 '11 at 15:52
Ok, let me give the hint in a different way. $$ \log_{12} 3 = \log_{12} \Big( \frac{12}{2 \cdot 2} \Big) = \ldots $$ Can you use the properties of logarithms to write that in terms of $m$? – Srivatsan Sep 21 '11 at 15:58
Yes, problem solved, thanks. – Kaeser Sep 21 '11 at 16:02
I will answer the question, sir. Thanks. – Kaeser Sep 21 '11 at 16:13
up vote 4 down vote accepted

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 might essentially be the epsilon-clever solution mentioned by Didier, but it just looked much simpler to me. – robjohn Sep 21 '11 at 19:17
This is at least square-root-of-epsilon-clever! And possibly more... I like the way you avoid the logs. – Did Sep 21 '11 at 20:17

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).

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