<s>[INST] Logarithm base transformation I am trying to solve a problem which, I think, revolves around base transformation of logarithms. It goes like this: 
$\log_5\,{\log_6\,{\frac{6x-1}{x+1}}} < \log_\frac{1}{5}\,{\log_\frac{1}{6}\,{\frac{x+1}{6x-1}}}$
I tried transforming "first" logarithms to base 5, yielding 
$\log_6\,{\frac{6x-1}{x+1}}< \frac{1}{\log_\frac{1}{6}\,{\frac{x+1}{6x-1}}}$ (if I am right, of course..) Further transformation to base 6 leaves me helpless with : 
$\log_6\,{\frac{6x-1}{x+1}}<- \frac{1}{\log_6\,{\frac{x+1}{6x-1}}}$
Thanks.
 [/INST] First of all, the existence conditions: the argument of a logarithm must always be $>0$. So we must impose
$$
\log_{6}\frac{6x-1}{x+1}>0\;\;,\;\;\log_{\frac16}\frac{x+1}{6x-1}>0
$$
But observe that $\log_{\frac16}\frac{x+1}{6x-1}=\log_{6}\frac{6x-1}{x+1}$ (from the changing base formula... see later!); hence we can work only on the first.
The same holds for this one: the arguments needs to be $>0$... but we want also this logarithm $>0$ so the conditions on the argument is simply
$$
\frac{6x-1}{x+1}>1\;,\;\;\mbox{i.e.}\;\;x>5/2
$$
Moreover $x$ must be different from $-1$ (the denominator has to be different from zero), but this is included in $x>5/2$.
Let's now go to the computations: you approached to
$$
\log_{6}\frac{6x-1}{x+1}<-\frac{1}{\log_{6}\frac{x+1}{6x-1}}
$$
using the change base formula for logarthims: $\log_ax=\frac{\log_bx}{\log_ba}$. And you're right.
Now note that
$$
-\frac{1}{\log_{6}\frac{x+1}{6x-1}}=\frac{1}{-\log_{6}\frac{x+1}{6x-1}}=\frac{1}{\log_{6}\frac{6x-1}{x+1}}$$
where the last equality follows from the basic properties of logarithms, i.e. $-\log_ax=\log_a{\frac1x}$.
So our inequality is now turned in the following one:
$$
\log_{6}\frac{6x-1}{x+1}<\frac1{\log_{6}\frac{6x-1}{x+1}}
$$
Now simply multiply every side for $\log_{6}\frac{6x-1}{x+1}$, so we have:
$$
\left(\log_{6}\frac{6x-1}{x+1}\right)^2<1
$$
i.e.
$$
-1<\log_{6}\frac{6x-1}{x+1}<1
$$
Then taking the power of $6$ of all sides we came to
$$
\frac16<\frac{6x-1}{x+1}<6
$$
that leads to $x>1/5$. But the existence conditions impose that $x>2/5$.
Hence the solution is $x>2/5$.
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