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.

You need to use the following two properties of the logarithm: $\log_b{y} = -\log_{1/b}{y}$ and $\log_b{y} = -\log_b{1/y}$.
Using these properties the right hand side becomes:
$\log_{1/5} \log_{1/6} \frac{x+1}{6x-1} = \log_{1/5} \log_{6} \frac{6x-1}{x+1} = -\log_5 \log_6 \frac{6x-1}{x+1}$.
So your inequality now becomes $\log_5 \log_6 \frac{6x-1}{x+1} < 0$ which is quite trivial to solve.
