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.
 A: 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.
A: 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 $x>2/5$.
A: Remember that for a real number $b > 0$, $log_b$ is such that $b^{log_b(x)}=x$ for $x > 0$. You usually define $log_b(x)=\frac{log(x)}{log(b)}$, and you can check that writing $b=10^{log(b)}$, $$b^{\frac{log(x)}{log(b)}}=(10^{log(b)})^{\frac{log(x)}{log(b)}}=10^{log(x)}=x$$ and this definition is consistent. Using our definition, $$log_{\frac{1}{b}}(x)=\frac{log(x)}{log(\frac{1}{b})}=-\frac{log(x)}{log(b)}=log_b(\frac{1}{x})$$
Moreover, you can check with this formula that indeed $(\frac{1}{b})^{log_b(\frac{1}{x})}=x$
So, doing this transformation, you get $$log_6(\frac{6x-1}{x+1})=log_{\frac{1}{6}}(\frac{x+1}{6x-1})$$. Using the formula once again : $$log_5(log_6(\frac{6x-1}{x+1}))=log_5(log_{\frac{1}{6}}(\frac{x+1}{6x-1}))=-log_{\frac{1}{5}}(log_{\frac{1}{6}}(\frac{x+1}{6x-1}))$$
So you need to show that $log_5(log_6(\frac{6x-1}{x+1}))$ is negative, that you can do studying the function $x \mapsto log_6(\frac{6x-1}{x+1})$ and show that (on a reasonable interval) it is bounded by $1$
