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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.

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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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  • $\begingroup$ @Irrational: thanks for having accepted my answer! But.. I thought I should earn the +100 reputation points automatically with the "accepted answer"... but maybe I didn't understand.. can you please explain this to me? Thanks a lot! :) $\endgroup$ – Joe May 12 '14 at 5:44
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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.

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  • $\begingroup$ $$\log_{1/x}\,y = (\forall z) \frac{\log_z\,y}{\log_z\,1/x} = \frac{\log_z\,y}{\log_z\,x^{-1}} = \frac{\log_z\,y}{-\log_z\,x} = -\frac{\log_z\,y}{\log_z\,x} = -\log_x\,y$$ $\endgroup$ – DanielV May 7 '14 at 7:57
  • $\begingroup$ Excellent! Thanks! $\endgroup$ – Transcendental May 7 '14 at 8:04
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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 $1$

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