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 :

$\log_6\,{\frac{6x-1}{x+1}}<- \frac{1}{\log_6\,{\frac{x+1}{6x-1}}}$



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 $x>2/5$.

  • $\begingroup$ @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

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.

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

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 $1$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.