Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Draw the set: $ S=\{(x,y): \log_{x}|y-2|-\log_{|y-2|}x>0\} $

We know that $x>0$ (base of the logarithm). Also, $$\log_{|y-2|}x=\frac{1}{\log_{x}|y-2|},$$ so we have $$\log_{x}|y-2| - \frac{1}{\log_{x}|y-2|}>0$$ and so $$((\log_{x}|y-2|)+1)((\log_{x}|y-2|)-1)>0\;.$$

What should I do next, though? $(\log_{x}|y-2|)+1>0$ or $(\log_{x}|y-2|)-1>0$?

share|cite|improve this question
if $x-\frac{1}{x}>0$, is it really true that $(x+1)(x-1)=x^2-1>0$? What happens if $x<0$? – robjohn Oct 23 '11 at 10:05

Hint: First $$((\log_{x}|y-2|)+1)((\log_{x}|y-2|)-1)(\log_{x}|y-2|)>0 .$$ Consider the regions where each of these factors is positive and negative.

share|cite|improve this answer
Sorry but I understand neither why this inequality should hold nor how it would help. – Did Oct 23 '11 at 10:25
@Didier: Multiply both sides of $\log_{x}|y-2|-\log_{|y-2|}x>0$ by $\left(\log_{x}|y-2|\right)^2$ to get the inequality I gave. Then all one needs to do is to look at the level curves of $\log_{x}|y-2|$ for $\{-1,0,+1\}$. – robjohn Oct 23 '11 at 10:53
Hmmm... clever. Maybe such a hint is more destined to the answerer and their kin than to the asker. – Did Oct 23 '11 at 12:35
@Didier: The asker erroneously got to $((\log_{x}|y-2|)+1)((\log_{x}|y-2|)-1)>0$, and I commented that that ignored the sign of $\log_{x}|y-2|$. I thought that perhaps they could put that together with my hint to get an answer. The question seemed like a homework problem, and, in any case, the asker was asking for help, not an answer, so I didn't want to give too much. – robjohn Oct 23 '11 at 12:48

I’d start by letting $u=y-2$, and asking when $\log_x |u|-\log_{|u|}x > 0$. Now convert the logs to the same base: $\log_a b = \frac{\ln b}{\ln a}$, so your inequality is $$\frac{\ln |u|}{\ln x} - \frac{\ln x}{\ln |u|} > 0\;.$$ Be a little careful in solving this: $\ln x$ and $\ln |u|$ can be negative.

share|cite|improve this answer

You can start by drawing the important contour lines of $f(x,y) = \log_x |y-2| - \log_{|y-2|} x$ in the plane.

There are several lines where $f$ has a discontinuity, namely $x=0$, $y=2$, $y=1$, and $y=3$. Do you see why?

Next, you can solve for the curves where $f(x,y)=0$. Use Didier's hint. You should get four different curves.

Once you have these lines and curves, plot them in the plane. Inside each region $f$ is either positive or negative; you can determine which one by testing a point.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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