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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 $(\log_{x}|y-2|)-1>0$?

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    if $x-\frac{1}{x}>0$, is it really true that $(x+1)(x-1)=x^2-1>0$? What happens if $x<0$?2011-10-23

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