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