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What is the graph of $x^{\log y}=y^{\log x}$? This question appears on GRE exam.

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6  
Take the log on both sides. –  Stefan Oct 29 '12 at 18:10
2  
All points in the first quadrant satisfy this. –  user17762 Oct 29 '12 at 18:10
3  
Not exactly, $x,y > 0$, otherwise $\log$ isnt defined. –  Stefan Oct 29 '12 at 18:11
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2 Answers

up vote 5 down vote accepted

The key observation here is that $$ x^{\log y}=(e^{\log x})^{\log y}=e^{\log(x)\log(y)}=(e^{\log y})^{\log x}=y^{\log x} $$ for any $x,y$ such that all of these operations are defined - namely, $\{(x,y)\mid x>0\text{ and }y>0\}$.

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The three comments together answer the question. To summarize, the graph is the open first quadrant $\{(x,y)\in \Bbb R^2:x,y>0\}$, because $x^{\log y}=y^{\log x}$ for all $(x,y)$ in this set. Note that this is true whatever base is considered for the logarithm.

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