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I know this might be an obvious question as we all know that the answers (beside $(x,y)=(1,1)$) are $(x,y)=(2,4)$ but the problem is, how is this exactly solved?

Tags might be inaccurate so feel free to edit my question at will.

By the way so far I tried this:

$x^y = y^x \rightarrow y\log x=x\log y $

Later on,I tried to draw their graphs and see where they intercept,but well ... no use.

A hint would be highly appreciated.

Ivo Terek
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Minuano
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    Prime factorization. Note that there's no addition in sight. – Qiaochu Yuan Jul 18 '14 at 00:29
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    Any ordered pair of the form $(r,r)$ for $r \in \mathbb{R}$ is a solution. – mweiss Jul 18 '14 at 00:29
  • @QiaochuYuan Sorry if I'm asking such a question,but what is a prime factorization?Actually I don't study math books in English and that's why I can't guess the meanings :D – Minuano Jul 18 '14 at 00:33
  • @FuriousMathematician please google it. – djechlin Jul 18 '14 at 00:37
  • @mweiss Obviously that's a trivial solution, but are there others that you can derive? – Vishwa Iyer Jul 18 '14 at 00:37
  • @IvoTerek Thanks for the edit.But I want an answer that somehow relates to the Number Theory the link you've included does not relate to such a thing.Actually I'm not good at limits and function so I'd really prefer an easier answer if possible – Minuano Jul 18 '14 at 00:42
  • @FuriousMathematician: Why ask for an Answer that "somehow relates to the Number Theory" if prime factors are such a mystery that you "can't guess the meanings"? I see the smiley emoticon, but it seems you having a laugh at our expense. – hardmath Jul 18 '14 at 00:49
  • But the behaviour of $\frac{\ln t}{t}$ is basic calculus, kid stuff. – André Nicolas Jul 18 '14 at 00:50
  • Instead of plotting $x^y$ and $y^x$, why not plot $x^{1/x}$ and, oh, $y^{1/y}$? – robjohn Jul 18 '14 at 01:48

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