The idea of this challenge is to investigate the equation(adsbygoogle = window.adsbygoogle || []).push({});

[tex]x^y = y^x[/tex]

Prove the following parts:

- If ##0<x<1## or if ##x=e##, then there is a unique real number ##y## such that ##y^x = x^y##.
- However, if ##x>1## and ##x\neq e##, then there is precisely one number ##g(x)\neq x## such that ##g(x)^x = x^{g(x)}##.
- If ##1<x<e##, then ##g(x)>e##. And if ##x>e##, then ##g(x)<e##.
- The function ##g## defined on ##1<x<e## is differentiable.

Please list all the sources you consulted when solving this question! Using google is considered cheating :tongue:

Points will be given as follows:

1) First person to post a correct solution to one of the above points will receive 3 points (So if you solved 3 and 4, you will receive 6 points).

2) Anybody to post a (original) correct solution to one of the above points will receive 1 point.

3) Anybody posting a (nontrivial) generalization of some of the above questions will receive 2 points

4) The person who posts a solution with the least advanced mathematical machinery will receive 1 extra point for his solution (for example, somebody solving this with basic calculus will have an "easier" solution than somebody using singular homology), if two people use the same mathematical machinery, then we will look at how complicated the proof is.

5) The person with the most elegant solution will receive 1 extra point for his solution (I decide whose solution is most elegant)

Please thank this post if you think this is an interesting challenge (no, I'm not doing this to get more thanks, I just want to see if I should post more things like this and the "thanks" system is the easiest way for this).

Private messages with questions, problem suggestions, etc. are always welcome!

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# Challenge 16: About the equation x^y = y^x

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