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$y^x = x^y$

I graphed this function. I have two question about it:

  1. Its graph consists of two curves, one of which is $x=y$ and the other curve looks like a hyperbola(in the first quadrant) and the curves meet at $(e,e)$. Why $(e,e)$?
  2. $y$ value seems to approach $1$ as $x$ approaches $\infty$, how to prove that?
Zehran Bashir
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    Zehran, what did you try in order to solve your exercise ? Read this thread and learn how to ask a good question on this site. Your post is being closed by moderators because it is a very low-quality question. Moreover, you should learn to use MathJax in order to write formulas in an appropriate manner. – Angelo Aug 30 '23 at 10:03
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    This looks like a decent question to me. – TonyK Aug 30 '23 at 10:07
  • https://math.stackexchange.com/questions/785784 – Gary Jan 23 '24 at 06:53

1 Answers1

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Parametric form for the "almost hyperbola" part is $$ x = v^{1/(v-1)} \\ y = v^{v/(v-1)} \tag1$$ with $0<v<+\infty$. (See HERE) Then take $v \to 0$ and $v \to +\infty$ to find the behavior in the two directions.

Also, to explain $(e,e)$, try to find $v$ in $(1)$ where $x=y$.

GEdgar
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