# Just how to address the formula $x^y=y^x=k$?

For an offered $k>0$ constant, thinking that $x,y>0$. Additionally this formula has a certain name, or some mathematician related to it?

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2019-05-04 17:39:35
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WLOG intend that $x \ge y$ and also allow $x = y^n$ for some $n \ge 1$. After that (overlooking $k$) the formula comes to be $y^{ny} = y^{y^n}$, or $ny = y^n$, or $y^{n-1} = n$, which offers
$\displaystyle y = n^{ \frac{1}{n-1} }, x = n^{ \frac{n}{n-1} }$.
Currently it just continues to be to locate the values of $n$ such that $x^y = k$, which you need to simply do numerically. (I presume one have to additionally make up the remedies where $x = y$ ; the chart $x^y = y^x$ has 2 parts which converge at $(e, e)$, which one obtains from the above by taking the restriction as $n \to 1$.)