# Misc Solution: Solve exponential equation $$x^y = y^x$$

Given the equation $$x^y = y^x$$, solve for all $$x\neq y$$ with $$x,y\in\mathbb{R}$$.

## Solution

When we parametrize $$y$$ such that $$y=qx$$, we can say

$\begin{array}{rrl}& x^y &= y^x\\\Leftrightarrow & x^{qx} &= (qx)^x\\\Leftrightarrow & x^{q} &= qx\\\Leftrightarrow & x^{q-1} &= q\\\Leftrightarrow & x &= q^{\frac{1}{{q-1}}}\\\end{array}$

Similarly, we now can set $$x=\frac{y}{q}$$ and set both versions of $$x$$ equal to say

$\begin{array}{rrl}& x &= q^{\frac{1}{{q-1}}}\\\Leftrightarrow & \frac{y}{q} &= q^{\frac{1}{{q-1}}}\\\Leftrightarrow & y &= q^{1+\frac{1}{q-1}}\\\Leftrightarrow & y &= q^{\frac{q}{q-1}}\\\end{array}$

This way we have a generating expression for the pair $$(x, y):= \left(q^{\frac{1}{{q-1}}}, q^{\frac{q}{q-1}}\right)$$ that satisfy the required condition of $$x^y = y^x$$ for all $$q\in\mathbb{R}\backslash\{1\}$$.

## Examples

Let $$q = 1$$, then we can calculate the limits $$x=\lim\limits_{q\to 1}q^{\frac{1}{q-1}} = e$$ and $$y=\lim\limits_{q\to 1}q^{\frac{q}{q-1}}=e$$, which results in the trivial case that was excluded with $$x\neq y$$.

Let $$q = 2$$, then $$x=2^{\frac{1}{2-1}} = 2$$ and $$y=2^{\frac{2}{2-1}} = 4$$. Therefore $$2^4 = 4^2$$.

Let $$q = 3$$, then $$x=3^{\frac{1}{3-1}} = \sqrt{3}$$ and $$y=3^{\frac{3}{3-1}} = \sqrt{3^3} = \sqrt{27}$$. Therefore $$\sqrt{3}^{\sqrt{27}} = \sqrt{27}^{\sqrt{3}}$$.

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