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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