The set of pairs of positive real numbers \(( x , y )\) with \(x ^ { y } = y ^ { x }\) is a union of two
smooth curves.
Find the point where they intersect.
Solution
Let \[ D\coloneqq\{(x,x):x>0\} \] be the diagonal branch.
1. Every solution with \(x\ne y\) has a unique parameter\[ t\coloneqq\frac yx>0, \qquad t\ne1, \]and is of the form\[ (x,y) = \left( t^{1/(t-1)}, t^{t/(t-1)} \right). \]
For positive \(x,y\), write \(y=tx\). Then \[ x^y=y^x \] becomes \[ x^{tx}=(tx)^x. \] Taking the positive \(x\)th root gives \[ x^t=tx, \] and hence, because \(x>0\), \[ x^{t-1}=t. \] If \(x\ne y\), then \(t\ne1\), so \[ x=t^{1/(t-1)}, \qquad y=tx=t^{t/(t-1)}. \] The parameter is uniquely \(t=y/x\).
2. Conversely, for every \(t>0\) with \(t\ne1\), the point\[ \gamma(t) \coloneqq \left( t^{1/(t-1)}, t^{t/(t-1)} \right) \]satisfies \(x^y=y^x\) and does not lie on \(D\).
For \(\gamma(t)=(x,y)\) one has \(y=tx\) and \(x^{t-1}=t\). Therefore \[ x^t=tx=y, \] so \[ x^y=x^{tx}=(x^t)^x=y^x. \] Since \(y/x=t\ne1\), one has \(x\ne y\), so the point is not on the diagonal.
3. The off-diagonal branch has the limiting point\[ \lim_{t\to1}\gamma(t)=(e,e). \]
The first coordinate satisfies \[ \log\left(t^{1/(t-1)}\right) =\frac{\log t}{t-1} \longrightarrow1 \] as \(t\to1\). Hence its limit is \(e\). For the second coordinate, \[ \log\left(t^{t/(t-1)}\right) =t\frac{\log t}{t-1} \longrightarrow1, \] so its limit is also \(e\).
4. The two branches intersect at exactly\[ \boxed{(e,e)}. \]
By step 2, no point \(\gamma(t)\) with \(t\ne1\) lies on the diagonal. Step 3 shows that the off-diagonal branch extends to \(t=1\) at \((e,e)\), which itself lies on \(D\) and satisfies \(e^e=e^e\). Therefore this is the unique intersection point of the two branches.
5. Q.E.D.
Step 4 gives the requested intersection point.