What are the technique to solve this type of exercises with closed form solution?

Question

I have the following statement:

Solve for $y$ in $x^y=y^x$ with $x=\sqrt3$.

When i tried to solve i got $y^y=\sqrt{3}^{\frac{1}{\sqrt3}}$ Clearly I have to transform the exponent in such a way that it remains a variable raised to itself, but is hard to me.

The closed-solution is $y=\sqrt{27}$.

What are the techniques to solve exercises involving $x^x=a^b$ with closed form solution?

Answer 1

No general solution without Lambert W as GEdgar said.

But you can try something like this to help you "guess" the 2nd solution:

let $y=kx$

$x^{kx} = {(kx)}^x$

take xth of both sides

$x^k = kx$

$x^{k-1} = k$

$(\sqrt{3})^{k-1} = k$

This seems a bit more manageable.

We know there are two solutions, because the two curves $y={(\sqrt{3}})^{x-1}$ and $y=x$ intersect in two points.

We already know $k=1$ is a solution. At this point without W lambert, I'd just guess and check for the second solution. With a bit of guessing we see $k=3$ works. That gives $y=\sqrt{3}$ or $y=3\sqrt{3}$. We can only get nice solutions because the problem is set up that way.

Answer 2

All possible rational solutions come from rational $u$ with $$ x = \left( 1 + \frac{1}{u} \right)^u \; , \; \; \; y = \left( 1 + \frac{1}{u} \right)^{1+u} $$

To get $x,y$ rational there are conditions on $u$ beyond being rational. On the other hand, failing the extra conditions still gives a solution to $x^y = y^x,$ now irrational.

The point is that $$ u = \frac{1}{2} $$ gives what you want.