Prove $t^{1\over t-1}$ is rational only when $t=2$ in the context of solving $x^y=y^x$ where $x\neq y$ Prove $t^{1\over t-1}$ is rational only when $t=2$ where $t$ is any real number.
This problem arises from me watching the blackpenredpen video solving $x^y=y^x$ where $x\neq y$. https://www.youtube.com/watch?v=PI1NeGtJo7s, in which he gives the solution $x=t^{1\over t-1}$.
First I considered the integer solutions where $x\neq y$ other than $2^4=4^2$ but realized it is trivial because $1&lt;t^{1\over t-1}&lt;2$ for any $t\geq 2$ and $-1&gt;t^{1\over t-1}&gt;-2$ for any $t\leq-2$.
The next question is then, are there any rational solutions other than the $2,4$ pair. It does not seem trivial to me as inequalities do not help.
My guess is that $t^{1\over t-1}$ cannot be rational other than when $t=2$ but I wan't able to prove.
The range of $f(t)=t^{1/(t-1)}$ is $(1,\infty)$ and the function is continuous and monotonic, hence invertible. So there is an infinite (but countable) number of solutions of
$$t^{1/(t-1)}=q$$ where $q$ is rational.

$$t=-\frac1{\log q}W\left(-\frac{\log q}q\right).$$
We have,
$$t-1=a^{t-1}-1, ~ \text{where} ~ a\in\mathbb {Q^{+}}$$
$$\implies a^x=x+1, x=t-1$$
Substitute $a^x=t \implies x\ln a=\ln t $
$$\implies  t=\dfrac {\ln t}{\ln a}+1$$
Substitute $\ln t=u$
$$\implies e^u=\dfrac{u+\ln a}{\ln a}$$
$$\implies e^{u+\ln a}=\dfrac {a}{\ln a}(u+\ln a)$$
$$\implies (u+\ln a)^{-1} e^{u+\ln a}=\dfrac {a}{\ln a}$$
$$\implies  \left((u+\ln a)^{-1} e^{u+\ln a}\right)^{-1}=\dfrac {\ln a}{a}$$
Substitue $z=u+\ln a$
$$\implies -ze^{-z}=-\dfrac {\ln a}{a}$$
$$\implies W(-ze^{-z})=W\left(-\dfrac {\ln a}{ a}\right)$$
$$\implies z=-W\left(-\dfrac {\ln a}{a}\right)$$
If we take the substitutes backwards
$$t=\dfrac{e^z}{a}$$
We have:
$$\implies t=\dfrac{e^{-W\left(-\frac {\ln a}{a}\right)}}{a}$$
So, $a \in\mathbb {Q^{+}}$, where $$\color {gold}{\boxed {\color{blue}{t=\dfrac{e^{-W\left(-\frac {\ln a}{a}\right)}}{a}}}}$$
Thus, your conjecture is not true.

Here $W(x)$ is a Lambert function