# 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 $$1 for any $$t\geq 2$$ and $$-1>t^{1\over t-1}>-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.

• Over what $t$ are you making the claim? Positive integers? Feb 10, 2021 at 15:29
• Over all real numbers. Feb 10, 2021 at 15:31
• So $t=0$ is possible? Feb 10, 2021 at 15:32
• Then that is not true. The function $f(t) = t^{1/(t-1)}$ is continuous on $(1, \infty)$. Note that $f(2) \neq f(4)$ and so, all the rational values (of which there are infinitely many) between $f(2)$ and $f(4)$ are achieved. Feb 10, 2021 at 15:33
• $t=0$ is possible but that gives $x=y=0$ so $x=y$. What I want is a pair generating $x\neq y$. Feb 10, 2021 at 15:33

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.

$$t=-\frac1{\log q}W\left(-\frac{\log q}q\right).$$

• Thank you for the answer. Is it possible to write down a rational solution, or is it just existential without being possible to be written down? Feb 10, 2021 at 15:37
• @cr001: possibly in terms of Lambert's $W$ function.
– user65203
Feb 10, 2021 at 15:38
• Can you elaborate a little bit how to apply the Lambert $W$ function to generate $q$ in order to guarantee $q$ is rational? Feb 10, 2021 at 15:43
• @cr001: as said, by solving the equation. en.wikipedia.org/wiki/Lambert_W_function#Solving_equations
– user65203
Feb 10, 2021 at 15:45
• What I got is ${\ln(t)\over t-1}e^{\ln(t)\over t-1}={q\ln(t)\over t-1}$ so ${\ln(t)\over t-1}=W({q\ln(t)\over t-1})$ but there is $t$ inside the $W$ while what we want is only $q$ inside $W$. Feb 10, 2021 at 15:52

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.

• Downvoters please at least say a valid argument.. Feb 25, 2021 at 19:00