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 $1<t^{1\over t-1}<2$ 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.

  • 1
    $\begingroup$ Over what $t$ are you making the claim? Positive integers? $\endgroup$ Feb 10, 2021 at 15:29
  • $\begingroup$ Over all real numbers. $\endgroup$
    – cr001
    Feb 10, 2021 at 15:31
  • $\begingroup$ So $t=0$ is possible? $\endgroup$ Feb 10, 2021 at 15:32
  • 1
    $\begingroup$ 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. $\endgroup$ Feb 10, 2021 at 15:33
  • $\begingroup$ $t=0$ is possible but that gives $x=y=0$ so $x=y$. What I want is a pair generating $x\neq y$. $\endgroup$
    – cr001
    Feb 10, 2021 at 15:33

2 Answers 2


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.

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

  • $\begingroup$ 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? $\endgroup$
    – cr001
    Feb 10, 2021 at 15:37
  • $\begingroup$ @cr001: possibly in terms of Lambert's $W$ function. $\endgroup$
    – user65203
    Feb 10, 2021 at 15:38
  • $\begingroup$ Can you elaborate a little bit how to apply the Lambert $W$ function to generate $q$ in order to guarantee $q$ is rational? $\endgroup$
    – cr001
    Feb 10, 2021 at 15:43
  • $\begingroup$ @cr001: as said, by solving the equation. en.wikipedia.org/wiki/Lambert_W_function#Solving_equations $\endgroup$
    – user65203
    Feb 10, 2021 at 15:45
  • $\begingroup$ 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 $W$. $\endgroup$
    – cr001
    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


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.

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

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