Solutions for $a^b = b^a$?

Question

How to solve the equation $$a^b = b^a$$ if $a ≠ b$. It doesn't have any whole solutions, but are there no solutions at all? And how to prove that?

Any help is appreciated.

Edit: Thanks to Henry & Bernard. It is correct that (2,4) is a solution, but is it the only one? Irracional solutions maybe?

Answer 1

• $a^b = b^a$ for positive real numbers if and only if $\dfrac{\log a}{a}=\dfrac{\log b}{b}$

• $f(x)= \dfrac{\log x}{x}$ is a decreasing function of $x$ when $x \gt e \approx 2.7183$ and an increasing function when $0 \lt x \lt e$ - consider the derivative

• So the only times $a^b = b^a$ can be possible for distinct positive integers is when one of $a$ or $b$ is $1$ or $2$ and the other is above $e$

  • $a=2$ leads to the solution $b=4$, and these can be swapped over
  • but $1^b \lt b^1$ when $b \gt 1$

• so the only solution in distinct positive integers is $(2,4)$ or $(4,2)$

Answer 2

Hints:

If $a,b>0$, take the logarithm of both sides: $$a^b=b^a\iff b\ln a=a\ln b\iff\frac{\ln a}a=\frac{\ln b}b,$$ which implies the function $\varphi(x)=\dfrac{\ln x}x$ is not injective.

Show that $\varphi$ defines an increasing bijection from $(0,\mathrm e]$ onto $(-\infty,1/\mathrm e]$ and a decreasing bijection from $[\mathrm e, +\infty)$ onto $(0,1/\mathrm e]$. Finally, observe there are only two natural numbers between $0$ and $\mathrm e$.