Identical bases and powers!

Consider the function $f(x) = x^x$ . It has derivative $f'(x) = x^x( \log x + 1)$ . If we set it equal to 0 we have $x^x (\log x + 1) = 0 \implies \log x + 1 = 0 \implies x = e^{-1}$ .

Let's compute $f(e^{-1}) = e^{-\frac{1}{e}} \approx 0.69220...$ . There are now many ways to proceed but to keep things simple let's use the hint given in the problem. Notice that $f \left( \frac{1}{2} \right) = \frac{\sqrt{2}}{2} = f \left( \frac{1}{4} \right)$ . Because $f$ is a continuous function this mean that in the interval $(e^{-1},\frac{1}{2}]$ the function takes all values between $e^{-\frac{1}{e}}$ and $\frac{\sqrt{2}}{2}$ but the same is true of the interval $[\frac{1}{4}, e^{-1} )$ . This means that for every $\alpha \in (e^{-1},\frac{1}{2}]$ there exists a $\beta \in [\frac{1}{4}, e^{-1})$ such that $\alpha^{\alpha} = \beta^{\beta}$ and as the sets are disjoint, it is trivial that $\alpha \neq \beta$ .