Find two numbers $a,b$ such that $a^a=b^b$ Playing around on Desmos with the equation $y=x^x$, I noticed that the function, for numbers $0&lt;x&lt;1$, has two $x$ values for every $y$ value. For instance, $\frac12^{\frac12}=\frac14^{\frac14}$, which we can rewrite as $\sqrt[2]\frac12=\sqrt[4]\frac14$.

What I'm trying to find is: Given some number $a$ such that $0&lt;a&lt;1$, find $b$ such that $\sqrt[a]\frac1a=\sqrt[b]\frac1b$.

Is there a general formula for this? I've tried using Newton's Method, but that just approximates the result; it doesn't give an exact answer.

I'm struggling with tagging this question properly, so I'd appreciate some help with that as well.
Let $t=a/b$ be a positive number other than $1$.  Then

$(bt)^{bt}=b^b$

$(bt)^t=b$

$bt=b^{1/t}$

$t=b^{(1-t)/t}$

$\color{blue}{b=t^{t/(1-t)}}$

From this we then get

$\color{blue}{a=bt=t^{1/(1-t)}}$

For instance, if $t=2$ then $b=2^{2/(1-2)}=1/4, a=2^{1/(1-2)}=1/2$.
Let $f(x)=x^x$
$$\ln(f(x))=x\ln x$$
$$\frac{f'(x)}{f(x)}=(1+\ln x)$$
$$f'(x)=f(x)\ln(ex)$$
$$\text{For }0&lt;ex&lt;1,f'(x)\lt0$$
$$\text{For }ex&gt;1,f'(x)\gt 0 $$
$$\text{Hence, if }a\in\left(0,\frac1e\right),b&gt;\frac1e$$