I was talking with a friend about how to find a more explicit formula for the relationship between the positive numbers $a$ and $b$ in the equation $$ \log_b a = (a/b)^{1/2} $$ after some rearranging we get that this is equivalent to $$ a^{a^{-1/2}}=b^{b^{-1/2}}. $$ The function $f(x)=x^{x^{-1/2}}$ achieves its global maximum of $e^{2/e}$ at $x=e^2$ and is strictly increasing on $(0,e^2)$ and strictly decreasing on $(e^2,\infty)$. Moreover $\lim\limits_{x\to-\infty}f(x)=-\infty,$ $\lim\limits_{x\to-\infty}f(x)=1$ and $f(1)=1$.
Hence for any $a \in (1,e^2)$ there exists exactly one other number $b=b(a) \in (e^2,\infty)$ so that $f(a)=f(b).$
Here is a plot for the number $a$ with $f(a)=2:$
$a$ is the $x$-value of the first intersection of the two lines, $b$ is the $x$-value of the second intersection.
The function $b:(1,e^2)\to(e^2,\infty)$ is differentiable, strictly decreasing and satisfies $\lim\limits_{a\downarrow 1}b(a)=\infty$ and $\lim\limits_{a\uparrow e^2}b(a)=e^2.$ I think it should be possible to find its derivative given what we have but I'm not sure how to do this.
Question: Is it possible to find an explicit expression for the function $b$?
Also, is there a name for this way of defining a function?
Welcome to the world of Lambert function !
The solutions are given by
$$b=\frac{4 a }{\log ^2(a)}W\left(\pm\frac{\log (a)}{2 \sqrt{a}}\right)^2$$ $$a=\frac{4 b}{\log ^2(b)} W\left(\pm\frac{\log (b)}{2 \sqrt{b}}\right)^2$$
The plot consists in a straight line for $0 \leq a \leq e^2$ corresponding to $b=a$ and for $a > e^2$ it is a decreasing function "looking like an hyperbola".