what is the inverse function of $f(x) = \frac{2^x}{x}$ for $x > 2$? For $x > 2$, the function $$f(x) = \frac{2^x}{x}$$ is continuous, and monotonically increasing. For any given output $f(x)$, I could always get a close approximation for $x$ with some kind of root-finding algorithm, but I'm wondering if there's a better way. Thanks.
 A: There's no inverse that only uses the standard functions from calculus and precalculus (i.e. logarithms, radicals, or inverse trig functions). But if you allow yourself the use of a certain special function called the Lambert W-function, it can be done.
The Lambert W-function is the inverse of the function $y = x e^x.$ In other words, if $y = W(x)$, then $y e^y = x$.
(Note: There are some technical restrictions on the domain of $W(x)$ that won't come into play for us, since $f(x) = 2^x/x$ is being inverted on a range where its outputs are $> 2$.)
Using this function, it's not hard (after some algebra manipulations) to get an inverse of $y = 2^x/x$. Start by writing the inverse as $x = 2^y/y$, then get $y$ by itself on one side using algebra operations and Lambert W:
\begin{align*}
2^y/y &= x \\
y/2^y &= 1/x \\
y e^{-y \ln(2)} &= 1/x \\
-y \ln(2) e^{-y \ln(2)} &= -\ln(2)/x \\
-y \ln(2) &= W(-\ln(2)/x) \\
y &= \frac{-W(-\ln(2)/x)}{\ln(2)} \
\end{align*}
So the desired inverse function is $$f^{-1}(x) = \frac{-W(-\ln(2)/x)}{\ln(2)},$$ and since $W$ does not have an expression in terms of radicals/logarithms/trig functions, neither does this inverse.
A: As already said in answers
$$y=\frac{2^x}x \implies x=-\frac{1}{\log (2)}W\left(-\frac{\log (2)}{y}\right)$$ which, in the real domain, is defined if
$$-\frac{\log (2)}{y}\geq-\frac 1 e \implies y \geq e \log(2)$$
Depending if $y$ is small or large, you can use the expansions given in the Wikipedia page.
A: Better how? Faster? More accurate? Easier to implement?
Whatever you mean, I think that just calling a standard root-finding function is a good approach. Convergence should be pretty fast, and accuracy can be as good as you want.
If you’re looking for some closed-form formula for the inverse, I doubt that one exists. And even if it did, it’s not clear to me that it would be much better than a root-finding solution. But, again, that depends on your criteria for “better”.
