How to find the positive solution to $\frac{x^d}{d} = 2^x$?

Question

I asked Wolfram, and it gave me a complicated answer involving the ProductLog function $W$. Is there a simpler, approximate answer?

Answer 1

Probably not possible without the $W$-function.

Here's how to get Wolfram's answer:

Writing $2^x=e^{x\log 2}$ taking the $d$th root, and doing some re-arranging, and you get:

$$xe^{-x\log 2/d} = d^{1/d}$$

Multiplying both sides by $-\frac{\log 2}d$ and letting $y=-\frac{x\log 2}d$ you get:

$$ye^y = -\frac{d^{1/d}\log 2}{d}$$

That means:

$$y=W\left(-\frac{d^{1/d}\log 2}{d}\right)$$

or:

$$x=\frac{-d}{\log 2}W\left(-\frac{d^{1/d}\log 2}{d}\right)$$

Since we are dealing with negative values, $W(u)$ has two real values for $-1/e<u<0$, called $W_0(u)$ and $W_{-1}(u)$. For $u$ small, we have that $W_0(u)=u-u^2+O(u^3)$. so we get:

$$\begin{align}x &= d^{1/d} +\frac{\log 2}{d^{1-2/d}} + O\left(\frac{1}{d^{2}}\right) \end{align}$$

The asymptotic for $W_{-1}(u)$ are more complicated, unfortunately, but they might be useful.