One of my friends posed the following question with a prize of free bubble tea for anyone who could find a closed-form solution to the problem (not necessarily in terms of elementary functions):
What is the maximum value of $c$ such that $(c + \frac{1}{x})^x=x$ has real solutions?
I've been easily able to whittle this problem down to solving the equation $r^r+\text{ln }r+1=0$ (where the solution to the problem is $c=r^r-r$) but from there I don't really know how to proceed – I've already tried a bit of variable shuffling using Lambert's W function, but with no luck. I've also tried to solve similar equations (like $r^r + \text{ln }r=0$) first to see if they could give me further insight to the problem – still no dice. Are there any other tools or approaches that could yield fruit? Or perhaps is there a way to show this problem isn't solvable neatly (for some definition of neatly) at all?
Best Answer
$$\left(c+\frac{1}{x}\right)^x=x \qquad \implies \qquad c=x^{\frac{1}{x}}-\frac{1}{x}$$ $$\frac {dc}{dx}=\frac{x^{\frac{1}{x}} (1-\log (x))+1}{x^2}$$ which is, in slightly different words, what you properly wrote.
So, the problem is to find the zero of function $$f(x)=x^{\frac{1}{x}} (1-\log (x))+1$$ Using inspection, we know that the solution is somewhere between $e^{3/2}$ and $e^2$ since $$f\left(e^{3/2}\right)= 1-\frac{1}{2} e^{\frac{3}{2 e^{3/2}}}=0.301$$ $$ f\left(e^{2}\right)=1-e^{\frac{2}{e^2}}=-0.311$$
Between these two values, the function is close to linearity. So, as a guess, use the midpoint $$x_0=\frac{1}{2} \left(e^{3/2}+e^2\right)$$ and perform one single iteration of any Newton-like method of order $n$ to obtain a nasty but fully explicit expression for $x_1^{(n)}$.
Converting the results to decimals, the results are (respectively for Newton and Halley methods) $$x_1^{(2)}=\color{red}{5.6}6977 \qquad \text{and} \qquad x_1^{(3)}= \color{red}{5.6773}3$$ while the "exact" solution is $x=5.67737$.
The number $$x=5.6773739131926818822733208953370077229026966798039\cdots$$ is not identified by inverse symbolic calculators but it is surprizingly close to the largest root of the quartic equation $$3 x^4+20 x^3+100 x^2-10000=0$$ which can be solved with radicals.
The absolute difference is $3.04\times 10^{-7}$.
Since we know $x$, just compute $c$.
Using the $ISC$ a good approximation is $$c_{\text{max}} \sim e^{-\sqrt{2}} \left(\tan ^{-1}\left(\frac{1}{2}\right)+\Gamma \left(\frac{5}{24}\right)\right)$$ which is an absolute error of $2.37\times 10^{-8}$.
Another approximation is $$c_{\text{max}} \sim 2 \,\Omega -\frac 1{100}\left(\psi\left(\frac{3}{16}\right)-\psi \left(\frac{7}{12}\right)+\psi \left(\frac{13}{14}\right)\right)$$which is an absolute error of $1.74\times 10^{-8}$.
Edit
Pushing the inspection, a very good guess is $$x_0=e^{\sqrt{3}}\qquad \implies \qquad f(x_0)=0.0056$$ which gives $$x_1^{(2)}=\color{red}{5.6773}01 \qquad \text{and} \qquad x_1^{(3)}= \color{red}{5.67737}4$$.