How to solve this algebraic equation?

Question

$$x^{\cfrac{1}{x}}=(1+x)^{\cfrac{1}{1+x}}$$ Domain: $(0,+\infty)$
I know a numerical solution of $x\approx 2.293$
Does it have any analytical solutions? If not, is it possible to prove that it doesn't have any analytical solutions?

Answer 1

I'd bet money that there is no way to express this solution in terms of elementary functions and constants. I don't know how to prove this.

However, I can prove that the solution is unique, which might be useful to you.

Note that $x$ is a solution to the equation iff $(x+1)\log(x) - x\log(x+1) = 0$. But now notice that $$\frac{\mathrm{d}}{\mathrm{d}x} ((x+1)\log(x) - x\log(x+1)) = \frac{2 x+1}{x^2+x}+\log (x)-\log (x+1)$$ so $$\frac{\mathrm{d}^2}{\mathrm{d}x^2} ((x+1)\log(x) - x\log(x+1)) = -\frac{x^2+x+1}{x^2 (x+1)^2}.$$ When $x > 0$, $-\frac{x^2+x+1}{x^2 (x+1)^2} < 0$, so $\frac{\mathrm{d}}{\mathrm{d}x} ((x+1)\log(x) - x\log(x+1))$ is a decreasing function. But also $$\lim_{x \to \infty} \frac{\mathrm{d}}{\mathrm{d}x} ((x+1)\log(x) - x\log(x+1)) = \lim_{x \to \infty} \left(\frac{2 x+1}{x^2+x}+\log (x)-\log (x+1)\right) = 0,$$ so we conclude that $\frac{\mathrm{d}}{\mathrm{d}x} ((x+1)\log(x) - x\log(x+1))$ is always positive. Thus, $(x+1)\log(x) - x\log(x+1)$ is an increasing function, so it has at most one zero.

Answer 2

If $\frac{log(x)}{x}= \frac{log(1+x)}{1+x}$ then $log(x)+xlog(x)=xlog(1+x)$ so $log(x) + x(log(x)-log(x+1)) = 0$ so $log(x) + xlog(\frac{x}{x+1}) = 0$. so $log(x) + log((\frac{x}{x+1})^x) = 0$ so $log(x(\frac{x}{x+1})^x) = 0$ so $(\frac{x}{x+1})^x = \frac{1}{x}$ so $$x^{x+1}= (x+1)^x$$ aswell. So like maybe another way to state it is that its the fixed point of the mapping $(1+\frac{1}{x})^x$ So like its well known that this limit approaches e, but idk about like fixed points. But hopefully this is helpful somewhat.

You might also notice that this is a contraction mapping around your point. So if you want to approximate it very quickly, recursively apply the function $f(x)= (1+\frac{1}{x})^x$ to itself over and over starting at x=2 (this is a result of banach fixed point thm). So like its equal to $$\lim_{i\to \infty} f^i(2)$$