Is it possible to solve $(x+1)^x = x^{x+1}$ for $x$ in a closed form?

Question

Is it possible to solve for $x$ in a closed form?

$$(x+1)^x = x^{x+1}$$

Inspiration is simply because my kid asked if it is possible to solve. We used Desmos to solve for it graphically. He has tried for a few days to solve algebraically. Wonder if there is any way to get to a closed-form solution?

Answer 1

$$(x+1)^x=x^{x+1}$$

Because this is not so obvious, we first want to show that the terms $(x+1)^x$ and $x^{x+1}$ are algebraically independent.

Let $x$ be a real number: $$(x+1)^\frac{x}{x+1}=x$$

Let $x$ be an algebraic irrational real number $a$: $$(a+1)^\frac{a}{a+1}=a\tag{1}$$ On the left-hand side of equation 1:
Because $a$ is algebraic irrational, $\frac{a}{a+1}$ is algebraic irrational.
Because the base $(a+1)$ is algebraic and not in $\{0,1\}$, and its exponent $\frac{a}{a+1}$ is algebraic irrational, Gelfond–Schneider theorem implies that $(a+1)^\frac{a}{a+1}$ is transcendental.
But the right-hand side of equation 1, $a$, is algebraic. Therefore the terms $(x+1)^x$ and $x^{x+1}$ are algebraically independent.

$$(x+1)^x-x^{x+1}=0$$

We see, your equation can be rearranged to a polynomial equation of more than one algebraically independent monomials ($(x+1)^x,x^{x+1}$) and with no univariate factor. We therefore don't know how to solve the equation by rearranging for $x$ by applying only finite numbers of elementary functions (elementary operations) we can read from the equation.

I don't know if your equation has solutions in the elementary numbers.

Also, your equation seems not to be in a form that can be solved in terms of Lambert W.

Answer 2

Consider that you look for the zero of fuction $$f(x)=x\log(x+1)-(x+1)\log(x)$$ Just by inspection, the root is slightly large than $2$ since $f(2)=\log \left(\frac{9}{8}\right)$.

Perform $\color{red}{\text{one single}}$ iteration of a Newton-like method of order $n$ starting with $x_0=2$. This gives an $\color{red}{\text{explicit}}$ closed form expression.

Playing with the order of the method $$\left( \begin{array}{cc} n & x_{(n)} & \text{method} \\ 2 & 2.275278763 & \text{Newton} \\ 3 & 2.293646382 & \text{Halley} \\ 4 & 2.293128131 & \text{Householder} \\ 5 & 2.293170097 & \text{no name} \\ 6 & 2.293165867 & \text{no name} \\ 7 & 2.293166336 & \text{no name} \\ 8 & 2.293166282 & \text{no name} \\ 9 & 2.293166288 & \text{no name} \\ \end{array} \right)$$