Solution of $x^x = (x-1)^{x+1}$

Question

Is there any way to solve this equation algebraically and give an exact form of the solution: $x^x = (x-1)^{x+1}$? WolframAlpha only finds the approximate solution 4.14.

Answer 1

Consider the function $$f(x) = \frac{(x-1)^{x+1}}{x^x}, \quad x > 1$$ and its logarithm $$g(x) = \log f(x) = (x+1) \log (x-1) - x \log x, \quad x > 1.$$ The original equality is satisfied whenever $g = 0$. A plot of $g$ is shown as follows: We can compute successive iterates of the Newton's method recursion of $g$ with an initial guess of $x_0 = 4$:

$$x_{n+1} = \frac{2x_n - (x_n - 1) \log (x_n - 1)}{2 + (x_n - 1)\log(1 - 1/x_n)}$$

gives $$\begin{array}{c|c} n & x_n \\ 0 & 4.00000000000000 \\ 1 & 4.13751482760659 \\ 2 & 4.14103935294361 \\ 3 & 4.14104152540996 \\ 4 & 4.14104152541079 \\ 5 & 4.14104152541079 \end{array}$$

When $x < 1$, we run into problems with defining a real-valued function for $(x-1)^{x+1}$, which is why the domain is restricted.