What is the non-trivial solution to $0 = (x-1) \ln x - \ln 2$?

Question

Consider the following logarithmic equation in $x$:

$$ 0 = (x-1) \ln x - \ln 2 $$

It is easy to see the trivial solution $x = 2$. There is a non-trivial solution $x \approx 0.346...$. Is it possible to express the non-trivial solution in closed form?

Answer 1

Transcendence checks show that $x\neq 2$ cannot be rational or algebraic irrational, therefore $x$ is transcendental.

Applying Ritt's theorem on elementary inverses of elementary functions (Ritt 1925) and Schanuel's conjecture yields that the elementary functions $x\to(x-1)\ln(x)-\ln(2)$ over non-discrete domains cannot have elementary partial inverses over non-discrete domains. Therefore the equation cannot be solved by simply rearranging it by applying only finite numbers of elementary functions (elementary operations) readable from the equation. It is not known if all solutions of the equation are elementary numbers (Lin 1983, Chow 1999).

$$(x-1)\ln(x)-\ln(2)=0$$ $x\to e^t$: $$(e^t-1)\ln(e^t)-\ln(2)=0$$ $$(e^t-1)(t+2k\pi i)-\ln(2)=0\ \ \ (\forall k\in\mathbb{Z})$$ $$(t+2k\pi i)e^t-t-\ln(2)-2k\pi i=0\ \ \ (\forall k\in\mathbb{Z})$$ $$(t+2k\pi i)e^t=t+\ln(2)+2k\pi i\ \ \ (\forall k\in\mathbb{Z})$$ $$\frac{t+2k\pi i}{t+\ln(2)+2k\pi i}e^t=1\ \ \ (\forall k\in\mathbb{Z})$$

We see, the equation cannot be solved in terms of Lambert W. But the equation can be solved in terms of Generalized Lambert W:

$$t=W(^{\ \ \ -2k\pi i}_{-\ln(2)-2k\pi i};1)=-W(^{\ln(2)+2k\pi i}_{\ \ \ 2k\pi i};1) $$

[Mezö 2017] Mezö, I.: On the structure of the solution set of a generalized Euler-Lambert equation. J. Math. Anal. Appl. 455 (2017) (1) 538-553

[Mezö/Baricz 2017] Mezö, I.; Baricz, A.: On the generalization of the Lambert W function. Transact. Amer. Math. Soc. 369 (2017) (11) 7917–7934 (On the generalization of the Lambert W function with applications in theoretical physics. 2015)

[Castle 2018] Castle, P.: Taylor series for generalized Lambert W functions. 2018

Answer 2

Not a close form but very good approximations.

By inspection, the solution is close to $\frac 1e$. So $$(x-1)\log(x)=\left(1-\frac{1}{e}\right)-e \left(x-\frac{1}{e}\right)+\sum_{n=2}^\infty (-1)^n\frac{ e^{n-1} (e (n-1)+1)}{(n-1) n} \left(x-\frac{1}{e}\right)^n \tag 1$$ Truncating to some order and using series reversion $$x=\frac{1}{e}-t+\frac{1+e}{2} t^2-\frac{e^2+5e+3}{6} t^3+\frac{e^3+17e^2+35e+15}{24} t^4+O\left(t^5\right)$$ where $t=\frac{e \log \left(\frac{2}{e}\right)+1}{e^2}$.

Using this very truncated series, then the approximate value $x=0.34632354$ to be compared with the "exact" solution given by Newton method $x=0.34632336$ (absolute error $=1.71\times 10^{-7}$).

Since we have the infinite expansion $(1)$, we could continue for ever the inversion process. Done with $10$ terms, the absolute error drops down to $3.99\times 10^{-14}$.