Inverse function of $y=\frac{\ln(x+1)}{\ln x}$ I've been wondering for a while if it's possible to find the inverse function of $y=\frac{\ln(x+1)}{\ln x}$ over the reals. This is the same as finding the positive real root of $x^y-x-1$. I realize that it's impossible with elementary functions, but is it possible with transcendental ones? There's certainly a pattern.
EDIT: I asked this question on MO, and I was given means to arrive at an answer, but I don't really understand it at all. I was told to go to college. Maybe some of you folks have already been, and can help explain this a little bit.
 A: Well, as you know, I can't actually give you a nice solution to put a box around, but I can give you some information about the function $f(x)$ satisfying $f(x)^x+f(x)+1=0$.
There is a pleasant, if not closed-form, asymptotic expansion for $f(x)$ for large $x$ in polynomials of $s=\log 2$:
$$\begin{align}
f(x)=1+\frac sx+&\frac1{2x^2}\left(s+s^2\right)+\frac1{24x^3}\left(6 s+15 s^2+4 s^3\right)+ \\
+&\frac1{48x^4} \left(6 s+27 s^2+18 s^3+2 s^4\right)+\cdots
\end{align}$$
This was obtained by series inversion on the series for $\dfrac{\log(x+1)}{\log x}$ about $x=1$. More specifically, letting $y=x-1$, the series for
$$\begin{align}
\frac1{f^{-1}(x)} & =\frac{\log x}{\log(x+1)}=\frac{\log(y+1)}{\log(y+2)}=\frac{\log(y+1)}{\log(y/2+1)+s} \\
& =\frac{-\sum_1^\infty(-y)^n/n}{s-\sum_1^\infty(-y/2)^n/n}:=\sum_1^\infty a_ny^n
\end{align}$$
can be solved for $a_n$ using the Cauchy product formula $\sum_0^\infty \alpha_ny^n\sum_0^\infty \beta_ny^n=\sum_{n=0}^\infty[\sum_{k=0}^n\alpha_k\beta_{n-k}]y^n$ to get
$$a_n=\frac{(-1)^{n+1}}{sn}+\sum_{k=1}^{n-1}\frac{(-2)^{-k}a_{n-k}}{sk}$$
(which yields coefficients in $\mathbb Q[1/s]$) and if the inverse of this is $f(x)=\sum_0^\infty b_nx^{-n}$, then
$$\begin{align}
y+1 & =f(f^{-1}(y+1))=\sum_{n=0}^\infty b_n\bigg(\sum_{k=1}^\infty a_ky^k\bigg)^n=\sum_{n=0}^\infty b_n\bigg(\sum_{k=1}^\infty a_k(x+1)^k\bigg)^n \\
& =b_0+(a_1b_1)y+(a_2b_1+a_1^2b_2)y^2+(a_3b_1+2a_1a_2b_2+a_1^3b_3)y^3+\cdots.
\end{align}$$
Unfortunately, the last sum has no nice closed form (although it can be done, using Faà di Bruno's formula), and the solution can only be written recursively anyway, so I'll stop here. This equation is supposed to be solved component-by-component to get the terms in the formula, so $1=b_0$, $1=a_1b_1\Rightarrow b_1=s$, $0=a_2b_1+a_1^2b_2\Rightarrow b_2=(s+s^2)/2$, etc. This is the source of the series quoted above.
The series is convergent for $|x|>1$. Here is a plot of the function, in the complex plane:

Red means near $1$, yellow-green is near $i$, blue is near $-i$, white is near $\infty$, black is near $0$. And here is a plot of the above asymptotic expansion, up to the $x^{-12}$ term:

A basic approximation of the function around $x=0$ is given by
$$x=\frac{\log(y+1)}{\log y}=\frac y{\log y}+O(xy)\Rightarrow -\frac1x=-\frac yxe^{-y/x}=y\Rightarrow y=-xW\Big(\!-\frac1x\!\Big),$$
where $W(x)$ is the Lambert W-function. The asymptotics of this function near $\infty$ are quite complex, and there is no proper power series, or even a generalized asymptotic series (involving terms other than $x^k$ for $k\in\mathbb Z$), because the truncated series, with error term $O\left(\big(\frac{\log\log x}{\log x}\big)^m(\log x)^{-k}\right)$, is never $O(1/x)$. The "basic" solution $y=-xW(-1/x)$ is however accurate to $O(x^2)$ (or so, proof or revision to come).
A: Your function $$(0,1)\cup(1,\infty)\to\mathbb{R},\ x\mapsto\frac{\ln(x+1)}{\ln(x)}$$ is injective and has an inverse function therefore.
(Each surjective non-injective function can be decomposed into injective pieces by decomposing the domain of the function. Because each of the injective pieces of your function is bijective, each of them has an inverse. The inverse of a bijective piece of a function is called a partial inverse of the original function.)
$$y=\frac{\ln(x+1)}{\ln(x)}$$
can be rearranged to the equation
$$x^y-x-1=0.$$
For rational $y$, this equation is related to an algebraic equation and we can use the known solution formulas and methods for algebraic equations.
For rational $y\neq 0,1$, the equation is related to a trinomial equation.
For real or complex $y\neq 0,1$, the equation is in a form similar to a trinomial equation. A closed-form solution can be obtained using confluent Fox-Wright Function $\ _1\Psi_1$ therefore, as is written in https://mathoverflow.net/questions/426543/is-it-possible-to-solve-for-y-in-this-equation.
$\ $
Szabó, P. G.: On the roots of the trinomial equation. Centr. Eur. J. Operat. Res. 18 (2010) (1) 97-104
Belkić, D.: All the trinomial roots, their powers and logarithms from the Lambert series, Bell polynomials and Fox–Wright function: illustration for genome multiplicity in survival of irradiated cells. J. Math. Chem. 57 (2019) 59-106
A: I may not be leading you anywhere, or I'm maybe saying something that you have checked already, but I learned that a function needs to be injective and surjective (hence bijective) to be an inversible one.
(Sorry for any grammar/english mistake)
As a start off point, i'm not looking very hard at the function, but i see that if x goes to 1, there will be a gap in there, thus not being injective. If that shows up to be true, then your equation wouldn't be inversible for real numbers.
