Is it possible to solve the equation $x - 1 = x^{-y}$ explicitly? I'm trying to solve the equation
$$
x - 1 = x^{-y}
$$
or to find the inverse of the function that is represented by this equation - both explicitly (symbolically).
However, I cannot find a way to do it. I've tried transforming it in multiple ways including:
$$y = \log_x \left( \frac{1}{x-1} \right)$$
$$e^{y \ln(x)} = \frac{1}{x-1}$$
Unfortunately, none of these equations have helped me to find a solution. Some mentioned it may not have an actual explicit inverse function, could this be the case?

You can find $y$ in terms of $x$ by taking (natural) logarithms:
$$
\log(x-1)=-y\log x
$$
and therefore
$$
y=-\frac{\log(x-1)}{\log x}
$$
which is defined for $x>1$. The function is decreasing: indeed
$$
y'=-\frac{\dfrac{\log x}{x-1}-\dfrac{\log(x-1)}{x}}{(\log x)^2}=
\frac{(x-1)\log(x-1)-x\log x}{x(x-1)(\log x)^2}
$$
and we want to study
$$
g(x)=(x-1)\log(x-1)-x\log x
$$
because this determines the sign of $y'$. The limit for $x\to1$ is $0$ and
$$
g'(x)=1+\log(x-1)-1-\log x=\log\frac{x-1}{x}=\log\Bigl(1-\frac{1}{x}\Bigr)
$$
which is negative for every $x>1$. Thus $g$ is decreasing and we have proved that $g(x)<0$ for every $x>1$.
Hence $y'<0$ for every $x>1$ and therefore the function is invertible.
That's as much as you can say, I'm afraid. Functions may be invertible, but their inverse has no expression in terms of “elementary functions”. Think to $f(x)=x+\sin x$, for instance. In some cases one can get away with some “special function” such as Lambert's $W$ or the hypergeometric functions or many others that have been studied.
