Is it possible to solve the equation $x - 1 = x^{-y}$ explicitly?

Question

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?

Answer 1

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.

Answer 2

Your equation has two parameters. Such equations can be a determination equation for a function and its inverse (or its partial inverses).

1)

We have

$$y=-\frac{\ln(x-1)+2k\pi i}{\ln(x)}.$$

$y$ is the function term of a function $f$ with $f(x)=y$. We see: $f$ is an elementary function. The function $f$ is a binary algebraic function in dependence of two non-constant elementary functions. A theorem of Ritt (1925) states that the elementary invertible elementary functions are finite compositions of unary algebraic functions, $\exp$ and/or $\ln$. Your function $f$ cannot be transformed to that form. Therefore your function doesn't have a non-constant partial inverse that is an elementary function. You cannot solve the equation by simply applying only elementary operations (elementary functions) you can read from the equation therefore.
It's an open question if the equation has solutions that are elementary numbers.

2)

$$x-1=x^{-y}$$

$$x-1=e^{-\ln(x)y}$$

LambertW cannot be applied either.

3)

$$x-1=x^{-y}$$

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 $y\neq -1,0$, the equation is related to a trinomial equation.
For real or complex $y\neq -1,0$, 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.

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

Answer 3

If $y$ was supposed to be "large" (that is to say $x$ small (just above $1$) $$y=-\frac{\log (x-1)}{\log (x)}=-\frac{\log (x-1)}{x-1}-\frac{1}{2} \log (x-1)+\frac{1}{12} (x-1) \log (x-1)+O\left((x-1)^2\right)$$ Using the first term only, then $$x_0=1+\frac{W(y)}{y}$$ Using Newton method $$x_{n+1}=\frac{x_n \left(x_n^y+y+1\right)}{x_n^{y+1}+y}$$ should converge quite fast and, by Darboux theorem, there will not be any overshoot of the solution since $f(x_0)\times f''(x_0) >0$.

Trying for $y=\pi$, these iterates will be $$\left( \begin{array}{cc} n & x_n \\ 0 & 1.34176 \\ 1 & 1.37044 \\ 2 & 1.37105 \end{array} \right)$$ Suppose that you use the first iterate of Halley method : it would be $1.37044$ and the first iterate of Householder method would be $1.37105$.

The first iterate of these methods is totally explicit in terms of $x_0$.

Edit

Using @Quanto idea, instead fo Taylor expansion, you could use $[n,n]$ Padé approximant $P_n$ built around $t=0$. This would give for $t (1+t)^y=1$ $$P_2=\frac {t+\frac{1}{3} (y+2)t^2} {1-\frac{2}{3} (y-1)t+\frac{1}{6} y\left(y-1\right)t^2 }=1$$ $$-1+\frac{1}{3} (2 y+1)t-\frac{1}{6} (y-4) (y+1)t^2=0\implies t=\frac{2 y+1-\sqrt{-2 y^2+22 y+25}}{(y-4) (y+1)}$$ For $y=\pi$, this would give $t=0.377185$ which is not bad (we just solved a quadratic equation. Notice that $P_2$ is $O(t^5)$.

Using $P_3$, we should need to solve the cubic $$-1+\frac{3 y-1}{5} t+\frac{-3y^2+17y+18}{20} t^2 +\frac{y^3+17 y+18}{60} t^3 =0$$

The discriminant is negative as soon as $y>1$; so only one real root to be solved by the hyperbolic method. For $y=\pi$, $t=0.371016$ which is almost the solution.

Answer 4

Observe that the root lies within $x\in (1,2) $, since $\lim_{y\to \infty} x(y)=1$ and $\lim_{y\to 0} x(y)=2$. So, let $x=1+t$, with $t\in(0,1)$, and solve instead $$t(1+t)^y=1$$ To approximate, expand $(1+t)^y= 1+y t +\frac12y(y-1)t^2+\cdots$, to get a cubic equation $$ \frac12y(y-1)t^3+y t^2+t-1=0$$ whose solution can be obtained analytically.