Inverse of $\frac{\log(a-x)}{\log(x)}$

Question

I was working with a series, and found that my problem include this function

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$f(x)=\dfrac{\log(a-x)}{\log(x)}$?

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Does anyone know what the inverse of this function is?

Answer 1

To get to a different representation, multiply by $\ln x$:

$$y=\frac{\ln(a-x)}{\ln x}$$ becomes $$y\ln x=\ln(a-x)$$ then apply $\exp$ at both sides $$x^y=\exp(y\ln x)=a-x$$

So you want to determine $x$ in terms of $y$ and $a$. For example, if $y>1$ is an integer, then you want roots of a polynomial of degree $y$. There is no closed form for this, and non-integral $y$ don't improve the situation.

There is question How to solve $x^y =ax-b$ which is basically the same problem$^1$ and some analysis in the answer. And there is no closed or simple solution either, of course.

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$^1$Just use that other question with $a=-1$ and $b=-a$.