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So, I have a function $y(x) = \log_{x}{\frac{x + 1}{2}}$, and have a task to find an inverse $x(y)$ of it.

Here is how I approached it. First I swapped $x$ and $y$ for my comfort. $$ x = \log_{y}{\frac{y + 1}{2}} $$

And did some transformations.

$$ y^x = y^{log_{y}{\frac{y + 1}{2}}} $$ $$ y^x = \frac{y + 1}{2} $$ $$ 2y^x = y + 1 $$ $$ 2y^x - y - 1 = 0 $$

And here I got stuck. This equation has two solutions for $y$: one trivial $\forall x\, y(x)=1$, which doesn't fit because it results in $\log_{1}$ in the original equation, and one non-trivial. And I am unable to find the non-trivial one. I have a feeling that $y(x)$, in this case, is probably transcendental and can not be described in algebraic functions, as well as in Lambert-W, but I can't just conclude that it's unsolvable solely based on the fact I can't solve it. I need some justification why it isn't.

So, my questions are

  1. Is it solvable?
  2. If no, why exactly? Is there a formal way to prove it's unsolvable?
  3. If yes, what is the solution?

I really appreciate any help you can provide.

Yuki Endo
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    You should probably state your question as Why does [...] have no inverse?. Or Why is [...] not invertible? I will leave it to the professionals to say whether the second form is OK. – Steven Thomas Hatton Dec 31 '24 at 12:08
  • Your intuition is right, this is not solvable. Even for the case where $x$ is a positive integer, you're considering polynomials of degree $x$, and there is no formula for solving polynomial equations of degree greater than $4$. The cases where $x$ is not an integer or at least rational will be even further out of reach. – jjagmath Dec 31 '24 at 13:00
  • If $0.672438<y<1$, you could do quite nice things – Claude Leibovici Dec 31 '24 at 13:44
  • One can use this general trinomial question – Тyma Gaidash Dec 31 '24 at 13:53
  • A function is defined from a set $A$ to a set $B$. What are $A$ and $B$ for your question? – MasB Dec 31 '24 at 14:22
  • See Wolfram Alpha with "plot log(z,(z+1)/2)" and "domain log(z,(z+1)/2)". Your function has a singularity at $z=1$, and its limit for $x\to\infty$ is $1$. The function is invertible over its real domain because it is bijective there. – IV_ Jan 03 '25 at 11:24
  • But you are asking for solutions in closed form. Your equation $\log_x\left(\frac{x+1}{2}\right)=y$ is a polynomial equation of two algebraically independent monomials. We therefore don't know how to solve the equation by rearranging by applying only finite numbers of elementary functions which we can read from the equation. The inverse function cannot be represented by elementary functions together with Lambert W because it doesn't have the required form. – IV_ Jan 03 '25 at 11:42

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