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I'm trying to solve for $y$ in terms of $x$ for the expression below.

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

First I multiplied both sides by $$ \frac{\ln(1-x)}{\ln(x)} $$

to get

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

but I don't see how to isolate $y.$ I tried using every technique I know including logarithm properties.

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    $\begingroup$ @TheMadcapLaughs: you should delete that at once ! $\endgroup$ – Yves Daoust Dec 23 '19 at 20:01
  • $\begingroup$ Uhm... thats not how logs work... $\endgroup$ – QC_QAOA Dec 23 '19 at 20:01
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The function $f(x)=\dfrac{\ln(1-x)}{\ln(x)}$ is monotonic in its domain $(0,1)$, hence it is invertible. So the relation between $x$ and $y$ is a bijection, and…

$$y=1-x.$$


Interestingly, the function is well approximated by $\left(\dfrac1x-1\right)^{-3/2}$, and a solution with $a$ in the RHS is approximately

$$\left(\dfrac1x-1\right)^{-3/2}=a\left(\dfrac1y-1\right)^{3/2},$$ or

$$y=\frac{1-x}{1+(a^{2/3}-1)x}.$$

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    $\begingroup$ +1, very clever $\endgroup$ – gt6989b Dec 23 '19 at 20:08
  • $\begingroup$ okay I just came to that conclusion myself, thanks $\endgroup$ – geocalc33 Dec 23 '19 at 20:08
  • $\begingroup$ just curious what happens when the $1$ on the RHS becomes $2$ or $3$? $\endgroup$ – geocalc33 Dec 23 '19 at 20:15
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    $\begingroup$ @Aqua: why should I prove that (which is false) ?? $\endgroup$ – Yves Daoust Dec 23 '19 at 23:05
  • $\begingroup$ Sorry, I meant this. How do you prove $\lim_{x→1−}f(x)=0$? It has nothing to do with the problem, but I'm interested since I want to draw a graph. $\endgroup$ – Aqua Dec 24 '19 at 7:34
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If it helps: $$\frac{\ln{(y)}}{\ln{(1-y)}} = \frac{\ln{(1-x)}}{\ln{(x)}}$$ $$e^{\frac{\ln{(y)}}{\ln{(1-y)}}} = e^\frac{\ln{(1-x)}}{\ln{(x)}}$$ $$ {{e^{\ln{(y)}}}^{\frac{1}{\ln{(1-y)}}}} = {{e^{\ln{(1-x)}}}^{\frac{1}{\ln{(x)}}}}$$ $$y^{\frac{1}{\ln{(1-y)}}} = (1-x)^{\frac{1}{\ln{(x)}}}$$ $$y^{\ln{(x)}} = (1-x)^{\ln{(1-y)}}$$


From here you can make a substitution: $1-y = t$.


And you get: $$(1-t)^{\ln{x}} = (1-x)^{\ln{(t)}}$$ Since both $\ln{x}$ and $e^x$ are injective it follows: $$x = t$$ And since $t = 1-y$, we conclude: $$y = 1-x $$

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