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

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.

• @TheMadcapLaughs: you should delete that at once ! – Yves Daoust Dec 23 '19 at 20:01
• Uhm... thats not how logs work... – QC_QAOA Dec 23 '19 at 20:01

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

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

• +1, very clever – gt6989b Dec 23 '19 at 20:08
• okay I just came to that conclusion myself, thanks – geocalc33 Dec 23 '19 at 20:08
• just curious what happens when the $1$ on the RHS becomes $2$ or $3$? – geocalc33 Dec 23 '19 at 20:15
• @Aqua: why should I prove that (which is false) ?? – Yves Daoust Dec 23 '19 at 23:05
• 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. – Aqua Dec 24 '19 at 7:34

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: $$y = 1-x$$