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

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

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

Tags:

Algebra Precalculus

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