equation of the transformation of a curve in the plane

Question

The following curve is given: (plot of curve on demos)

$$\frac{1}{4}\log(x)\log(y)=\log(1-x)\log(1-y). $$

A map $f:\Bbb R^2\to\Bbb R^2$ with $f(x,y)=(\log(x),\log(y))$ transforms points on the above curve.

What is the equation for the image of the given curve?

After some algebra, I came up with $$ \frac{1}{4}\log(x)\log(y)=\log(\log(1-e^x))\log(\log(1-e^y)). $$

Is this correct? For some reason the curve won't show up on a plot, so I'm not sure if it's the correct equation.

Answer 1

As @vonbrand wrote, after rearranging, the face the problem of the intersection of $$f(x)=\frac{\log (x)}{4 \log (1 - x)} \qquad \text{and} \qquad g(x)=\frac{\log(1 - x)}{\log (x)}$$ which means $$\log(x)=2\log(1-x) \implies x=(1-x)^2\implies x=\frac{1}{2} \left(3-\sqrt{5}\right)$$

Answer 2

If you rearrange:

$\begin{align*} \frac{\log x}{4 \log (1 - x)} &= \frac{\log(1 - y)}{\log y} \end{align*}$

you see that this defines (at most) a single point, as the left hand side is strictly decreasing while the right hand side is strictly increasing (differentiate!).

Answer 3

The correct equation is $$\frac{1}{4}xy=\log(1-e^x)\log(1-e^y)$$

using the maps $x\mapsto e^x$ and $y\mapsto e^y$ on the original equation:

$$\frac{1}{4}\log(x)\log(y)=\log(1-x)\log(1-y).$$