How to plot implicit function $\log_{10}{(xy-1)} = \log_{10}{((1-x)(1-y))}$

Question

I'm inspecting various function and try to plot their graph based on my observations and compare with graphing tools afterwards.

Given an implicit function $\log_{10}{(xy-1)} = \log_{10}{((1-x)(1-y))}$ how do i plot it?

This comes really hard to me since I have strong doubts about the domain of the function. I can't just reduce the $\log$ since it comes with some constraints.

Basically what i think is that the above reduces to:

$$ |xy-1|=|(1-x)(1-y)|, $$ since logarithm argument is greater than $0$ (at least for $x\in \mathbb R$). But now how do i find the explicit form of $y=f(x)$ considering the constraints implied by the logarithm?

upd:

Here is how the graph looks, build based on the answers below:

Intersection of black an purple areas is the domain of the function, and red line is $y=2-x$. Those parts never cross hence no solution in reals is present. Based on that i conclude that the graph is just an empty plane

Answer 1

The given equation implies $xy-1=1-x-y+xy$ i.e. $x+y=2$. For real variables, the maximum obtainable $xy=1-(1-x)^2$ is $1$, i.e. we cannot have $xy>1$. There are therefore no real $x,\,y$ that give both sides of the equation the same finite real value.

Answer 2

Since $\log x$ is injective and defined for $x>0$ we have

$$\log_{10}{(xy-1)} = \log_{10}{((1-x)(1-y))} \implies xy-1 = (1-x)(1-y) \implies x+y-2=0$$

with the conditions

• $xy-1>0$
• $(1-x)(1-y)>0$