The domain of the function $f(x) = \log_e\frac{x}{1-x}$ is
1. (0, 2) ~ {1}
2. (0, 1)
3. (1, $\infty$)
4. (0, $\infty$)
What I've tried by now..
I thought of since the log function can't be less than $0$ or equal to $0$.
Therefore, $\frac{x}{1-x} > 0$, where $x\neq1$.
Therefore, the domain is $(1, \infty)$.
I want to know the correct answer out of these options and also why that answer, since I don't have solutions.
On
As Michael Rozenberg already said, you've done everything correctly up to a point where you write $$\frac x{1-x}>0,\ x\neq 1.$$
I'm not exactly sure how you got to the conclusion that the answer is $(1,+\infty)$. One of the common mistakes is to multiply the inequality by $1-x$, but that would get you $x>0$, so this is not what you've done.
But here's the thing, why exactly is multiplying by $1-x$ wrong and how can we remedy that?
Well, we know that $1>0$ and if we multiply it by $2$, we will get $2>0$ which is correct, but if multiply it by $-2$, we must change the inequality symbol, so we get $-2<0$. And there lies the problem with $1-x$. Is it positive or negative? Do we need to change the inequality symbol, or not? We don't know.
The trick is to multiply it by $(1-x)^2$ instead which we know is positive (we excluded $x=1$ already) so the inequality becomes $x(1-x) > 0$ which is easy to solve now, because $x(1-x)$ describes parabola with roots $0$ and $1$ and negative leading coefficient, so the solution is interval $(0,1)$ (if you are not convinced, draw the parabola in question).
If you don't like this geometric argument, there is another approach to the problem. We can ask ourselves when is fraction $a/b$ positive? Well, there are two cases, either both $a$ and $b$ are positive, or both $a$ and $b$ are negative. So write down the cases: $x>0$ and $1-x>0$ or $x<0$ and $1-x<0$. The first case will give you $0<x<1$, while the second will give you $1<x<0$, which is impossible.
You answer is not true, but the start is right.
$$\frac{x}{1-x}>0$$ or $$0<x<1$$ by the intervals method.