Is a function $f(x)=\ln({x^2-1})$ even and symmetric

Question

We have a function: $$ f(x)=\ln(x^2-1) $$ The function is symmetric because: $D_f=(-\infty,-1) \ \cup\ (1,\infty)$ I understand this as if we would multipy this by $-1$ we would get the same $D_f$

I don't quite understand the following:

EDIT* $$g(x)=\ln(x-1)+\ln(x+1)=\ln((x-1)(x+1))=\ln(x^2-1)$$

Now, this function is not symmetric, because the domain of this function is only $D_g=(1,\infty)$. So because of that its also not an even function.

But why is that so since the way I see it: $g(x)=f(x)$? How can both functions have different domains? Do I understand the theory behind symmetry and even, odd functions correctly?

Symmetry as I written in my notes is: $D \subseteq \mathbb R $ is symmetric if: EDIT** $$ x \in D \implies -x \in D$$

Answer 1

Unless explicitly restricted, the domain of $$g(x)=\ln((x-1)(x+1))$$ is $$D_g=(-\infty,-1) \ \cup\ (1,\infty).$$

$f$ and $g$ are rigorously the same.

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Update:

The question was changed after this answer.

In the new version,

$$f(x)\ne g(x).$$