Consider $f(x)=\log(x^2)$. Clearly, the domain of $f$ is $D_f=\mathbb{R}-\{0\}$, since $x^2>0$ for any $x<0$. However, by the fundamental properties of logarithms:
$$ f(x)=\log(x^2)=2\log(x)\equiv g(x) $$
where I abuse notation to give separate names $f$ and $g$ to two equivalent (but different in appearance) expressions. Clearly $f(x)\equiv g(x)$ but the domain of $g$ is $G_g=\{x\in\mathbb{R} : x>0\}\neq G_f$. How is this possible? What am I missing in the definition of a logarithm?
Note that this comes from a high-school exercise book on derivatives, so I don't think it should involve the definition of $\log(z)$ for $z\in\mathbb{C}$.
My take is that we define:
$$ \int\frac{1}{x}dx = \ln (|x|) + c $$
so one must consider the argument of the logarithm as positive and/or enclosed in an absolute value operation. However, this doesn't look rigorous to me.
I have not found any similar question on this website, even though it's hard to believe no one asked it before. If I missed any useful threads, I will be happy to be redirected.
Apologies for the abuse of notation and for poor language; I am not a native English speaker and I am relatively new to mathematical concepts. Thanks in advance for your kind answers.
The property $$\log(a^b)=b\log(a)$$ is only valid for $a>0$, precisely because $\log(a)$ may not even be defined.
For this reason, $f(x)=\log(x^2)$ and $g(x)=2\log(x)$ are not the same function, but they do agree on the set $\mathbb{R}^+$ (which is the domain of $g$).