$x^2 =(-x)^2, \;\forall x \in \mathbb{R}^+$
$$\begin{align*} \therefore \ln(x^2) &=\ln(-x)^2\\ 2\ln(x)&=2\ln(-x)\\ \ln(x)&=\ln(-x) \end{align*}$$
If the statement above is correct, then compare between: $\ln(2)$ and $\ln(-2)$?
My try is, I think the statement is wrong, because $x$ must be positive? any help?
The problem is that the statement $2 \ln a = \ln (a^2)$ only works for $a > 0$.
For insight on why this is, this post explores that.