Finding the domain of $ln(x-3)^2$ 
Find the domain, in $\mathbb{R}$, of  $ln(x-3)^2$

To solve this I did:


*

*Since this is a logarithm, the input cannot be negative. Therefore, 
$$(x-3)^2 > 0 \Leftrightarrow x - 3 > \sqrt{0} \Leftrightarrow x > 3$$


So the solution should be 
$$\{x \in \mathbb{R} : x > 3\}$$ or
$$]3; +\infty [$$
But my book says the solution is $\mathbb{R}\setminus\{3\}$
What did I do wrong?

Note that solution of this inequation is  $\mathbb{R}\setminus\{3\}$ because  $$(x-3)^{ 2 }>0\quad $$ LHS is always positive except at $x=3$ point
