Why does $\ln(1)+\ln(2)+\ln(3)=\ln(1+2+3)$?

We know that by the proprieties of $\ln(x)$: $$\ln(n!)=\sum_{i=1}^{n}\ln(i)$$ Let $n=3$, we have that: $3!=1+2+3=6$, so: $$\ln(3!)=\ln(1)+\ln(2)+\ln(3)$$

Note that is a very special case because it is only verified for $n=\{1,2,3\}$. Namely: $$1+2+\dots+n\leq n!\,\,\,\forall n \geq 4$$

The statement that $\log(x+y) \neq \log(x) + \log(y)$ is actually not true. There are actually infinitely many choices of $x$ and $y$ such that $\log(x+y) = \log(x)+\log(y).$ But for most choices we have an inequality. And that is the important: if you get $\log(x+y),$ do not think that it can be rewritten as $\log(x)+\log(y).$

In the same way, $\log(x+y+z)\neq\log(x)+\log(y)+\log(z)$ for most choices of $x,y,z.$ But as you have found, there are some choices for which we have equality, for example for $x=1,\ y=2,\ z=3,$ since both sides evaluate to $\log(6).$

Do you see something there ?

$\begin{array}{l} \ln(1+2+3)&=\ln(1)+\ln(2)+\ln(3)\\ \ln(1+1+2+4)&=\ln(1)+\ln(1)+\ln(2)+\ln(4)\\ \ln(1+1+1+2+5)&=\ln(1)+\ln(1)+\ln(1)+\ln(2)+\ln(5)\\ \ln(1+1+1+1+2+6)&=\ln(1)+\ln(1)+\ln(1)+\ln(1)+\ln(2)+\ln(6)\end{array}$

In fact this exploits the fact that $\ln(1)=0$ to get to $$\ln(2a)=\ln(2)+\ln(a)$$