Why does $\ln(1)+\ln(2)+\ln(3)=\ln(1+2+3)$? I was in a math class today and we were learning about logarithms. The teacher explained that:
$$\log(x+y) \neq \log(x) + \log(y)$$
And to prove this, decided to solve these two equations for the class:
$$\ln(1) + \ln(2) + \ln(3)$$
and
$$\ln(1+2+3).$$
For some reason, these two ln equations exactly equal each other. I divided one by the other with my calculator and the answer was 1. Is this pure coincidence? Or is there something interesting going on under the hood?
 A: 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:
$$1+2+\dots+n\leq n!\,\,\,\forall n \geq 4$$
A: 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 $\log(6).$
A: 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 $$\ln(2a)=\ln(2)+\ln(a)$$
A: Your teacher wanted to show you that it can be dangerous to generalize based on a few cases. This example is made on purpose, based on the exceptional property $1+2+3=1\cdot2\cdot3$. (He could as well have used any solution to $x+y=xy$, but none are integer.)
