logarithm of a product VS its expanded form let $f(x) = \log{x} + \log{(x - 2)}$
let $g(x) = \log{(x^{2} - 2x)}$
if
$f(x) = g(x)$
then why in the RHS, x can be -1 and in the LHS not ? There's a restriction that i do not know to when its valid to expand the logarithm of a product ?
 A: This is easier to see if you plug in the actual number for both functions.
$$g(-1) = \log((-1)^2-2(-1)) = \log(1+2) = \log(3)$$
$$f(-1) = \log(-1) + \log(-3)$$
Now notice that $-1*-3 = 3$. So $\log(3) = \log(-1*(-3))$. But you can't actually split the product into a sum of two logarithms here, because then you'd be putting invalid numbers into the logarithm (if you want the result to be real).
Basically, $\log(ab) = \log(a) + \log(b)$ iff $a$ and $b$ are both positive if we're talking about real number valued logarithms. Which means that $f(x) = g(x)$ only if $x-2 > 0$, i.e. $x>2$.
However, for the part of $g(x)$ where $x < 0$, you can extend $f(x)$ by defining it as equal to $g(x)$ in that region.
A: In terms of functions, the Domain of $f(x) ≠ g(x)$. Never simplify graphical functions unless the question specifies it. One can relate the two functions, $Range_f \in Range_g$ and $Domain_f \in Domain_g$
You can see this through their graphs:
$f:$

$g:$

Its the same with: $\frac{x^2}{x} ≠ x$ for $x = 0$.
D[$\frac{x^2}{x}$] = $\mathbb R$ \ {$0$}, however D[$x$] = $\mathbb R$.
