Domain of the natural logarithm of a factorisable quadratic polynomial

Question

Suppose I have a function that is the natural logarithm of a quadratic polynomial, which can be factorised: $$ f(x) = \ln(x^2+2x-8)$$ The domain of $f$ is $x \lt -4$ and $x \gt 2$. However, if I factorise $f$ and apply the product rule, we have $$f(x) = \ln[(x-2)(x+4)] = \ln(x-2) + \ln(x+4).$$ The domain of each term is $x \gt 2$ and $x \gt -4$ respectively, giving an overall domain of $x \gt 2$. Here, the subdomain $x < -4$ is lost. Can anyone show me what I am missing or have gone wrong?

Answer 1

The domain of $\log(x^2)$ is $\Bbb R\setminus\{0\}$, but $\log(x^2)=2\log(x)$ and the domain of $\log$ is $(0,\infty)$. What happens here is that is if $\log$ is undefined at a certain point $x$, it may well be defined at $x^2$. Your situation is similar.

Answer 2

The comment and other answer have already pointed out the reason.
(That $\ln(ab) = \ln a + \ln b$ is valid only for $a, b > 0$.)

Here's how you should split the $\ln$:

For $x > 2,$ it is the way you did it because both factors are then positive.

For $x < -4$, you can rewrite it as $$\ln[(x - 2)(x + 4)] = \ln[(-x + 2)(-x - 4)].$$ Now, both the factors are positive in the interval $(-\infty, -4)$ and thus, you can write $$\ln[(x - 2)(x + 4)] = \ln(2 - x) + \ln(-x-4).$$