How do you find the domain of ln(x^2-9)?

Oct 2, 2016

$x < - 3 , x > 3$, $\left(x \in \mathbb{R}\right)$

Explanation:

$\ln x$ is defined for all $x > 0$

$\therefore \ln \left({x}^{2} - 9\right)$ is defined for $\left({x}^{2} - 9\right) > 0$

${x}^{2} > 9$

$\left\mid x \right\mid > \sqrt{9} \to \left\mid x \right\mid > 3$

Hence the domain of $\ln \left({x}^{2} - 9\right)$ is $x < - 3 , x > 3$, $\left(x \in \mathbb{R}\right)$

We can get a sense of this from the graph of $\ln \left({x}^{2} - 9\right)$ below:

graph{ln (x^2-9) [-12.66, 12.66, -6.32, 6.34]}