# Hi everyone I would really appreciate some help with finding the domain of this logarithmic function f(x)=ln((x^2)+3x-4) is this something I can use the quadratic equation to solve?

Apr 12, 2018

$\left(- \infty , - 1 - \sqrt{5}\right) \cup \left(\sqrt{5} - 1 , \infty\right)$

#### Explanation:

This is something you can use the quadratic equation to solve, at least partially.

The function $g \left(j\right) = \ln \left(j\right)$ has a domain of $\left(0 , \infty\right)$; it is only defined for positive numbers. Your function is $g \left({x}^{2} + 2 x - 4\right) = f \left(x\right) = \ln \left({x}^{2} + 2 x - 4\right)$. Thus, it is only defined when ${x}^{2} + 2 x - 4 > 0$. (When it is positive.)

We wish to factor ${x}^{2} + 2 x - 4$ and we do this with the help of the quadratic formula.

$x = \frac{- 2 \pm \sqrt{4 - 4 \left(1\right) \left(- 4\right)}}{2}$
$x = - 1 \pm \frac{\sqrt{20}}{2}$
$x = - 1 \pm \frac{2 \sqrt{5}}{2} = - 1 \pm \sqrt{5}$

So our factored form is $\left(x + 1 + \sqrt{5}\right) \left(x + 1 - \sqrt{5}\right)$.
We wish to find when this equation is positive. To be positive, either both terms need to be negative or both terms need to be positive.

If both terms are negative, then $x + 1 + \sqrt{5} < 0 \to x < - 1 - \sqrt{5}$ and $x + 1 - \sqrt{5} < 0 \to x < \sqrt{5} - 1$. Both terms are negative when $x < - 1 - \sqrt{5}$.

Both terms are positive if $x + 1 + \sqrt{5} > 0 \to x > - 1 - \sqrt{5}$ and $x + 1 - \sqrt{5} > 0 \to x > \sqrt{5} - 1$. Both terms are positive when $x > \sqrt{5} - 1$.

Thus, ${x}^{2} + 2 x - 4 > 0$ when $x < - 1 - \sqrt{5}$ and $x > \sqrt{5} - 1$. So our domain is $\left(- \infty , - 1 - \sqrt{5}\right) \cup \left(\sqrt{5} - 1 , \infty\right)$