# How do you solve ln(x^2-20) = ln5?

Nov 30, 2015

The equation has 2 solutions: $5$ and $- 5$.

#### Explanation:

First we have to find the possible values for $x$ (the domain).
Since $\log$ is only defined for positive arguments we have to solve inequality:

${x}^{2} - 20 > 0$

${x}^{2} > 20$

$\left\mid x \right\mid > \sqrt{20}$

x in (-oo;-2sqrt(5)) uu (2sqrt(5);+oo)

Now we can solve the equation:

Since the base of the logarythm is the same on both sides we can write this equation as:

${x}^{2} - 20 = 5$

${x}^{2} = 25$

$\left\mid x \right\mid = 5$

$x = 5 \vee x = - 5$

Both $5$ and $- 5$ are in the domain, so the equation has 2 solutions.

The equation has 2 solutions: $5$ and $- 5$.