# How do you solve ln(x+6)+ln(x-6)=0?

Jul 8, 2015

I found: $x = \sqrt{37} = 6.082$

#### Explanation:

You can start by using the rule of logs:
$\log a + \log b = \log \left(a \cdot b\right)$

In your case you get:
$\ln \left[\left(x + 6\right) \cdot \left(x - 6\right)\right] = 0$

Now you can use the definition of log as:
${\log}_{a} x = b \to x = {a}^{b}$ remembering that the natural log is: $\ln x = {\log}_{e} x$

so:
$\left(x + 6\right) \left(x - 6\right) = {e}^{0}$
$\left(x + 6\right) \left(x - 6\right) = 1$
rearranging:
${x}^{2} - 36 = 1$
${x}^{2} = 37$
$x = \pm \sqrt{37} = \pm 6.082$
The negative $x$ cannot be accepted (substituted back it gives a negative argument in the original logs).