# How do you solve ln 4 + 2 ln x = 0?

May 13, 2016

I found $x = \frac{1}{2}$

#### Explanation:

We can first operate on the $2$ of the second $\ln$ to write:
$\ln \left(4\right) + \ln {\left(x\right)}^{2} = 0$
we now use the property of the sum to write:
$\ln \left(4 {x}^{2}\right) = 0$
we use the definition of log to write:
$4 {x}^{2} = {e}^{0}$
$4 {x}^{2} = 1$
${x}^{2} = \frac{1}{4}$
$x = \pm \sqrt{\frac{1}{4}} = \pm \frac{1}{2}$
where we use only the positive one.