# How do you solve logx + log(x-3) = 1?

Sep 27, 2015

I found $x = 5$

#### Explanation:

You csn use a rule of logs and get:
$\log x \left(x - 3\right) = 1$
If your logs are in base $10$ you can write:
$x \left(x - 3\right) = {10}^{1}$
${x}^{2} - 3 x = 10$
${x}^{2} - 3 x - 10 = 0$
Using the Quadratic Formula:
${x}_{1 , 2} = \frac{3 \pm \sqrt{9 + 40}}{2} = \frac{3 \pm 7}{2}$
So:
${x}_{1} = 5$
${x}_{2} = - 2$ No