# How do you solve Log(x+2)+log(x-1)=4?

May 3, 2018

$x = \setminus \frac{- 1 \setminus \pm \sqrt{9 + 4 {e}^{4}}}{2}$

#### Explanation:

Use the rule that states

$\log \left(a\right) + \log \left(b\right) = \log \left(a b\right)$

to get

$\log \left(\left(x + 2\right) \left(x - 1\right)\right) = 4$

exponential to both sides:

$\left(x + 2\right) \left(x - 1\right) = {e}^{4}$

Expand the parenthesis and bring everything to left side:

${x}^{2} + x - 2 - {e}^{4} = 0$

Now this is a standard quadratic equation $a {x}^{2} + b + c = 0$, with $a = 1$, $b = 1$ and $c = - 2 - {e}^{4}$.

Pull these values into the formula

${x}_{1 , 2} = \setminus \frac{- b \setminus \pm \setminus \sqrt{{b}^{2} - 4 a c}}{2 a}$