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

Nov 22, 2016

$x \cong 99.51$

#### Explanation:

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

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

i.e. $\left(x + 2\right) \left(x - 1\right) = {10}^{4} = 10000$

or ${x}^{2} + x - 10002 = 0$

and using quadratic formula $x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

$x = \frac{- 1 \pm \sqrt{1 + 40008}}{2}$

Now, we cannot use $-$ sign as that makes $\log \left(x + 2\right)$ or $\log \left(x - 1\right)$ not possible.

Hence $x = \frac{- 1 + \sqrt{1 + 40008}}{2} \cong \frac{- 1 + 200.02}{2} = \frac{199.02}{2} = 99.51$