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

Feb 29, 2016

Use the log rule ${\log}_{a} m + {\log}_{a} n = {\log}_{a} \left(m \times n\right)$

#### Explanation:

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

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

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

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

Convert to exponential form.

${x}^{2} + x - 2 = {10}^{1}$

${x}^{2} + x - 12 = 0$

$\left(x + 4\right) \left(x - 3\right) = 0$

$x = - 4 \mathmr{and} 3$

Since the log of a negative number is undefined the solution is x = 3.

Hopefully this helps.