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

Aug 2, 2015

Take exponent of both sides to get: $\left(x + 2\right) \left(x - 1\right) = 10$

Solve the quadratic and discard the spurious solution to get $x = 3$

#### Explanation:

Note that we require $x + 2 > 0$ and $x - 1 > 0$ in order that $\log \left(x + 2\right)$ and $\log \left(x - 1\right)$ be defined.

This boils down to requiring $x > 1$.

Take exponent of both sides to find:

$10 = {10}^{1} = {10}^{\log \left(x + 2\right) + \log \left(x - 1\right)} = {10}^{\log} \left(x + 2\right) \cdot {10}^{\log} \left(x - 1\right)$

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

Subtract $10$ from both ends to get:

$0 = {x}^{2} + x - 12 = \left(x + 4\right) \left(x - 3\right)$

Which gives us $x = - 4$ or $x = 3$

Discard the spurious solution $x = - 4$ since $\log \left(x + 2\right)$ and $\log \left(x - 1\right)$ are undefined for $x = - 4$.

So $x = 3$