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

Apr 15, 2018

see below

#### Explanation:

Using the addition property of logs you know that
this equals log(x+2)(x-1)=1
and using the common log you know that (x+2)(x-1) needs to equal 10 for the equation to be true, (x+2)(x-1)=10 and
this evaluates to (x+4)(x-3)=0, but -4 can't be a solution because that would make one of the original logs undefined.

Apr 15, 2018

 log(x+2)+log(x−1)=1

=> log(x+2)(x−1)=log10 color(white)(x=3 $\left[\text{as } \log a + \log b = \log a b\right]$

=> (x+2)(x−1)=10 color(white)(xwwww3 $\left[\text{taking anti log on both sides}\right]$

=> x^2+2x-x−2=10

=> x^2+x−12=0

=> x^2+4x-3x−12=0

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

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

Therefore, $x = 3$ or $x = - 4$

since, $x$ can't be negative, color(teal)(x=3