ln(x+2) + ln(x-1) = ln10
ln(x+2) + ln(x-1) - ln10 = 0
ln((x+2)(x - 1)/10) = 0
e^0 = ((x+2)(x - 1)/10
1 = ((x+2)(x - 1)/10
1*10 = (x+2)(x - 1)
x^2 +2x - x - 2 = 10
x^2 + 2x - x - 2 - 10 = 0
x^2 + x - 12 = 0
( x + 4)(x - 3) = 0
x + 4 = 0 ⇒ x = - 4, it is not a solution because there is not ln of a negative number
x - 3 = 0 ⇒ x = 3 is a solution