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

Hopefully log is base 10. I will assume this.

log(x+2)+log(x-1)=1
log((x+2)(x-1))=1
log(x^2+x-2)=1
10^1 = x^2+x-2
x^2+x-2=10
x^2+x-12=0
This is easily factored to (x-3)(x+4)=0
x=3, -4

But try inserting -4 into the original equation...You'll have trouble.
So we can say the only solution is x=3.

first use log a + log b = log ab
so log (x+2) + log (x-1) = log ((x+2)(x-1))=1
take anti log of both sides
(x +2)(x -1) =10
x^2 +x - 12 =0
(x+4)(x -3) = 0
x = -4 or 3
however log (x +2) is not defined for x = -4
So x =3

log(x + 2) + log(x - 1) = 1

log(x² + x - 2) = 1

10 = x² + x -2

x² + x - 12 = 0

(x + 4)(x - 3) = 0

x = -4(reject) and x = 3

yeah