log12(x+1) + log12x = 1
log12[x(x+1)] = 1
x(x+1) = 12^1
x^2 +x = 12
x^2 +x - 12=0
- 12 = 4( -3)
4-3 = 1
(x+4)(x -3) = 0
x = -4 and x = 3
x = -4 is not an answer, there is not log of a negative number
log12(x^2+x) = 1
log 12(3^2 +3) = 1
log12(12) = 1
1 = 1, x = 3 is athe solution