Question 198549
Apply "log rules":
log3(x-3)+log3(x+2)=log3(6)
log3(x-3)(x+2)=log3(6)
(x-3)(x+2)=6
x^2+2x-3x-6 = 6
x^2-x-12 = 0
(x-4)(x+3) = 0
x = {-3,4}
We can toss out the negative solution leaving us with:
x = 4