First of all, this equation is defined on ]3,+oo[]3,+∞[ because you need x+3 > 0x+3>0 and x - 3 > 0x−3>0 at the same time or the log won't be defined.
The log function maps a sum into a product, hence log(x+3) + log(x-3) = 27 iff log [(x+3)(x-3)] = log 27log(x+3)+log(x−3)=27⇔log[(x+3)(x−3)]=log27.
You now apply the exponential function on both sides of the equation : log [(x+3)(x-3)] = log 27 iff (x+3)(x-3) = 27 iff x^2 - 9 = 27 iff x^2 - 36 = 30log[(x+3)(x−3)]=log27⇔(x+3)(x−3)=27⇔x2−9=27⇔x2−36=30. This is a quadratic equation that has 2 real roots because Delta = -4*(-36) = 144 > 0
You know apply the quadratic formula x = (-b+- sqrtDelta)/2a with a = 1 and b = 0, hence the 2 solutions of this equation : x = ± 6
-6 !in ]3,+oo[ so we can't keep this one. The only solution is x = 6.
