Solve the following equation: \(log_3(x+3)+log_3(x-5)=2\)
Log3 ((x+3)(x-5) ) = 2
log3 (x^2-2x-15) = 2
x^2 - 2x - 15 = 3^2
x^2 -2x-24 =0
(x-6)(x+4) = 0 x = 6 or -4 (throw out....cannot have log of a negative )
Oh it makes sense now! I only had issues when it came to the part when there was the 3^2, so I must've forgotten a log rule there.