Log(x-6) +Log(x+3)=1
I assume we are solving this equation ?
by rules of logs ....
log( (x-6)(x+3) ) = 1
10^1 = (x-6)(x+3)
x^2 - 3x - 18 = 10
x^2 - 3x - 28 = 0
(x-7)(x+4) = 0
x = 7 or x = -4 , but to be defined in log(x-6), x > 6
so x = 7