color(red)((1) log_aX=log_aY=>X=Y)
color(red)((2) log_aM+log_aN=log_a(MN))
So
log_3(x-3+x^2-3x)=log_3(x-1)+log_3(x-3),.. use (2)
log_3(x^2-2x-3)=log_3((x-1)(x-3))
x^2-2x-3=(x-1)(x-3),..use(1)
x^2-2x-3=x^2-4x+3
-2x-3=-4x+3
4x-2x=3+3=>2x=6=>x=3
Check:
LHS=log_3(3-3+9-9)=log_3(0) toundefined
RHS=log_3(3-1)+log_3(3-3)=log2+log_3(0)!=LHS
x=3 does not satisfy the equation.
OR
(x-3)+x^2-3x=(x-3)+x(x-3)=(x-3)(x+1)=>log_3(x-3)(x+1))=log_3(x-1)+log_3(x-3)=>log_3(x-3)+log_3(x+1)=log_3(x-1)+log_3(x-3)=>log_3(x+1)=log_3(x-1)
=>x+1=x-1=>1=-1, which is not possible.