smc

From the given we have $$\log (x+3)+\log (x-1)=\log (x^2-2x-3)$$ $$\log (x^2+2x-3)=\log (x^2-2x-3)$$ $$x^2+2x-3=x^2-2x-3$$ $$x=0$$ However substituing into $\log (x-1)$ gets a negative argument, which is impossible $\boxed{\text{no real values of}x}$.