[tex]\log_8(x^2-4x+3)<1\\\\x^2-4x+3>0\\x^2-x-3x+3>0\\x(x-1)-3(x-1)>0\\(x-3)(x-1)>0\\x\in(\infty,1)\cup(3,\infty)\\\\\log_8(x^2-4x+3)<1\\\log_8(x^2-4x+3)<\log_88\\x^2-4x+3<8\\x^2-4x-5<0\\x^2+x-5x-5<0\\x(x+1)-5(x+1)<0\\(x-5)(x+1)<0\\x\in(-1,5)\\\\x\in(-1,5)\wedge x\in(\infty,1)\cup(3,\infty)\\x\in(-1,1)\cup(3,5)[/tex]
