2log2(x-1)+log2(x-3)^2=0 Ребяяят, пожалуйста!

ОДЗ: x > 1; x ≠ 3

$log_2(x-1)^2+log_2(x-3)^2=0\\\\ log_2((x-1)(x-3))^2=0\\\\ ((x-1)(x-3))^2=2^0=1\\\\ 1)(x-1)(x-3)=1\\\\ x^2-4x+3=1\\\\ x^2-4x+2=0\\\\ D=16-8=8\\\\ x_1=\frac{4-2\sqrt{2}}{2} =2-\sqrt{2} <1\\\\ x_2=\frac{4+2\sqrt{2}}{2} =2+\sqrt{2} \\\\ 2)(x-1)(x-3)=-1\\\\ x^2-4x+3=-1\\\\ x^2-4x+4=0\\\\ (x-2)^2=0\\\\ x_3=2$

Ответ: 2 и 2+√2