\(log_2(1-x) + log_2(3-x) = 3\)
\(log_2((1-x)(3-x)) = 3\)
\((1-x)(3-x) = 2^3\)
\(3 - x - 3x + x^2 = 8\)
\(x^2 - 4x + 3 = 8\)
\(x^2 - 4x - 5 = 0\)
\(D = b^2 - 4ac\)
\(D = (-4)^2 - 4 \cdot 1 \cdot (-5) = 16 + 20 = 36\)
\(x_1 = \frac{-b + \sqrt{D}}{2a} = \frac{4 + \sqrt{36}}{2 \cdot 1} = \frac{4 + 6}{2} = 5\)
\(x_2 = \frac{-b - \sqrt{D}}{2a} = \frac{4 - \sqrt{36}}{2 \cdot 1} = \frac{4 - 6}{2} = -1\)
1) \(x_1 = 5\)
\(1 - x = 1 - 5 = -4 < 0\) (не подходит)
2) \(x_2 = -1\)
\(1 - x = 1 - (-1) = 2 > 0\)
\(3 - x = 3 - (-1) = 4 > 0\) (подходит)
Ответ: \(x = -1\)