Question

помогите решить уравнение

Answer 1

$log_{3} ^{2} (x) - 2 log_{3}(x) = 3 \\ odz \: \: \: x > 0 \\ log_{3}(x) =y \\ {y}^{2} - 2y = 3 \\ {y}^{2} - 2y - 3 = 0 \\ d = 4 - 4 \times ( - 3) = = 4 + 12 = 16 \\ y1 = \frac{2 + \sqrt{16} }{2} = \frac{2 + 4}{2} = \frac{6}{2} = 3 \\ y2 = \frac{2 - \sqrt{16} }{2} = \frac{2 - 4}{2} = - \frac{2}{2} = - 1 \\ - - - - - \\ log_{3}(x) = y1 \\ log_{3}(x) =3 \\ x1 = 27 \\ - - - - - - \\ log_{3}(x) =y2 \\ log_{3}(x) = - 1 \\ x2 = \frac{1}{3} \: \: \: \\ odz \: \: \: \: \: x > 0$