Question

Помогите
Log7(x2+3x-1)= Log7 9

Answer 1

Ответ:

ОДЗ:

$\frac{-3 - \sqrt{13}}{2} > x > \frac{-3 + \sqrt{13}}{2}$

$\log_7 (x^2 + 3x -1) = \log_7 9\\x^2 + 3x - 1 = 9\\x^2 + 3x - 10 = 0\\D = 9 - 4 * (-10) = 49\\x_1 = \frac{-3 + 7}{2} = 2\\x_2 = \frac{-3 - 7}{2} = -5$