Question

помогите, пожалуйста. Логарифмы

Answer 1

$log_{2}(x - 5) \times (x + 2) = 3 \\ log_{2}(x {}^{2} + 2x - 5x - 10) = 3 \\ x {}^{2} + 2x - 5x - 10 = 2 {}^{3} \\ x {}^{2} - 3x - 10 = 8 \\ x {}^{2} - 3x - 18 = 0 \\ d = b {}^{2} - 4ac = 9 + 72 = \sqrt{81} = 9 \\ x1 = 6 \\ x2 = - 3$

1) Ответ x= 6, т.к. -3 не подходит

$log_{ \frac{1}{3} }(2 - x) > 1 \\ 2 - x < \frac{1}{3} \\ - x < \frac{1}{3} - 2 \\ - x < - \frac{5}{3} \\ 2 > x > \frac{5}{3}$

3) {5/3 ; 2}

$log_{2}(x - 4) < 1 \\ x - 4 < 2 \\ x < 2 + 4 \\ 6 > x > 4$

2) {4 ; 6}