Question

нужно решить логарифмические уравнения ​

Answer 1

1.

$log_{ \frac{1}{3} }(2x - 1) + log_{ \frac{1}{3} }(x - 2) = - 3$

ОДЗ:

$\left \{ {{2x - 1 0} \atop {x - 2 0}} < = \right.\left \{ {{x \frac{1}{2} } \atop {x 2}} \right. \\ = x 2$

$log_{ \frac{1}{3} }((2x - 1)(x - 2)) = - 3 \\ (2x - 1)(x - 2) = {( \frac{1}{3}) }^{ - 3} \\ 2 {x}^{2} - 4x - x + 2 = 9 \\ 2 {x}^{2} - 5x - 7 = 0 \\ D = 25 + 56 = 81 \\ x_1 = \frac{5 + 9}{4} = \frac{14}{4} = \frac{7}{2} = 3.5 \\ x_2 = - 1 \\$

корень -1 не подходит.

ответ: 3.5

$log_{ \sqrt{2} }( {x}^{2} - 3x) = 4$

ОДЗ:

${x}^{2} - 3x 0 \\ x(x - 3) 0 \\ \\ znaki \\ + \: \: \: \: \: \: - \: \: \: \: \: + \\ - - - 0 - - 3- - -$

$x\in( - \infty; 0)U(3 ;+ \infty )$

${x}^{2} - 3x = {( \sqrt{2}) }^{4} \\ {x}^{2} - 3x = 4 \\ {x}^{2} - 3x - 4 = 0 \\ D = 9 + 16 = 25 \\ x_1 = \frac{3 + 5}{2} = 4 \\ x_2 = - 1$

ответ: -1; 4.