Question

Помогите решить 2,4 и 8

Answer 1

2.
$log_{2}(x) + log_{2}(10 - x) = 4 \\log_{2}(x\times (10 - x)) = 4 \\ 10x - {x}^{2} = {2}^{4} \\ - {x}^{2} + 10x - 16 = 0 \\ {x }^{2} - 10x + 16 = 0$
по теореме Виета найдем корни квадратного уравнения:
$x_{1} + x_{2} = 10 \\ x_{1} \times x_{2} = 16 \\ x_{1} = 2 \\ x_{2} = 8$
ОДЗ:
$x > 0 \\ 10 - x > 0 \\ - x > - 10 \\ x < 10 \\ x ( - \infty ; \: 10)$
Ответ: (2; 8)

4.
$log_{2} log_{4}(5 - x) = - 1 \\ log_{4}(5 - x) = {2}^{ - 1} \\ log_{4}(5 - x) = \frac{1}{2} \\ 5 - x = {4}^{ \frac{1}{2} } \\ x = 5 - 2 \\ x = 3$
5-х>0
-х>-5
х<5

Ответ: х=3

8.
$log_{4}(x) - \frac{1}{ log_{4}(x) } = \frac{3}{2} \\ \frac{ log_{4}^{2} (x) - 1}{ log_{4}(x) } = 1.5\\ log_{4}^{2} (x) - 1 = 1.5log_{4}(x) \\ log_{4}^{2} (x) - 1.5log_{4}(x) - 1 = 0$
замена:
$log_{4}(x) = a$

получаем квадратное уравнение:

${a}^{2} - 1.5a - 1 = 0 \\ 2 {a}^{2} - 3a - 2 = 0 \\ d = 9 - 4 \times 2 \times ( - 2) = \\ = 9 + 16 = 25 \\ a_{1} = \frac{3 - 5}{2 \times 2} = \frac{ - 2}{4} = - \frac{1}{2} \\ a_{2} = \frac{3 + 5}{2 \times 2} = \frac{ 8}{4} =2$
$log_{4}(x) = - \frac{1}{2} \\ x = {4}^{ - \frac{1}{2} } \\ x = \frac{1}{ \sqrt{4} } \\ x = \frac{1}{2}$
х>0

$log_{4}(x) = 2 \\ x = {4}^{2} \\ x =16$
Ответ: 1/2; 16