Question

Помогите решить 2,4 и 8

Answer 1

2.
$log_{2}(x) + log_{2}(10 - x) = 4 \log_{2}(xtimes (10 - x)) = 4 \ 10x - {x}^{2} = {2}^{4} \ - {x}^{2} + 10x - 16 = 0 \ {x }^{2} - 10x + 16 = 0$
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