Нужна с решениями логорифмами 1)3-log3x=log(x-6) 2)log7(5x+4)≤2 3){log4x-log4 y=1/2 {3x-y=10

1)3-log(3) x=log(3) (x-6)
ОДЗ х > 0 и х - 6>0
x > 6
log(3) 27 - log(3) x = log(3) (x - 6)
log(3) (27/x) = log(3) (x - 6)
27/x = x - 6
27 = x^2 - 6x
x^2 - 6x - 27 = 0
D = 36 + 108 = 144
x1 = (6+12)/2 = 18/2 = 9
x2 = (6 - 12)/2 = -6/2 = -3 - лишний корень
ответ 9
2)log(7) (5x+4)≤2
ОДЗ 5х + 4 > 0
x > -0.8
log(7) (5x + 4) ≤ log(7) 49
5x + 4≤ 49
5x ≤ 45
x ≤ 9
-0.89
ответ x ∈(-0.8; 9]
3){log4x-log4 y=1/2
{3x-y=10
ОДЗ x>0 и y>0
y = 10 - 3x
log(4) x - log(4) (3x - 10) = 1/2
log(4) (x/(3x - 10)) = log(4) √4
x/(3x - 10) = 2
x = 2(3x- 10)
x = 6x - 20
-5x = -20
x = 4
y = 3*4 - 10 = 12 - 10 = 2
ответ: (4; 2)