Нужна срочно помощь 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 _____-0.8////////////9____ 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)