Решите пожалуйста срочноо!!!) log2(x-3)=0

2*log(x) 3 - 3*log(9/x) 3 + 2*log(3x) 3 >= 0 ___ log(b) a = 1/log(a) b <===> Для наглядности обозначим log(3) x=a 2*log(x) 3 = 2/log(3) x = 2/a 3*log(9/x) 3 = = 3/log(3) 9/x = 3/[log(3) 9 - log(3) x] = 3/[2 - log(3) x] = 3/(2-a) 2*log(3x) 3 = 2/log(3) 3x = 2/[log(3) 3 + log(3) x] = 2/[1 + log(3) x] = 2/(1+a) ==> 2/a - 3/(2-a) + 2/(1+a) >= 0 2*(2-a)*(1+a) - 3*a*(1+a) + 2*a*(2-a) >= 0 4 - 2a + 4a - 2a^2 - 3a - 3a^2 + 4a - 2a^2 >= 0 -7a^2 +3a +4 >= 0 7a^2 - 3a - 4 <= 0 Реши уравнение 7a^2 - 3a - 4=0 Корни будут: а1=1 и а2= -4/7 => log(3) x= a log(3) x = a1 =1 => x1=3^1=3 log(3) x = a2 = -4/7 => x2=3^(-4/7) 11*log(13) [x^2 - 4x - 5] <= 12 + log(13) [(x+11)^11 / (x-5)] log(13) [x^2 - 4x - 5]^11 <= log(13) 13^12 + log(13) [(x+11)^11 / (x-5)] log(13) [x^2 - 4x - 5]^11 <= log(13) 13^12 * [(x+11)^11 / (x-5)] [x^2 - 4x - 5]^11 <= 13^12 * [(x+11)^11 / (x-5)] x^2 - 4x - 5=0 Реши уравнение, корни будут х1=5, х2=-1 => [(x-5)(x+1)]^11 <= 13^12 * (x+1)^11 / (x-5) (x-5)^11 * (x+1)^11 <= 13^12 * (x+1)^11 / (x-5) (x-5)^12 <= 13^12 x-5 = 13 x = 13+5=18