Не могу решить 2logx(3) - 3log9/x(3) +2log3x(3)> =0

Решение: 

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