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 </span>
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)] </span>
log(13) [x^2 - 4x - 5]^11 <= log(13) 13^12 + log(13) [(x+11)^11 / (x-5)] </span>
log(13) [x^2 - 4x - 5]^11 <= log(13) 13^12 * [(x+11)^11 / (x-5)] </span>
[x^2 - 4x - 5]^11 <= 13^12 * [(x+11)^11 / (x-5)] </span>
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) </span>
(x-5)^11 * (x+1)^11 <= 13^12 * (x+1)^11 / (x-5) </span>
(x-5)^12 <= 13^12 </span>
x-5 = 13
x = 13+5=18