Помогите пожалуйста, не могу решить 2logx(3) - 3log9/x(3) +2log3x(3)>=0

Решение: 

<span>2*log(x) 3 - 3*log(9/x) 3 + 2*log(3x) 3 >= 0 </span>

<span>___ log(b) a = 1/log(a) b <===> </span>
<span>Для наглядности обозначим log(3) x=a </span>
<span>2*log(x) 3 = 2/log(3) x = 2/a </span>
<span>3*log(9/x) 3 = </span>
<span>= 3/log(3) 9/x = 3/[log(3) 9 - log(3) x] = 3/[2 - log(3) x] = 3/(2-a) </span>
<span>2*log(3x) 3 = 2/log(3) 3x = 2/[log(3) 3 + log(3) x] = 2/[1 + log(3) x] = 2/(1+a) </span>
<span>==> </span>
<span>2/a - 3/(2-a) + 2/(1+a) >= 0 </span>
<span>2*(2-a)*(1+a) - 3*a*(1+a) + 2*a*(2-a) >= 0 </span>
<span>4 - 2a + 4a - 2a^2 - 3a - 3a^2 + 4a - 2a^2 >= 0 </span>
<span>-7a^2 +3a +4 >= 0 </span>
<span>7a^2 - 3a - 4 <= 0 </span>
<span>Реши уравнение 7a^2 - 3a - 4=0 </span>
<span>Корни будут: а1=1 и а2= -4/7 => </span>
<span>log(3) x= a </span>
<span>log(3) x = a1 =1 => x1=3^1=3 </span>
<span>log(3) x = a2 = -4/7 => x2=3^(-4/7) </span>

<span>11*log(13) [x^2 - 4x - 5] <= 12 + log(13) [(x+11)^11 / (x-5)] </span>
<span>log(13) [x^2 - 4x - 5]^11 <= log(13) 13^12 + log(13) [(x+11)^11 / (x-5)] </span>
<span>log(13) [x^2 - 4x - 5]^11 <= log(13) 13^12 * [(x+11)^11 / (x-5)] </span>
<span>[x^2 - 4x - 5]^11 <= 13^12 * [(x+11)^11 / (x-5)] </span>
<span>x^2 - 4x - 5=0 </span>
<span>Реши уравнение, корни будут х1=5, х2=-1 => </span>
<span>[(x-5)(x+1)]^11 <= 13^12 * (x+1)^11 / (x-5) </span>
<span>(x-5)^11 * (x+1)^11 <= 13^12 * (x+1)^11 / (x-5) </span>
<span>(x-5)^12 <= 13^12 </span>
<span>x-5 = 13 </span>
<span>x = 13+5=18 </span>