1. log2^2(x-1)+log2(x-1)6+log1/2(3x+1) помогите!

$log_{2}^{2}(x-1)+log_{2}(x-1)<2$

ОДЗ : x - 1 > 0 ⇒ x > 1

Сделаем замену :

$log_{2}(x-1)=m\\\\m^{2}+m-2<0\\\\(m+2)(m-1)<0$

+ - +

_________₀____________₀___________

- 2 1

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-2\\\\x-1>0,25\\\\x>1,25\\\\log_{2}(x-1)<1\\\\x-1<2\\\\x<3" alt="log_{2}(x-1)>-2\\\\x-1>0,25\\\\x>1,25\\\\log_{2}(x-1)<1\\\\x-1<2\\\\x<3" align="absmiddle" class="latex-formula">

Ответ : x ∈ (1,25 ; 3)

6+log_{\frac{1}{2} }(3x+1)" alt="2)log_{\frac{1}{2} }^{2} (3x+1)>6+log_{\frac{1}{2} }(3x+1)" align="absmiddle" class="latex-formula">

ОДЗ : 3x + 1 > 0 ⇒ x > - 1/3

Сделаем замену :

0\\\\(m-3)(m+2)>0" alt="log_{\frac{1}{2} }^{2}(3x+1)=m\\\\m^{2}-m-6>0\\\\(m-3)(m+2)>0" align="absmiddle" class="latex-formula">

+ - +

__________₀___________₀___________

- 2 3

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4\\\\3x>3\\\\x>1\\\\log_{\frac{1}{2} }(3x+1)>3\\\\3x+1<\frac{1}{8} \\\\3x<-\frac{7}{8} \\\\x<-\frac{7}{24}" alt="log_{\frac{1}{2} }(3x+1)<-2\\\\3x+1>4\\\\3x>3\\\\x>1\\\\log_{\frac{1}{2} }(3x+1)>3\\\\3x+1<\frac{1}{8} \\\\3x<-\frac{7}{8} \\\\x<-\frac{7}{24}" align="absmiddle" class="latex-formula">

Ответ : (- 1/3 ; - 7/24) ∪ (1 ; + ∞)