Log2(3x-1)- log2(5x+1)< log2(x-1)-2

$log_{2} (3x-1)- log_{2}(5x+1)\ \textless \ log_{2} (x-1)-2$

ОДЗ:

$\left \{ {{3x-1\ \textgreater \ 0} \atop {5x+1}\ \textgreater \ 0} \atop {x-1}\ \textgreater \ 0\right.$

$\left \{ {{x\ \textgreater \ \frac{1}{3} } \atop {x}\ \textgreater \ - \frac{1}{5} } \atop {x}\ \textgreater \ 1\right.$

$x$ ∈ $(1;+$ ∞ $)$

$log_{2} (3x-1)- log_{2}(5x+1)\ \textless \ log_{2} (x-1)- log_{2} 4$

$log_{2} (3x-1)+ log_{2} 4\ \textless \ log_{2} (x-1)+ log_{2}(5x+1)$

$log_{2}(4 (3x-1))\ \textless \ log_{2} ((x-1)(5x+1))$

$log_{2}(12x-4)\ \textless \ log_{2} (5 x^{2} +x-5x-1)$

$log_{2}(12x-4)\ \textless \ log_{2} (5 x^{2} -4x-1)$

$12x-4\ \textless \ 5 x^{2} -4x-1$

$-5x^2+16x-3\ \textless \ 0$

$5x^2-16x+3\ \textgreater \ 0$

$D=(-16)^2-4*5*3=256-60=196=14^2$

$x_1= \frac{16+14}{10}=3$

$x_2= \frac{16-14}{10}=0.2$

+ - +
---------------(0.2)----------(3)------------
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С учётом ОДЗ получаем

Ответ: $(3;+$ ∞ $)$