Question

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

Answer 1

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

ОДЗ:

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

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

$x$ ∈ $(1;+$ ∞ $)$

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

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

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

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

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

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

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

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

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

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

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

       +                     -               +
---------------(0.2)----------(3)------------
////////////////                       ///////////////

С учётом ОДЗ получаем

Ответ: $(3;+$ ∞ $)$