Log2(3x-1)=5-log2(x+1)

$log_2(3x-1)=5-log_2(x+1)$

ОДЗ:

0} \atop {x+1>0}} \right. " alt=" \left \{ {{3x-1>0} \atop {x+1>0}} \right. " align="absmiddle" class="latex-formula">

1} \atop {x>-1}} \right. " alt=" \left \{ {{3x>1} \atop {x>-1}} \right. " align="absmiddle" class="latex-formula">

\frac{1}{3}} \atop {x>-1}} \right. " alt=" \left \{ {{x>\frac{1}{3}} \atop {x>-1}} \right. " align="absmiddle" class="latex-formula">

$x$ ∈ $(\frac{1}{3} ; +$ ∞ $)$

$log_2(3x-1)+log_2(x+1)=5$

$log_2[(3x-1)*(x+1)]=log_22^5$

$log_2(3x^2+2x-1)=log_232$

$3x^2+2x-1=32$

$3x^2+2x-1-32=0$

$3x^2+2x-33=0$

$D=2^2-4*3*(-33)=4+396=400=20^2$

$x_1=\frac{-2+20}{6} =3$

$x_2=\frac{-2-20}{6} =-\frac{11}{3} =-3\frac{2}{3}$ ∉ ОДЗ

Ответ: $3$