Question

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

Answer 1

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

ОДЗ:

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

$left { {{3x>1} atop {x>-1}} right.$

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

$x$ ∈ $(frac{1}{3} ; +$ ∞ $)$

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

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

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

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

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

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

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

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

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

Ответ: $3$