Question

2log3 от (х-2) + log 3 от (х-4)^2=0

Answer 1

 

$2log_3(x-2)+log_3(x-4)^2=0, \ left { {{x-2>0,} atop {(x-4)^2>0,}} right. \ left { {{x>2,} atop {x-4neq0,}} right. \ left { {{x>2,} atop {xneq4,}} right. \ 2log_3(x-2)+2log_3(x-4)=0, \ log_3(x-2)+log_3(x-4)=0, \ log_3(x-2)(x-4)=log_31, \ (x-2)(x-4)=1, \ x^2+6x+7=0, \ D_1=3^2-7=2, \ x_1=-3-sqrt2<0<2,\ x_2=-3+sqrt2<0<2;$

действительных корней нет.