Question

решите уравнение log2(x^+3)=1+log(x+3)

Answer 1

log2(x^2+3)=1+log2(x+3)
...................................
ОДЗ:
{x^2+3>0; x e R
{x+3>0; x>-3
...................................
log2(x^2+3) = log2(2)+ log2(x+3)
log2(x^2+3)=log2(2x+6)
x^2+3=2x+6
x^2+3-2x-6=0
x^2-2x-3=0
D=(-2)^2-4*1*(-3)=16
x1=(2-4)/2=-1
x2=(2+4)/2=3
Ответ: {-1;3}