Răspuns :
[tex]\log_2(x+3)+\log_2x=2, \ x\in (0, \ +\infty) \\ \log_2(x(x+3))=2 \\ \log_2(x^2+3x)=2 \\ x^2+3x=2^2=4 \\ x^2+3x-4=0 \\ x^2+4x-x-4=0 \\ x(x+4)-(x+4)=0 \\ (x+4)(x-1)=0 \\ \Rightarrow \left \{ {{x+4=0} \atop {x-1=0}} \right. \Longleftrightarrow \ \left \{ {{x=-4} \atop {x=1}} \right. , \ x \in (0, \ +\infty) \\ \Rightarrow \boxed{\bold{x=1}}[/tex]