[tex]\ Condi\c{\it t}ii\ de\ existen\c{\it t}\breve a\ a\ ecua\c{\it t}iei :\\ \\ \\ \left.\begin{aligned}x+1>0 \Rightarrow x>-1\\ \\ x-2>0 \Rightarrow x>2\end{aligned}\right\}\Rightarrow domeniul\ de\ existen\c{\it t}\breve a\ este\ D=(2,\ \ \infty)[/tex]
[tex]\it log_2(x+1)+log_2(x-2)=2 \Rightarrow log_2(x+1)(x-2)=2 \Rightarrow (x+1)(x-2)=2^2 \Rightarrow \\ \\ \Rightarrow x^2-2x+x-2=4 \Rightarrow x^2-x-6=0 \Rightarrow x^2-3x+2x-6=0 \Rightarrow \\ \\ \Rightarrow x(x-3)+2(x-3)=0 \Rightarrow (x-3)(x+2)=0 \Rightarrow \begin{cases}\it x-3=0 \Rightarrow x=3\in D\\ \\ \it x+2=0\Rightarrow x=-2 \not{\in} D\end{cases}[/tex]
Deci, ecuația dată admite soluția unică x=3
