Решите log3(x-2)+log3(x+2)=log3(2x-1)

[latex]log_3(x-2)+log_3(x+2)=log_3(2x-1)\\ log_3\frac{x-2}{x+2}=log_3(2x-1)\\ \left \{ {{\frac{x-2}{x+2}=2x-1} \atop {\frac{x-2}{x+2}\ \textgreater \ 0}; \ \ \ 2x-1\ \textgreater \ 0} \right. \\ \left \{ {{x-2=(2x-1)(x+2)} \atop { \left \{ {{x-2\ \textgreater \ 0} \atop {x+2\ \textgreater \ 0}} \right. }; \ \ \ \left \{ {{x-2\ \textless \ 0} \atop {x+2\ \textless \ 0}} \right. ; \ \ \ x\ \textgreater \ \frac{1}{2}} \right. \\ \left \{ {{0=2x^2-x+4x-2-x+2} \atop { \left \{ {{x-2\ \textgreater \ 0} \atop {x+2\ \textgreater \ 0}} \right. }; \ \ \ \left \{ {{x-2\ \textless \ 0} \atop {x+2\ \textless \ 0}} \right. ; \ \ \ x\ \textgreater \ \frac{1}{2}} \right. \\[/latex] [latex] \left \{ {{2x^2+2x=0} \atop { \left \{ {{x\ \textgreater \ 2} \atop {x\ \textgreater \ -2}} \right. }; \ \ \ \left \{ {{x\ \textless \ 2} \atop {x\ \textless \ -2}} \right. ; \ \ \ x\ \textgreater \ \frac{1}{2}} \right. \\ \left \{ {{2x(x+1)=0} \atop x\ \textgreater \ 2\right. \\ \left \{ \left \{ {{x=0} \atop {x+1=0}} \right. } \atop x\ \textgreater \ 2\right. \\ \left \{ \left \{ {{x=0} \atop {x=-1}} \right. } \atop x\ \textgreater \ 2\right. \\ [/latex] Получаем, что уравнение не имеет решений