Log_2(3^x-1)=1-log_2(3^x-2) решите пожалуйста

[latex]log_2(3^x-1)=1-log_2(3^x-2)[/latex] ОДЗ:  [latex] \left \{ {{3^x-1\ \textgreater \ 0} \atop {3^x-2\ \textgreater \ 0}} \right. [/latex] [latex] \left \{ {{3^x\ \textgreater \ 1} \atop {3^x\ \textgreater \ 2}} \right. [/latex] [latex] \left \{ {{3^x\ \textgreater \ 3^0} \atop {3^x\ \textgreater \ 3^{log_32}} \right. [/latex] [latex] \left \{ {{x\ \textgreater \ 0} \atop {x\ \textgreater \ {log_32}} \right. [/latex] [latex]x[/latex] ∈ [latex](log_32;+[/latex] ∞ [latex])[/latex] [latex]log_2(3^x-1)+log_2(3^x-2)=1[/latex] [latex]log_2[(3^x-1)(3^x-2)]=log_22[/latex] [latex](3^x-1)(3^x-2)=2[/latex] [latex]3^{2x}-2*3^x-3^x+2-2=0[/latex] [latex]3^{2x}-3*3^x=0[/latex] [latex]3^x(3^{x}-3)=0[/latex] [latex]3^x-3=0[/latex]       или    [latex]3^x=0[/latex] [latex]3^x=3^1[/latex]                           ∅ [latex]x=1[/latex] Ответ: 1