Log5 (22-x)= log5 (2-x)+1 найти корень

[latex]log_5 (22-x)= log_5 (2-x)+1[/latex] ОДЗ: [latex] \left \{ {{22-x\ \textgreater \ 0} \atop {2-x\ \textgreater \ 0}} \right. [/latex] [latex] \left \{ {x\ \textless \ 22} \atop {x\ \textless \ 2}} \right. [/latex] [latex]x[/latex] ∈ [latex](-[/latex] ∞ [latex];2)[/latex] [latex]log_5 (22-x)= log_5 (2-x)+log_5 5[/latex] [latex]log_5 (22-x)= log_5 (5(2-x))[/latex] [latex]log_5 (22-x)= log_5 (10-5x)[/latex] [latex]22-x=10-5x[/latex] [latex]5x-x=10-22[/latex] [latex]4x=-12[/latex] [latex]x=-3[/latex] Ответ: -3