Log_2(x+3)+Log2(x-3)=Log_2 7

[latex]log_2(x+3)+log_2(x-3)=log_27\\ \left \{ {{x+3\ \textgreater \ 0} \atop {x-3\ \textgreater \ 0} \right. \\\\ \left \{ {{x\ \textgreater \ -3} \atop {x\ \textgreater \ 3}}=\ \textgreater \ x\ \textgreater \ 3 \right.\\\\ log_2(x+3)(x-3)=log_27\\\\(x+3)(x-3)=7\\x^2-9=7\\x^2=16[/latex] x(1)=-4∉(3;+∞) x(2)=4∈(3;+∞) Ответ: 4