Найдите сумму целых решений неравенства [latex] log_{2}(x-2)\ \textless \ log_{4}(x+10)[/latex]

[latex]\begin{cases} x-2\ \textgreater \ 0 \\ x+10\ \textgreater \ 0 \\(x-2)^2 \ \textless \ x+10 \end{cases} \begin{cases} x\ \textgreater \ 2 \\ x\ \textgreater \ -10 \\x^2-5x-6 \ \textless \ 0 \end{cases} \begin{cases} x\ \textgreater \ 2 \\ (x+1)(x-6) \ \textless \ 0 \end{cases}[/latex] [latex]\begin{cases} x\ \textgreater \ 2 \\ -1\ \textless \ x\ \textless \ 6 \end{cases} =\ \textgreater \ \ x \in (2;6)[/latex] сумма целых 3+4+5=12. Ответ: 12.

[latex]log_{2}(x-2)\ \textless \ log_{4}(x+10) [/latex] ОДЗ:  [latex] \left \{ {{x-2\ \textgreater \ 0} \atop {x+10\ \textgreater \ 0}} \right. [/latex] [latex] \left \{ {{x\ \textgreater \ 2} \atop {x\ \textgreater \ -10}} \right. [/latex] [latex]x[/latex]∈[latex](2;+[/latex]∞[latex])[/latex] [latex]log_{2}(x-2)\ \textless \ log_{2^2}(x+10) [/latex] [latex]log_{2}(x-2)\ \textless \ \frac{1}{2} log_{2}(x+10) [/latex] [latex]2log_{2}(x-2)\ \textless \ log_{2}(x+10) [/latex] [latex]log_{2}(x-2)^2\ \textless \ log_{2}(x+10) [/latex] [latex]}(x-2)^2\ \textless \ (x+10) [/latex] [latex] x^{2} -4x+4-x-10\ \textless \ 0[/latex] [latex] x^{2} -5x-6\ \textless \ 0[/latex] [latex]D=25+24=49[/latex] [latex]x_1=6[/latex] [latex]x_2=-1[/latex]  решаем методом интервалов и, учитывая ОДЗ , получаем x∈(2;6) целые решения:  3, 4, 5 3+4+5=12 Ответ: 12