[m]\left\{\begin {matrix}x^2+6x-7>0\\\frac{(x-1)^3}{x+7}>0\end {matrix}\right.[/m][m]\left\{\begin {matrix}(x+7)(x-1)>0\\\frac{(x-1)^3}{x+7}>0\end {matrix}\right.[/m]
[red]x ∈ (- ∞ ;-7)U(1;+ ∞ )[/red]
[m]log_{6}(x^2+6x-7) ≤ 4+log_{6}\frac{(x-1)^3}{x+7}[/m]
Применяем свойства логарифмов:
[m]log_{6}(x^2+6x-7)=log_{6}(x-1)+log_{6}(x+7)[/m]
[m]log_{6}\frac{(x-1)^3}{x+7} =log_{6}(x-1)^3-log_{6}(x+7)=3log_{6}(x-1)-log_{6}(x+7)[/m]
[m]3(log_{6}(x-1)+log_{6}(x+7) )≤4+3log_{6}(x-1)-log_{6}(x+7)[/m]
[m]3log_{6}(x-1)+3log_{6}(x+7)≤4+3log_{6}(x-1)-log_{6}(x+7)[/m]
[m]3log_{6}(x+7)≤4-log_{6}(x+7)[/m]
[m]4log_{6}(x+7)≤4[/m]
[m]log_{6}(x+7)≤1[/m]
[m]x+1 ≤ 6[/m]
С учетом ОДЗ получаем ответ
[b]x ∈ (- ∞ ;-7)U(1;5)[/b]