Log x-1(12x-x^2-19)=3.

Решите задачу:

$log_{x-1}(12x-x^2-19)=3\; ,\\\\ODZ:\; \left \{ {{12x-x^2-19\ \textgreater \ 0} \atop {x-1\ \textgreater \ 0\; ,\; x-1\ne 1}} \right. \; \left \{ {{(x-6+\sqrt{17})(x-6-\sqrt{17})\ \textless \ 0} \atop {x\ \textgreater \ 1\; ,\; x\ne 2}} \right. \\\\x\in (6-\sqrt{17}\, ;\, 2)\cup (2\, ;\, 6+\sqrt{17})\\\\log_{x-1}(12x-x^2-19)=log_{x-1}(x-1)^3\\\\12x-x^2-19=x^3-3x^2+3x-1\\\\x^3-2x^2-9x+18=0\\\\x^2(x-2)-9(x-2)=0\\\\(x-2)(x^2-9)=0\\\\(x-2)(x-3)(x+3)=0\\\\x_1=-3\; ,\; x_2=2,\; x_3=3\\\\x_1\; ,\; x_2\notin ODZ\\\\Otvet:\; \; x=3\; .$