Алгебра: Log x^3-9x^2+27x-27(9-x)≥0 нужно подробно!)

$log_{ x^{3} - 9 x^{2} +27x-27}(9 - x) \geq 0 \\ x^{3} - 9 x^{2} +27x-27 = (x-3)^{3} =\ \textgreater \ \\ log_{ (x-3)^{3}}(9 - x) \geq 0 \\ \frac{1}{3} log_{ (x-3)}(9 - x) \geq 0 \\$
$log_{ (x-3)}(9 - x) \geq 0 \\ log_{ (x-3)}(9 - x) - 0 \geq 0 \\ log_{ (x-3)}(9 - x) - log_{ (x-3)}1 \geq 0 \\$

$\left \{ {{(x-3 -1)(9-x-1)\geq 0 \\} \atop {x-3\ \textgreater \ 0, x-3 \neq 1,9 - x\ \textgreater \ 0} } \right. \\ \left \{ {{(x-4)(8-x)\geq 0 \\} \atop {x\ \textgreater \ 3, x \neq 4,x\ \textless \ 9} } \right. \\ \left \{ {{(x-4)(x-8) \leq 0 \\} \atop {x\ \textgreater \ 3, x \neq 4,x\ \textless \ 9} } \right. \\$
$\left \{ {{4 \leq x \leq 8 \\} \atop {x\ \textgreater \ 3, x \neq 4,x\ \textless \ 9} } \right. \\ 4\ \textless \ x \leq 8 \\$

Ответ: ( 4 ; 8 ].