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 ].