Решите кто-то log1/4(3x-8)<-2

$log_{ \frac{1}{4}} (3x-8)\ \textless \ -2$
ОДЗ:
$3x-8\ \textgreater \ 0$
$3x\ \textgreater \ 8$
$x\ \textgreater \ \frac{8}{3}$
$x\ \textgreater \ 2 \frac{2}{3}$

$log_{ \frac{1}{4}} (3x-8)\ \textless \ log_{ \frac{1}{4}} 16$
$3x-8\ \textgreater \ 16$
$3x\ \textgreater \ 16+8$
$3x\ \textgreater \ 24$
$x\ \textgreater \ 8$

------------(2 2/3)-----------------
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-------------------------(8)--------
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Ответ: (8; +∞)