Пожалуйста помогите решить неравенство

$log_{ \frac{1}{8} }(4x-32)\ \textgreater \ log_{ \frac{1}{8} }(2x)$

ОДЗ:
$\left \{ {{4x-32\ \textgreater \ 0} \atop {2x\ \textgreater \ 0}} \right.$

$\left \{ {{4x\ \textgreater \ 32} \atop {x\ \textgreater \ 0}} \right.$

$\left \{ {{x\ \textgreater \ 8} \atop {x\ \textgreater \ 0}} \right.$

$x$ ∈ $(8; +$ ∞ $)$

$4x-32\ \textless \ 2x$

$4x-2x\ \textless \ 32$

$2x\ \textless \ 32$

$x\ \textless \ 16$

------------(8)----------------------
///////////////////////
------------------------(16)---------
///////////////////////////

Ответ: $(8;16)$