Решите неравенство: log1/6(4x − 32) > log1/6(3x).

$log_{\frac{1}{6}}(4x-32)\ \textgreater \ log_{\frac{1}{6}}3x\\0\ \textless \ \frac{1}{6}\ \textless \ 1\to\\\left\{{{4x-32\ \textless \ 3x}\atop{4x-32\ \textgreater \ 0}}\right.\left\{{{4x-3x\ \textless \ 32}\atop{4x\ \textgreater \ 32}}\right.\left\{{{x\ \textless \ 32}\atop{x\ \textgreater \ 8}}\right.\\8\ \textless \ x\ \textless \ 32$
Ответ: x∈(8; 32)