\(\log _4^4\left( {x-1} \right) \cdot \left( {1-{{\log }_3}x} \right) \le 0.\)
ОДЗ: \(\left\{ {\begin{array}{*{20}{c}}{x-1 > 0,}\\{x > 0\,\,\,\,\,\,\,\,\,\,}\end{array}\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{x > 1,\,}\\{x > 0\,}\end{array}\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,x\, \in \,\left( {1;\infty } \right).} \right.} \right.\)
Решим данное неравенство методом интервалов:
\(\log _4^4\left( {x-1} \right) \cdot \left( {1-{{\log }_3}x} \right) = 0\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{\log _4^4\left( {x-1} \right) = 0,}\\{1-{{\log }_3}x = 0\,\,\,\,\,\,}\end{array}\,\,\,\,\,\,\, \Leftrightarrow } \right.\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{{{\log }_4}\left( {x-1} \right) = 0,}\\{{{\log }_3}x = 1\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\end{array}\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{x = 2,}\\{x = 3.}\end{array}} \right.} \right.\)
Таким образом, решением исходного неравенства является: \(x \in \,\left\{ 2 \right\} \cup \left[ {3;\infty } \right).\)
Ответ: \(\left\{ 2 \right\} \cup \left[ {3;\infty } \right).\)