Заметим, что \(25{x^2}-180x + 324 = {\left( {5x-18} \right)^2}.\)
\(\dfrac{{{{\log }_4}\left( {{4^x}-128} \right)-{{\log }_4}{2^{x + 3}}}}{{25{x^2}-180x + 324}} \le 0\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\dfrac{{{{\log }_4}\left( {{4^x}-128} \right)-{{\log }_4}{2^{x + 3}}}}{{{{\left( {5x-18} \right)}^2}}} \le 0.\)
Тогда последнее неравенство равносильно следующей системе:
\(\left\{ \begin{array}{l}{4^x}-128 > 0,\\{2^{x + 3}} > 0,\\5x-18 \ne 0,\\{\log _4}\left( {{4^x}-128} \right)-{\log _4}{2^{x + 3}} \le 0\end{array} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\left\{ \begin{array}{l}{2^{2x}} > {2^7},\\x \in R,\\x \ne 3,6,\\{\log _4}\left( {{4^x}-128} \right) \le {\log _4}{2^{x + 3}}\end{array} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\left\{ \begin{array}{l}x > 3,5,\\x \ne 3,6,\\{4^x}-128 \le 8 \cdot {2^x}.\end{array} \right.\)
Решим неравенство \({4^x}-128 \le 8 \cdot {2^x}.\) Пусть \({2^x} = t.\) Тогда:
\({t^2}-128 \le 8t\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,{t^2}-8t-128 \le 0\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,-8 \le t \le 16\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,-8 \le {2^x} \le 16\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,x \le 4.\)
Следовательно,
\(\left\{ \begin{array}{l}x > 3,5,\\x \ne 3,6,\\{4^x}-128 \le 8 \cdot {2^x}\end{array} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\left\{ \begin{array}{l}x > 3,5,\\x \ne 3,6,\\x \le 4\end{array} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,x \in \left( {3,5;3,6} \right) \cup \left( {3,6;4} \right].\)
Ответ: \(\left( {3,5;3,6} \right)\, \cup \left( {3,6;4} \right].\)