\(\log _2^2{x^2}-15{\log _2}x-4 \le 0.\)
ОДЗ: \(\left\{ {\begin{array}{*{20}{c}}{{x^2} > 0,\,\,\,\,}\\{x > 0\,\,\,\,\,\,\,\,}\end{array}\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{x \ne 0,}\\{x > 0}\end{array}\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,x\, \in \,\left( {0;\infty } \right).} \right.} \right.\)
\(\log _2^2{x^2}-15{\log _2}x-4 \le 0\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,{\left( {2{{\log }_2}x} \right)^2}-15{\log _2}x-4 \le 0\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,4\log _2^2x-15{\log _2}x-4 \le 0.\)
Пусть \({\log _2}x = t\). Тогда:
\(4{t^2}-15t-4 \le 0\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,-\dfrac{1}{4} \le t \le 4.\)
Вернёмся к прежней переменной:
\(-\dfrac{1}{4} \le {\log _2}x \le 4\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,{\log _2}\dfrac{1}{{\sqrt[4]{2}}} \le {\log _2}x \le {\log _2}16\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\dfrac{1}{{\sqrt[4]{2}}} \le x \le 16\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,x\, \in \,\left[ {\sqrt[4]{{0,5}};16} \right].\)
Ответ: \(\left[ {\sqrt[4]{{0,5}};16} \right]\).