Логарифмические неравенства, сводящиеся к простейшим заменой неизвестного. Задача 5

ОТВЕТ: \(\left( {0;\;\dfrac{1}{{16}}} \right) \cup \left( {16;\;\infty } \right).\)

\(\log _2^2x > 16.\)   ОДЗ:  \(x > 0.\)

Пусть \({\log _2}x = t\). Тогда:

\({t^2} > 16\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{t < -4,}\\{t > 4.\,\,\,}\end{array}} \right.\)

Вернёмся к прежней переменной:

\(\left[ {\begin{array}{*{20}{c}}{{{\log }_2}x < -4,}\\{{{\log }_2}x > 4\,\,\,\,}\end{array}} \right.\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{{{\log }_2}x < {{\log }_2}\dfrac{1}{{16}},}\\{{{\log }_2}x > {{\log }_2}16\,\,\,}\end{array}\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{0 < x < \dfrac{1}{{16}},}\\{x > 16\,\,\,\,\,\,\,\,\,\,\,}\end{array}\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,} \right.} \right.x\, \in \,\left( {0;\dfrac{1}{{16}}} \right) \cup \left( {16;\infty } \right).\)

Ответ:  \(\left( {0;\dfrac{1}{{16}}} \right) \cup \left( {16;\infty } \right).\)