Question

16．已知$f（x）=\left\{\begin{array}{l}-2，0＜x＜1\\ 1，x≥1\end{array}\right.$，則不等式${log_2}x-（{{{log}_{\frac{1}{4}}}4x-1}）f（{{{log}_3}x+1}）≤5$的解集為（$\frac{1}{3}$，4]．

Answer 1

分析 根據分段函數f（x）的解析式，把不等式${log_2}x-（{{{log}_{\frac{1}{4}}}4x-1}）f（{{{log}_3}x+1}）≤5$化為等價的不等式組，求出解集即可．

解答 解：不等式${log_2}x-（{{{log}_{\frac{1}{4}}}4x-1}）f（{{{log}_3}x+1}）≤5$
$?\left\{\begin{array}{l}{log_3}x+1≥1\\{log_2}x-（{{{log}_{\frac{1}{4}}}4x-1}）≤5\end{array}\right.$或$\left\{\begin{array}{l}0＜{log_3}x+1＜1\\{log_2}x+2（{{{log}_{\frac{1}{4}}}4x-1}）≤5\end{array}\right.$，
解得1≤x≤4或$\frac{1}{3}＜x＜1$；
∴原不等式的解集為（$\frac{1}{3}$，4]．
故答案為：$（\frac{1}{3}，4]$．

點評 本題考查了分段函數與對數不等式的解法與應用問題，是基礎題．