题目内容

7.若log2a(5a-2)>0,则实数a的取值范围为$a>\frac{3}{5}$或$\frac{2}{5}<a<\frac{3}{5}$.

分析 log2a(5a-2)>0,可得$\left\{\begin{array}{l}{2a>1}\\{5a-2>1}\end{array}\right.$,或$\left\{\begin{array}{l}{0<2a<1}\\{0<5a-2<1}\end{array}\right.$,解出即可得出.

解答 解:∵log2a(5a-2)>0,∴$\left\{\begin{array}{l}{2a>1}\\{5a-2>1}\end{array}\right.$,或$\left\{\begin{array}{l}{0<2a<1}\\{0<5a-2<1}\end{array}\right.$,
解得$a>\frac{3}{5}$或$\frac{2}{5}<a<\frac{3}{5}$.
则实数a的取值范围为$a>\frac{3}{5}$或$\frac{2}{5}<a<\frac{3}{5}$.
故答案为:$a>\frac{3}{5}$或$\frac{2}{5}<a<\frac{3}{5}$.

点评 本题考查了对数函数的单调性、分类讨论方法,考查了推理能力与计算能力,属于中档题.

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