设函数f(x)=lg
|
f(x)=lg
1+2x+4xa |
4 |
等式f(x)>(x-1)lg4即为
lg(1+2x+4xa)>xlg4=lg4x.
1+2x+4xa>4x.
将a分离得出a>
4x-2x -1 |
4x |
令g(x)=
4x-2x -1 |
4x |
4x-2x -1 |
4x |
1 |
2x |
1 |
4x |
易知g(x)在[1,3]上单调递增,最小值为g(1)=1-
1 |
2 |
1 |
4 |
1 |
4 |
所以a>
1 |
4 |
故答案为:a>
1 |
4 |
设函数f(x)=lg
|
f(x)=lg
1+2x+4xa |
4 |
等式f(x)>(x-1)lg4即为
lg(1+2x+4xa)>xlg4=lg4x.
1+2x+4xa>4x.
将a分离得出a>
4x-2x -1 |
4x |
令g(x)=
4x-2x -1 |
4x |
4x-2x -1 |
4x |
1 |
2x |
1 |
4x |
易知g(x)在[1,3]上单调递增,最小值为g(1)=1-
1 |
2 |
1 |
4 |
1 |
4 |
所以a>
1 |
4 |
故答案为:a>
1 |
4 |