Question

设f(x)=log以a为底(2x+1)为对数在区间(-1/2,0)上满足f(x)>0

1.求实数a的取值范围 2.求函数f(x)的单调区间 3.解不等式f(x)＞1

Answer 1

1)设f(x)=log以a为底(2x+1)为对数在区间(-1/2,0)上满足f(x)>0
所以f(x)在(-1/2,0)单调减
0<a<1
2)f(x)在(-1/2,正无穷)单调减
3)f(x)＞1
即有:
f(x)＞1=loga(a)
又因为f(x)在(-1/2,正无穷)单调减
所以:2x+1<a
即有:x<(a-1)/2