Question

满足不等式2（log0.5 x)^2+9log0.5 x+9≤0的x的范围

当x在（1）中范围变动时，函数f（x）=log2(x/2)*log2(x/4)的最大值和最小值

Answer 1

t=log0.5 x
不等式2（log0.5 x)^2+9log0.5 x+9≤0可化为：
2t^2+9t+9<=0
(2t+3)(t+3)<=0
-3<=t<=-3/2
即：-3<=log0.5 x<=-3/2
0.5^(-3/2)<=x<=0.5^(-3)
2根2<=x<=8

3/2<=t=log2(x)<=3

f（x）=log2(x/2)*log2(x/4)=
(log2(x)-1)(log2(x)-2)
=(t-1)(t-2)
=t^2-3t+2
对称轴是t=3/2
g(t)==t^2-3t+2在t>=3/2时递增，
最大值g(3)=2
最小值g(3/2)=-1/4