2(log1/2x)^2+7log1/2x+3≤0,
令t=log1/2(x),有
2t^2+7t+3≤0,
(2t+1)(t+3)≤0.
-3≤t≤-1/2.
當(dāng)t=-3時,log1/2(x)=-3,x=(1/2)^-3=8.
當(dāng)t=-1/2時,log1/2(x)=-1/2,x=(1/2)^(-1/2)=√2.
∵log1/2(x),是減函數(shù),有
√2≤X≤8.
又∵log2(x)為增函數(shù),
∴1/2≤log2(x)≤3.
f(x)=[log2(x/4)]*[log2(x/2)]
=[log2(x)-log2(4)]*[log2(x)-log2(2)]
=[log2(x)-2]*[log2(x)-1]
=[log2(x)]^2-3[log2(x)]+2
=[log2(x)-3/2]^2-1/4
由于log2(x)屬于[1/2,3]
則：當(dāng)log2(x)=3/2時,f(x)取最小值為-1/4.此時x=2√2.
當(dāng)log2(x)=3時,f(x)有最大值2.此時x=8.