y=[log2(8)+log2(x)][log2(4)+log2(x)]
=[3+log2(x)][2+log2(x)]
令a=log2(x)
2<=x<=8
log2(2)<=log2(x)<=log2(8)
所以1<=a<=3
y=(3+a)(2+a)=a^2+5a+6=(a+5/2)^2-1/4
1<=a<=3,y开口向上,对称轴a=-5/2
所以定义域在对称轴的右边,是增函数
所以a=1,y最小=12
a=3,y最大=30