(1)当)x=时,log2x=,代入y==(log2x-2)•(log2x-1)可得答案;
(2)若t=log2x,(2≤x≤4),则1≤t≤2,代入y==(log2x-2)•(log2x-1)可得y关于t的函数关系式.
(3)分析y=t2-3t+2的图象形状,结合1≤t≤2,由二次函数的图象和性质,可求出函数的最值,进而得到函数的值域.
【解析】
(1)x==时,log2x=
∴y=
=(log2x-log24)•(log2x-log22)
=(log2x-2)•(log2x-1)
=-•=-
(2)若t=log2x,(2≤x≤4)
则1≤t≤2,
则y=
=(log2x-2)•(log2x-1)
=(t-2)•(t-1)
=t2-3t+2(1≤t≤2)
(3)∵y=t2-3t+2的图象是开口朝上,且以t=为对称轴的抛物线
又∵1≤t≤2
∴当时,
当t=1或2时,ymax=0
故函数的值域是