题目内容：

已知函数y=loga(a^2x).loga^2(ax),当x∈[2,4]时,y范围为[-1/8,0],求a

优质解答

y=(log1/a(a^2)+log1/a(x))*(log1/a^2(a)+log1/a^2(x))
=[-(2+loga(x))]*[-(1/2+1/2loga(x))]
=1/2(loga(x))^2+3/2loga(x)+1
令t=loga(x)
则y=(1/2)t^2+(3/2)t+1,为t的二次函数,
-b/2a=-3/2 ,(4ac-b^2)/4a=-1/8
画出y--t图象（自己画下吧）,图象开口向上,当-1/8<=y<=0时,-2<=t<=-1
当y=0时,0=(1/2)t^2+(3/2)t+1,解得t1=-1,t2=-2
(1)loga(x)=-1,x=2
则a=1/2
当t=loga(4)=-2时,y=0,符合题意
(2)loga(x)=-1,x=4
则a=1/4
当t=loga(2)=-1/2时,y>0,不合题意
(3)loga(x)=-2,x=2
则a=1/sqrt(2) (sqrt是开方)
当t=loga(4)=-4时,y>0,不合题意
(4)loga(x)=-2,x=4
则a=1/2
当t=loga(2)=-1时,y=0,符合题意
综上所述,a=1/2