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问题: 跪求一道高中对数题的解答!!

已知 y=(log以1/a为底数,a的平方乘以x为真数)*(log以1/a的平方为底数,ax为真数) 2≤x≤4 最大值为0 最小值为-8/1 求a的值

解答:

y=[-loga(a^2x)][-loga^2(ax)]
=[2+loga(x)]*1/2[1+loga(x)]

2<=x<=4 y最大值为0
y=[2+loga(2)]*1/2[1+loga(2)]<=0
-2<=loga(2)<=-1
若a>1,loga(2)>0,无解
若0<a<1,
-2<=1/log2(a)<=-1
-1<=log2(a)<=-1/2
1/2<=a<=√2/2

[2+loga(4)]*1/2[1+loga(4)]<=0
-2<=loga(4)<=-1
若a>1,loga(4)>0,无解
若0<a<1,
-2<=1/log4(a)<=-1
-1<=log4(a)<=-1/2
1/4<=a<=1/2

∴a=1/2,y有最大值0

y=[2+loga(x)]*1/2[1+loga(x)]=1/2[loga(x)+3/2]^2-1/8
当a=1/2,log1/2(x)=-3/2,x=(1/2)^(-3/2)=2√2,y最小值=-1/8
2√2∈[2,4],符合题意

∴a的值为1/2