题目有误,最大值≠0.
设2^x=t,则t-20t+64≤0, 4≤t≤16, 4≤2^x≤16, ∴ 2≤x≤4.
再设log<a>x=m, 则f(m)=(2+m)(0.5+0.5m)=0.5(m^+3m+2)=0.5(m+1.5)^-0.125, 对称轴m=-0.15,
a>1时, 0<log<a>(2)≤m≤log<a>(4), f(min)=log<a>(2)=-1/8,a=1/2^8<1, ∴ a∈Φ
0<a<1时, log<a>(4)≤m≤log<a>(2)<0,
若log<a>(4)≤-0.5≤log<a>(2),即1/16≤a≤1/4时,log<a>(4)=0,或log<a>(2)=0，无解
若-0.5≤log<a>(4),即0<a≤1/16时,log<a>(4)=-1/8, log<a>(2)=0,无解.
若-0.5≥log<a>(2),即0<a≤1/4时,log<a>(2)=-1/8, log<a>(4)=0，无解.
若把最大值改为3/8,仿上可得a=1/4,把最大值改为21/32,仿上可得a=1/16,