f(x)=2(log2^X)^2+2alog2^1/x+b=f(x)=2(log2^X)^2-2alog2^x+b,
令t=log2^x,
f(t)=2t^2-2at+b
f(t)=2(t-a/2)^2+b-a^2/2
x=1/2时,f(x)最小值为-8
从而t=log2^(1/2)=-1时,f(t)有最小值-8
即 -1-a/2=0, a=-2时f(t)有最小值b-a^2/2=-8,b=-6
f(x)=2(log2^X)^2+4log2^x-6
f(x)>0, 2(log2^x-1)(log2^x+3)>0
log2^x1
即x的取值范围：0