2^x≤256=2^8x≤8log2x≥1/2=log2(√2)2≥√2设log2x= t （1/2≤t≤3）log根号2根号x/2=log2(x/2)=log2 x -1所以,所求=t/2(t-1)=1/2 *(t-1+1)/(t-1)=1/2 *(1+1/(t-1))所以 -1/2≤t-1≤2所以 1/(t-1)≥1/2 或1/(t-1)...