Question

（图不清楚处：题中2的指数为k-1)

Answer 1

由于：log2(x)+log2[3*2^(k-1)-x] >=(2k-1) -----[1]

则由真数大于0

得：x>0,且3*2^(k-1)-x>0

设t=2^k (t>0)

则：0=log2[2^(2k-1)]

即：log2[(3/2)tx-x^2]>=log2[(1/2)t^2]

则：(3/2)tx-x^2>=(1/2)t^2

2x^2-3tx+t^2