Therefore, to make (a-a? )/(2^x)？ & gt2a has been established.

I. When a>0 o'clock

(a-a？ )/2a & gt； (2^x)？

Namely (1-a)/2 >; four

Solution: a

Two. When a=0, it does not hold;

Three. When a man

(a-a？ )/2a & lt； (2^x)？

that is

(a-a？ )/2a & lt； 1

solve

a & gt- 1

To sum up, the range of real number A is

(- 1,0)