2<3<4所以1所以f[log2(3)适用f(x)=f(x+3)f[log2(3)]=f[log2(3)+3]则4即符合x>4时的f(x0=(1/2)^xlog2(3)+3=log2(3)+log2(2^3)=log2(24)所以f[log2(3)]=(1/2)^[log2(24)]=1/2^[log2(24)]=1/24