log2(2)=1<log2(3)<2=log2(4)
∴4<log2(3)+3=log2(3)+log2(8)=log2(24)
f(log2(3))=f(log2(3)+1)=f(log2(3)+2)=f(log2(3)+3)=(1/2)^log2(24﹚
=1/[2^log2(24)]=1/24
∵1
= (1/2)^(log2(3)+3)
=1/(3*2^3)
=1/24
f(log2(3))
=f(log2(3)+1+1+1)
=f(log2(3)+log2(8))
=f(log2(3*8))
=f(log2(24))
=f(log1/2(1/24))
=(1/2)^log1/2(1/24)
=1/24
(log2(3)+2<4