设函数y=f（x）的定义域为R+,且f（xy）=f（x)+f(y),f(8)=3,则f（2）是多少?

f(8)
=f(4×2)
=f(4)+ f(2)
f(4)
= f(2×2)
= f(2) + f(2)
所以f(8) = f(2) + f(2) + f(2) = 3
3f(2) = 3
f(2) =1