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

f(xy)=f(x)+f(y)所以f(8)=f(2√2*2√2)=f(2√2)+f(2√2)=3f(2√2)=3/2f(2√2)=f(2)+f(√2)=3/2f(2)=f(√2*√2)=f(√2)+f(√2)所以f(√2)+f(√2)+f(√2)=3/2f(√2)=1/2