Question

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)=

要过程..速度，谢谢诶~

Answer 1

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

f(2√2)=f(2)+f(√2)=3/2
f(2)=f(√2*√2)=f(√2)+f(√2)
所以f(√2)+f(√2)+f(√2)=3/2
f(√2)=1/2

Answer 2

f（8）=f(√2)=f(4√2)=f(√2)=f(4)=f(√2)=f(2√2)=f(√2)=f(2)=f(√2)=f(√2)=3