Question

若函数f(x)满足对任意a,b属于R,有f(ab)=f(a)+f(b)

Answer 1

(1)令a=b=1,得f(1)=2f(1),所以f(1)=0

令a=2,b=1/2,得f(2)+f(1/2)=f(1)=0

(2)

f(x²)=f(x)+f(x)=2f(x)

(3)

f(72)

=f(8)+f(9)

=f(2)+f(2)+f(2)+f(3)+f(3)

=3m+2n

谢谢