Question

已知函数fx的定义域为(0,正无穷),且满足f(2)=1,f(xy)=f(x)+f(y)

Answer 1

因为f(xy)=f(x)+f(y)

所以f(1×2)=f(1)+f(2)

f(1)=f(2)-f(2)=0

f(4)=f(2)+f(2)=1+1=2

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