Question

已知f(x)是定义在(0,+无穷大)上的增函数,且满足f(xy)=f(x)+f(y),f(2)=1.(1)求证:f(8)

Answer 1

1.f(xy)=f(x)+f(y),

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

2.f(x)-f(x-2)>3=f(8)

f(x)>f(8)+f(x-2）=f（8（x-2))

f（x）在(0,+无穷大)单调增

x>8（x-2)>0

2