# How to solve a functional equation involving log?

It's given that $$f(xy)=\frac {f (x)}{y}+\frac {f (y)}{x}$$ Also $$x,y>0$$ and $$f(x)$$ is differentiable for $$x>0$$ such that $$f(e)=\frac{1}{e}$$. By the look of the functional equation I am sure it does involve log at some point . By common substitutions I have been able to deduce that f(1)=0 and f(1/e)=-e but I am not sure how to proceed . Any hint is appreciated.

• Substitute $$y=1$$ Mar 20, 2019 at 14:13
• I got f(1)=0 from that Mar 20, 2019 at 14:36

Multiplying with $$xy$$ yields $$(xy)f(xy) = xf(x) + yf(y).$$ Define $$g(x) := xf(x)$$ to get $$g(xy) = g(x) + g(y)$$. Dou you know a/the function that satisfies this?

• Great answer! Thank you for sharing it :). Mar 20, 2019 at 14:17
• got it . There is a log there. and $f(x)=\frac {logx}{x}$ Mar 20, 2019 at 14:48

Let $$g(x)=xf(x)$$ so we have $$g(xy) = g(x)+g(y)$$

Let $$h(x)=g(e^{x})$$, then $$h(x+y) = g(e^{x+y})=g(e^x\cdot e^y)= g(e^x)+g(e^y)=h(x)+h(y)$$

So $$h$$ is Cauchy function and since it is differentiable it is linear, so $$h(x)=ax$$ for some real $$a$$.

Since $$h(1)= g(e)= ef(e)= e{1\over e} = 1$$ so $$a=1$$ and thus $$x = g(e^x)\implies g(x) = \log x$$

• My I ask on what bases you decide some answer is better than other? Mar 20, 2019 at 14:51
• So you are not going to answer me? Mar 20, 2019 at 15:51