Let $f:\mathbb{R^+}\rightarrow \mathbb{R} $ a differentiable function such that $f'(x) = \frac{1}{x}$ and $f(1)=0$ . Show that $f(xy) = f(x) + f(y)$ for all $x,y \in \mathbb{R^+}$

It seems to be the logarithm function but I am unable to prove this statement without the fundamental theorem of calculus, is there any other way to prove this fact?

Thanks for your help


3 Answers 3


From the definition $f'(x) = \frac{1}{x}$ with $f(0)=1$ we have $$f(x) \equiv \int_1^x \frac{dt}{t}$$

By a simple change of variables $z = ty$ to the integral above we get the result

$$f(x) = \int_y^{xy}\frac{dz}{z} = \int_1^{xy}\frac{dz}{z} - \int_1^y \frac{dz}{z} = f(xy) - f(y)$$

Note that we can also prove the logarithmic property $f(x^r) = r f(x)$ by applying the change of variables $z = t^{1/r}$. Taking it further we can also quite easily show that $f$ has an unique inverse $E(x)$ satisfying $E'(x) = E(x)$ and $E(x+y) = E(x)E(y)$ so $E(x) = e^x$ where $e = E(1)$ is defined by $\int_1^e\frac{dt}{t} = 1$. This gives an alternative definition of the exponential function and of Euler's constant $e$.


Consider $f(xy)-f(x)$. Differentiating with respect to $x$ yields $yf'(xy)-f'(x)=\frac{y}{xy}-\frac{1}{x}=0$, meaning that $f(xy)-f(x)=C$, where $C$ is a constant. Plug in $x=1$, $f(y)-f(1)=C$ and hence $f(y)=C$. The result follows.

  • $\begingroup$ I like your solution, really simple. $\endgroup$ Jun 13, 2015 at 0:00

Consider $G:\Bbb{R}^+\to\Bbb{R}$ such that $G(x,y)=f(xy)-f(x)-f(y)$.

One has

$${\partial{G(x,y)}\over\partial{x}}={y\over xy}-{1\over x}=0$$ $${\partial{G(x,y)}\over\partial{y}}={x\over xy}-{1\over y}=0$$

Therefore $G$ is constant. $G(x,y)=G(1,1)=0$


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