Functional Equation: $f(x\cdot y)=f(x)+f(y)~~~\forall x, y>0$

Question

So I have the question:

$$f(x\cdot y)=f(x)+f(y)~~~\forall x, y>0$$

We haven't learn these in class and I'm assuming it has a logarithmic answer. Could anyone please help?

Answer 1

Let $g(x)=f(e^{x})$ for all $x \in \mathbb R$. Then $g(x+y)=g(x)+g(y)$. [Because $g(x+y)=f(e^{x+y})=f(e^{x}e^{y})=f(e^{x})+f(e^{y})=g(x)+g(y)$]. There are many 'bad' solutions but the only continuous functions satisfying this equation are of the form $g(x)=cx$ where $c$ is a constant. So $f(x)=c\, log \,x$ if $f$ is continuous.

Answer 2

Functional equations are solved using methods such as @Kavi has presented in his anwer. Here I there is another approach for this problem using algebra.

The exponential function is a group homomorphism of $(\mathbb R,+)$ into $(\mathbb R^*,\cdot)$. furthermore, we can restrict ourselves to the subgroup of positive numbers of $\mathbb R^*$ and then $e^{cx}$ becomes a group isomorphism, $c>0$. Then, there exists an inverse function $f(x)$ that is also a group isomorphism of $(\mathbb R^+,\cdot)$ onto $(\mathbb R,+)$. The $f$ is the function we are looking for. A simple computation shoes that $f(x)=\frac{1}{c}\log(x)$.

Remark. As it has been discussed in the comments, I have obtained a family of solutions for the functional equation focus on a particular group homomorphisms (isomorphism). Moreover, it corresponds to the unique family of continuous functions. But there are more functions satisfying the functional equation $f(xy)=f(x)+f(y)$. They also are group homomorphisms of $(\mathbb R^+,\cdot)$ into $(\mathbb R,+)$ but its expression can be very difficult.

Answer 3

From the comments, we discover that you have 5 choices,

[...] This is a mutli choice with the 5 options being: $x, 2^x, \log(x)/\log(5), \sin(x)$ and $\cos(x)$.

So you can check that there's only one of these that works.

• $x$ doesn't work: $1=1\times 1 \neq 1+1=2$
• $2^x$ doesn't work: $1=2^{0\times0} \neq 2^0 + 2^0=2$
• $\sin x$ doesn't work: $\sin((\pi/2)^2)<1$ but $\sin(\pi/2)+\sin(\pi/2)=2>1$
• $\cos x$ doesn't work: $\cos(0\times 0)=1\neq \cos(0)+\cos(0) = 2$

The $\log$ one, you can check does work, using the properties of logs you learned in class. Or, if you trust that there is one correct answer, we are done.