Find all possible choices of real-valued functions $f(x)$ and $g(y)$ such that $f (x) + g(y) = log (1 + x + x y + y)$ for all positive $x$ and $y$.

Question

Question:

Find all possible choices of real-valued functions $f(x)$ and $g(y)$ such that $$f(x) + g(y) = \log (1 + x + x y + y)$$ for all positive $x$ and $y$.

Attempt:

Note that $$f(x)+g(y)=\log(1+x)+\log(1+y)\tag{1}$$

Plug in $x=0$, to get

$$f(0)+g(y)=\log(1+y)\tag{2}$$

Plug in $y=0$, to get

$$f(x)+g(0)=\log(1+x)\tag{3}$$

Plug in $x=0$ and $y=0$, to get

$$f(0)+g(0)=0\tag{4}$$

Now, I am stuck because I know by trial and error that the answer is $f(x)=\log(1+x)$ and $g(y)=\log(1+y)$.

Answer 1

In $(1)$ group whatever depends on $x$ in one side of the equality and whatever depends on $y$ on the other side of the equality.

The resulting expression will imply that:

$$ f(x) - log(1+x) = c $$ $$ g(y) - log(1+y) = -c $$

where $c$ is a constant. Think about why.

So, the functions that satisfy your condition are undefined by an arbitrary constant.