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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 $g(y)=\log(1+y)$.

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1 Answers1

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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.

  • Yes, you are right that the family of solutions will be $f(x)=\log(1+x)+c$ and $g(y)=\log(1+y)-c$. But how do I prove it? – user330477 May 10 '17 at 13:55
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    @user330477 rearrange the equality as I state in my answer. Observe that either side of the equation will depend on an individual variable. That means that if you change $x$ it will not affect the value of the other side. That implies that each side is constant. –  May 10 '17 at 13:58
  • Can you please me a bit more rigorous. What you are saying is intuitive, but not obvious. – user330477 May 10 '17 at 14:01
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    Well, I think that what I say stems from the fact that $x$ and $y$ are independent variables which is a strong enough condition. If you want more rigour, you can try to get to a contradiction by saying that as $x$ varies the other side of the equality varies as well, but I doubt that it will add value to your answer. Sorry, the details are up to you. –  May 10 '17 at 14:08
  • What do you mean by "will add value to your answer"? – user330477 May 10 '17 at 14:09
  • @user330477 That it's fine and rigorous as it is. There's not much space for rigour. Cheers! –  May 10 '17 at 14:10