If $f(xy) = f(x) + f(y)$ for all positive reals $x,y$, what can $f$ be? If given the starting conditions $f(10) = 14$ and $f(40) = 20$. We know $f(40) = f(10) + f(4)$ giving $f(4) = 6$, then $f(4) = 2f(2)$, so $f(2) = 3$, $f(5) = f(10) - f(2)$ so $f(5) = 11$, then $f(500) = f(5) + f(100), f(500) = f(5) + 2f(10)$ giving $11 + 28 = 39$.
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This is a duplicate: https://math.stackexchange.com/q/2843879/ . – Xander Henderson Jul 31 '23 at 18:27
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As noted here, the only continuous solutions are functions of the form $f(x) = k\ln(x)$, where $k$ is a constant. – Arturo Magidin Jul 31 '23 at 18:36
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Your edit has invalidated an existing answer. This is not good. Please refrain from doing that. – user1551 Jul 31 '23 at 18:37
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See also https://math.stackexchange.com/q/2056332/ and https://math.stackexchange.com/q/1408909/ . – Xander Henderson Jul 31 '23 at 18:45
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Thanks Xander, there is another question I would like to ask about functions but am I allowed to ask it here? (its not about the $f(xy) = f(x) + f(y)$ question)/ – Nav Bhatthal Jul 31 '23 at 18:50
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@NavBhatthal If you want to ask a new, unrelated question, use the "Ask Question" button at the top of the page to ask a new question. – Xander Henderson Jul 31 '23 at 19:20
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Edit - this answer is to a previous version of the question that specified $f(xy) = f(x) + f(y)$ for all $x,y\in \mathbb{R}$
This isn't rubbish --- it tells you something useful! In particular it says that $f(1) = 0$, ie $f$ maps one to zero.
Can you figure out something similar about $f(0)$?
Spoiler - don't click this until you've worked it out.
For any $x$, we must have $f(0) + f(x) = f(0\cdot x) = f(0)$, which implies that $f(x) = 0$ for all $x$.
Neal
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