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Is there any real-valued function, $f$, which is not a logarithm, such that $∀ x,y$ in $ℝ$ , $f(x*y) = f(x) + f(y)$?

So far, all I can think of is $z$ where $z(x) = 0$ $∀ x$ in $ℝ$

EDIT:

Functions having a domain of $ℝ^+$ or a domain of $ℝ$/{0} are acceptable as well.

What are examples of functions, $f$, from $ℝ$/{0} to $ℝ$ which are not logarithms, such that
$∀ x,y$ in $ℝ$, $f(x*y) = f(x) + f(y)$?

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    $\begingroup$ If you take a closer look, the constant zero function is also the only function $f:\Bbb R\to\Bbb R$ such that $f(xy)=f(x)+f(y)$, period. Logarithms do not satisfy that property. $\endgroup$ – Saucy O'Path Sep 22 '18 at 11:41
  • $\begingroup$ Probably you want functions $\mathbb R_{>0} \rightarrow \mathbb R$. $\endgroup$ – lisyarus Sep 22 '18 at 11:43
  • $\begingroup$ See Overview of basic facts about Cauchy functional equation under the related equations section. $\endgroup$ – dxiv Sep 23 '18 at 19:46
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$f(0\times 0)=f(0)+f(0)$ so $f(0)=0$. Put $x=0$ to get $f(0)=f(0)+f(y)$. Hence $f \equiv 0$. Note that this is the answer when the domain is the whole real line.

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  • $\begingroup$ I already gave that example in the original posting. I wrote, "So far, all I can think of is $z$ where$z(x)=0$ $∀x$ in $R$" $\endgroup$ – IdleCustard Sep 23 '18 at 19:31
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    $\begingroup$ I am not giving an example. I am proving that there is no other solution if the equation is to hold for all $x,y \in \mathbb R$. $\endgroup$ – Kavi Rama Murthy Sep 23 '18 at 23:26
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Yes, there are, at least if you assume the axiom of choice. Then there are functions $g\colon\mathbb{R}\longrightarrow\mathbb{R}$ which are not linear but which satisfy Cauchy's functional equation:

$g(x+y)=g(x)+g(y)$.

Now, define

$f\colon(0,+\infty)\longrightarrow\mathbb R$
by
$f(x)=g\bigl(\log(x)\bigr)$.

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  • $\begingroup$ OP clearly says 'for all $x,y \in \mathbb R$', so the domain is not $(0,\infty)$. $\endgroup$ – Kavi Rama Murthy Sep 22 '18 at 11:47
  • $\begingroup$ @KaviRamaMurthy But the OP also clearly mentions logarithms, so he probably may have mistaken saying "for all $\in \mathbb R$". $\endgroup$ – lisyarus Sep 22 '18 at 12:01
  • $\begingroup$ Quite possible. Let us see how he reacts. $\endgroup$ – Kavi Rama Murthy Sep 22 '18 at 12:03

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