$\log{(a+b)} = \log{(a)} ♤ \log{(b)} $

Question

Is there any operator ♤ such that the above operation is true for any numbers a, b, and a+b such that they make good logarithmands (their logs are defined)? I am guessing that if ♤ exists, then it is unique (at lest for such values); can you give a proof/counterexample of this assertion? What is this operator ♤ in terms of the more common arithmetic operators; can you describe this new operator to me? (I guess that we could describe it in terms of Taylor series.) Is ♤ actually useful for anything? Last, can we repeat this process where we find an operator ♧ that splits a logarithm of a quantity given by ♤?

Answer 1

$$x♤y=\log(e^x+e^y)$$ has this property. It's the only such operator.

If you mean "does it have a name," no, it doesn't. Unless you want to call it the pullback of $+$ on $\mathbb R^+$ through the function $e^x:\mathbb R\to\mathbb R^+$. It is almost certainly useless, but who knows?

Proving it is unique is pretty easy. Let $a=e^x$ and $b=e^y$. Then if $♤$ has this property, then $$\log(e^x+e^y)=\log(a+b)=\log (a)♤\log(b)=x♤y$$

In general, if $\star$ is a binary operator on $\mathbb R^+$ we can define a binary $♧$ on $\mathbb R$ as:

$$x ♧y=\log(e^x\star e^y)$$

The key question is, is $♤$ an operation on $\mathbb R^+$? That is to say, if $x,y\in\mathbb R^+$ then is $x♤y\in\mathbb R^+$? I'll leave that to you to answer.