Is there an operator below + and above (power of) Just a question that came up during the math lessons, don't kill me if it's stupid:
I take the logarithms to explain the basic of my question.
assume:
$$\log_b(x  y) = \log_b(x)  + \log_b(y)$$
$$\log_b(x+y) = \log_b(x) ? \log_b(y)$$
what would the ? operator be if it would exists?
Same for this:
$$\log_b(x^y) = y\log_b(x) $$
$$\log_b(x ? y) = \left(\log_b(x)\right) ^ y$$

I'll do a bit of  the case $b=e$ for simplicity. 
You can define your $?$ operator (let me call it $\boxdot$ to have some notation) by 
$$
a\boxdot b=\log(e^a+e^b). 
$$
Then 
$$
\log x\boxdot\log y=\log(e^{\log x}+e^{\log y})=\log(x+y). 
$$
This is not what I would call a nice operation; it is associative and commutative, but not too nice other than that. It has cancellation: if $a\boxdot b=a\boxdot c$, then $b=c$. 
On the bad side, you have things like 
$$
\log 2=\log (1+1)=\log 1\boxdot\log 1=0\boxdot0. 
$$
So $\boxdot$ cannot be seen as any kind of product (as it behaves badly with respect to the usual multiplication). More generally, 
$$
a\boxdot a=a+\log 2.
$$
There is no neutral element: the equality $a\boxdot b=b$ can be written as $e^a+e^b=e^b$, which would require $e^a=0$. 
