TITLE: The operation before addition
QUESTION [5 upvotes]: I asked a question about operations and one comment puzzled me.
Given a binary operation $\ast$ on integers at least $2$, define $\ast'$ by $$m\ast' n = \overbrace{m\ast m\ast \cdots \ast m}^{n\text{ times}}.$$
Example : 

if $*$ is $+$ , $*'$ is $×$. Multiplications are a lot of additions.
if $*$ is $×$ , $*'$ is $^$. Exponential are a lot of additions.

The question is what's $*$ when $*'$ is $+$ ?
Additions are a lot of... what?

REPLY [3 votes]: You can define operation $*_1 = +$,
$$x *_{n+1} y = e^{\log x *_n \log y},$$
and
$$x *_{n-1} y = \log\left(e^x *_n e^y\right).$$
Then we have $*_1 = +$, $*_2 = \times$, and $x *_3 y = x^{\log y}$ which isn't quite what you're looking for but is close. We can continue forever in both directions. (Be careful about the domains of these operations.)
In particular, the operation you're interested in, the predecessor of $+$, is $*_0$, defined as:
$$x *_0 y = \log\left(e^x + e^y\right).$$
I'm not aware of a name for this operation but you can feel free to explore it.
As an aside, note that all of these operations are commutative. Specifically, $*_3$ is like a commutative version of exponentiation.