Why are addition and multiplication commutative, but not exponentiation? We know that the addition and multiplication operators are both commutative, and the exponentiation operator is not.  My question is why.
As background there are plenty of mathematical schemes that can be used to define these operators. One of these is hyperoperation 
where
$H_0(a,b) = b+1$   (successor op)
$H_1(a,b) = a+b$   (addition op)
$H_2(a,b) = ab $   (multiplication op)
$H_3(a,b) = a^b$   (exponentiation op)
$H_4(a,b) = a\uparrow \uparrow b$   (tetration op: $a^{(a^{(...a)})}$ nested $b$ times )
etc.  
Here it is not obvious to me why $H_1(a,b)=H_1(b,a)$ and $H_2(a,b)=H_2(b,a)$ but not $H_3(a,b)=H_3(b,a)$
Can anyone explain why this symmetry breaks, in a reasonably intuitive fashion?
Thanks.

Edit. Okay, it turns out this idea has been studied before. See here.

I basically got this idea from user52541's answer. So no claim to originality.
Anyway. Define a sequence of operations $\langle n \rangle : (\mathbb{R}^+)^2 \rightarrow \mathbb{R^+}$ as follows.


*

*For all $x,y \in \mathbb{R}^+$, define $x\langle 0\rangle y = x+y.$

*For all $x,y \in \mathbb{R}^+$ and all $n \in \mathbb{N}$, define $x\langle n+1\rangle y=\exp(\log x\langle n\rangle\log y).$


Then $\langle 0 \rangle$ is addition, and $\langle 1 \rangle$ is multiplication. But $\langle 2 \rangle$ is not exponentiation. Furthermore, we can prove that for all $n$ it holds that $\langle n \rangle$ is both commutative and associative.
Remark. To prove commutativity, the functions $\exp$ and $\log$ don't even need to be inverses of one another. They can just be arbitrary functions. To prove associativity, we need a bit more. In particular, we require that $\exp$ and $\log$ are inverses. However, we're still not really using any of their properties, like $\log(xy) = \log x + \log y$ etc.
Question 1. Do we have existence of inverses at each level (we have negatives at level $0$, reciprocals at level $1$, and WHAT, if anything, at level $2$)?
Question 2. Does a form of distributivity hold?
