# Operation that Turns Powers into Products Like How Logarithms Turn Products into Sums

$$\newcommand{\pow}{\mathop{\vcenter{\huge{\text{E}}}}\limits}$$

The $$\sum$$ operator can be defined recursively as $$\sum_{i = a}^b f(i) = f(a) + \sum_{i = a + 1}^b f(i).$$

Likewise, the $$\prod$$ operator can be defined as $$\prod_{i = a}^b f(i) = f(a) \prod_{i = a + 1}^b f(i).$$

One can define an operator $$\pow$$ via $$\pow_{i = a}^b f(i) = f(a)^{\pow_{i = a + 1}^b f(i)}.$$

Logarithms have a well-known property of being able to turn products into sums $$\log \prod = \sum \log$$ which is actually what majorly prompted their study in the first place.

Is there analogous function $$\psi$$ that turns powers into products $$\psi \pow = \prod \psi?$$

I imagine that $$\phi$$ is either the super-logarithm or at the very least closely related to it, but I have not been able to verify this property for myself (I find it hard to wrap my mind around tetration).

The only such functions are constants, essentially because multiplication is commutative while exponentiation is not: for any $$x$$, $$\psi(x) = \psi(x^1) = \psi(x)\psi(1) = \psi(1)\psi(x) = \psi(1^x) = \psi(1).$$ Indeed this shows that the only two possible constant functions are $$\psi(x)=1$$ and $$\psi(x)=0$$.
• Oh, yeah. I just realized that for the most part, you cannot turn a noncommutative function into a commutative one with an operator because $\phi(a^b) = \phi(a)\phi(b) = \phi(b^a)$ in the case of this function. Commented Dec 14, 2022 at 5:33