# Does there exist a function which converts exponentiation into addition?

A useful property of the logarithm is that it can "convert" multiplication into addition, as in

$$\ln(a)+\ln(b)=\ln(ab) \text{ for all } a, b \in \mathbb{R}^+$$

Does there exist a function $$f$$, which holds a similar property for exponentiation?

$$f(a)+f(b)=f(a^b) \text{ for all } a, b \in \mathbb{R}^+$$

If so, are there any closed-form expressions for such a function?

• It can't exist as $f(a)+f(b)$ is symmetrical with respect to swap of $a$ and $b$, while $f(a^b)$ is not.
– user700480
Jun 19, 2022 at 8:02
• E.g. $f(x)=f(x^1)=f(x)+f(1)=f(1)+f(x)=f(1^x)=f(1)$ so $f$ is constant, $f(x)=c$ and then $c=f(1)=f(1^1)=f(1)+f(1)=c+c$, i.e. $c=0$.
– user700480
Jun 19, 2022 at 8:08
• (so, technically speaking it does exist - a constant function $f(x)=0$ - but I doubt that is what you are looking for.)
– user700480
Jun 19, 2022 at 11:28
– Jojo
Jun 19, 2022 at 18:38
• Viewed in another way, the difference is that multiplication is commutative while exponentiation isn't. Jun 19, 2022 at 20:53

As Stinking Bishop mentions in the comments, if $$f(a)+f(b)=f(a^b)$$, then it follows (by interchanging $$a$$ and $$b$$) that $$f(b)+f(a)=f(b^a)$$. Hence, $$f(a^b)=f(b^a)$$ for all $$a,b\in\mathbb R^+$$. Setting $$b=1$$, we see that $$f(a)=f(1)$$ for all $$a\in\mathbb R^+$$. Now $$f(1)+f(1)=f(1^1)$$, so $$f(1)=0$$. Hence, the only function $$\mathbb R^+\to\mathbb R$$ with the desired property is the zero function.

While your question as stated has been answered, I'd like to give a satisfying answer to a slight alteration of your question. To avoid the issue of commutivity, we can instead look at commutative hyperoperations $$F_n(a,b)$$. A few of them are:

\begin{align*} F_1(a,b)&=a+b=b+a\\ F_2(a,b)&=ab=ba\\ F_3(a,b)&=a^{\ln b}=b^{\ln a}\\ \end{align*}

Define $$\ln^{(k)}(x):=\underbrace{\ln(\ln(\dots\ln(x)))}_{k\text{ times}}$$. It is true that, in general:

$$\ln^{(k)}(F_n(a,b))=F_{n-k}(\ln^{(k)}(a),\ln^{(k)}(b))\tag{*}$$

for all integers $$n\geq 1$$ and $$0\leq k < n$$.

In other words, if we want to "go down" $$k$$ operations, then we need to use the $$\ln$$ function $$k$$ times.

Indeed, this aligns with the standard rule that $$\ln(ab)=\ln a +\ln b$$, or in our commutative hyperopration notation, $$\ln(F_2(a,b)) = F_1(\ln a,\ln b)$$.

So, for (my slight alteration of) your question, we want a function $$f$$ that satisfies

$$f(F_3(a,b))=f(a^{\ln b})=f(a)+f(b)=F_1(f(a),f(b)).$$

Here, $$k=2$$, so we can set $$f(x)=\ln^{(2)}(x)=\ln(\ln(x))$$.

$$\ln(\ln(a^{\ln b}))=\ln(\ln(a)\cdot\ln(b))=\ln(\ln(a))+\ln(\ln(b))$$

Proof of $$(*)$$:

The commutative hyperoperations are defined recursively by

$$F_n(a, b) = \exp(F_{n-1}(\ln(a), \ln(b))).$$

Therefore,

$$F_n(a, b) = \exp^{(k)}(F_{n-k}(\ln^{(k)}(a), \ln^{(k)}(b))).$$

Hence,

$$\ln^{(k)}(F_n(a, b)) = F_{n-k}(\ln^{(k)}(a), \ln^{(k)}(b)).$$