What is the relationship between a function's identities and those of its inverse? A while ago I was thinking about what are known in my area as "the laws of exponentiation" and the "the laws of logarithms", a set of identities for the functions of logarithms and exponentiation. A examples from the set of each include $e^{a+b}=e^{a} \times e^{b}$ and $\ln(a \times b)=\ln(a) + \ln(b)$. I found it interesting to notice that the two functions, $e^x$ and $\ln(x)$, seem to have identities that are in some way the reverse of each other. More specifically, adding the inputs for $e^x$ is equivalent to multiplying the outputs and multiplying the inputs for $\ln(x)$ is equivalent to adding the outputs. The pattern that seems to exist is that the operators, addition and multiplication, swap places between the two identities.
Does a relationship like this exist between all functions' identities and their inverses' identities? If so, what is the relationship in general and how can it be proven?
 A: To my surprise and enjoyment there is a general relationship between the identities, and the proof of such takes only a few steps! First, if $f$ is the function for which the identity is known, its identity can be written as
$$f(a \oplus b) = f(a) \otimes f(b)$$
where $\oplus$ and $\otimes$ represent some sort of binary operators. The other thing needed to construct proof is the the fact that $f$'s inverse, $f^{-1}$, by definition relates to it so that $f(f^{-1}(x))=f^{-1}(f(x))=x$. The proof follows and starts with the given identity for $f$,
$$f(a \oplus b) = f(a) \otimes f(b)$$
applying the inverse to both sides results in
$$f^{-1}(f(a \oplus b)) = f^{-1}(f(a) \otimes f(b))$$
which by the defining relationship of $f^{-1}$ and $f$ leads to
$$a \oplus b = f^{-1}(f(a) \otimes f(b)).$$
defining two new variables $c$ and $d$ as $c=f(a)$ and $d=f(b)$ allows the previous to be rewritten as
$$f^{-1}(c) \oplus f^{-1}(d)=f^{-1}(c \otimes d)$$
since, using the defining relationship of $f^{-1}$ again, $f^{-1}(a)=c$ and $f^{-1}(b)=d$.
With that final line of $f^{-1}(c) \oplus f^{-1}(d)=f^{-1}(c \otimes d)$ the relationship hypothesized was proven.

The proof can also be generalized to situations where the identity involves any number variables. The only change is that the identity is written as
$$f\left(\bigoplus x_n\right) = \bigotimes f(x_n)$$
where $\bigoplus$ and $\bigotimes$ are operators that take $n$ inputs (that is, have arity $n$) and $x_n$ are a set of $n$ variables. The proof's steps themselves are
$$f\left(\bigoplus x_n\right) = \bigotimes f(x_n)$$
$$f^{-1}\left(f\left(\bigoplus x_n\right)\right) = f^{-1}\left(\bigotimes f(x_n)\right)$$
$$\bigoplus x_n = f^{-1}\left(\bigotimes f(x_n)\right)$$
($y_n=f(x_n), f^{-1}(y_n)=x_n$)
$$\bigoplus f^{-1}(y_n) = f^{-1}\left(\bigotimes y_n\right)$$
