Asymmetry in Exponentiation

Question

So for addition, there are two operations, addition and its inverse, subtraction. \begin{equation} x + y = z \end{equation} \begin{equation} y + x = z \end{equation} \begin{equation} z - y = x \end{equation} \begin{equation} z - x = y \end{equation} Same for multiplication, multiplication and division. \begin{equation} x * y = z \end{equation} \begin{equation} y * x = z \end{equation} \begin{equation} z / y = x \end{equation} \begin{equation} z / x = y \end{equation} But for exponentiation, there appears to be three operations, exponentiation, root, and logarithm. \begin{equation} x^y = z \end{equation} \begin{equation} \sqrt[y]{z} = x \end{equation} \begin{equation} \log_{x} z =y \end{equation} Why is this the case? And does this continue for "higher" operations (for example, tetration)?

Answer 1

This is actually a really nice observation!

You're right that the inverse operation to addition is subtraction. But let's be a little more precise, because the sense of "inverse" here is really the inverse of a function—so let's be more precise about what function we're talking about.

When we have $x+y=z$, we can think of $y$ as fixed and that we are applying a function, the "adding $y$ function" $f_y$, to $x$, to yield $f_y(x) = x+y = z$. The inverse function $f_y^{-1}$ is definitely the "subtracting $y$ function", so that $f_y^{-1}(z)=z-y=x$.

Alternatively, we can think of $x$ as fixed and that we are applying a function, the "adding $x$ function $f_x$", to $y$, to yield $f_x(y) = y+x=z$. Since addition is commutative, this works exactly the same, and we get $f_x^{-1}(z) = z-x=y$.

Similarly, when we have $xy=z$, we can think of $y$ as fixed and that we are applying a function, the "multiplying by $y$ function" $g_y$, to $x$, to yield $g_y(x) = xy = z$. The inverse function $g_y^{-1}$ is definitely the "dividing by $y$ function", so that $g_y^{-1}(z)=z/y=x$. Alternatively, we can think of $x$ as fixed and that we are applying the "multiplying by $x$ function $g_x$" to $y$, to yield $g_x(y) = yx=z$. Since multiplication is commutative, this works exactly the same, and we get $g_x^{-1}(z) = z/x=y$.

Finally, when we have $x^y=z$, we can think of $y$ as fixed and that we are applying a function, the "raising to the $y$th power function" $r_y$, to $x$, to yield $r_y(x) = x^y = z$. The inverse function $r_y^{-1}$ is definitely the "taking the $y$th root function", so that $r_y^{-1}(z) = \sqrt[y]z=x$.

Alternatively, we can think of $x$ as fixed and that we are applying a function, the "exponentiating with base $x$ function" $e_x$", to $y$, to yield $e_x(y) = x^y=z$. However, exponentiation is not commutative—$x^y$ and $y^x$ are not the same. So the inverse function $e_x^{-1}$ is not another "taking a root" function, but rather the "base $x$ logarithm function", yielding $e_x^{-1}(z) = \log_x z=y$.

In summary: the difference between exponentiation and addition/multiplication is that it is not commutative; this means that it matters which variable we treat as the "function" when we take its inverse. (I see Andrew Chin mentioned this very point in a comment.)