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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)?

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  • $\begingroup$ Root is exponentiation: $\sqrt[n]{a^m}=a^{m/n}$. $\endgroup$ – Andrew Chin Mar 21 at 1:53
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    $\begingroup$ The first thing I can think of is that if you are trying to write the relationship between three variables, the reason why there is an extra equation is because addition and multiplication are commutative while exponentiation is not. $\endgroup$ – Andrew Chin Mar 21 at 1:55
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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.)

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  • $\begingroup$ But why is exponentiation not commutative? $\endgroup$ – Joshiepillow Mar 21 at 14:25
  • $\begingroup$ Most binary operations are not commutative. Why should we expect exponentiation to be commutative? Addition and multiplication are very special operations. $\endgroup$ – Greg Martin Mar 21 at 17:20

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