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I'm not sure if this is more educational or more "pure math", but:

For multiplication and addition, there is exactly one inverse operation, namely division and subtraction.

For exponentiation, we have both logarithms and roots.

Why is that the case? Maybe one could argue that exponentiation is "even less commutative" than division and there is a difference in solving for x in $x/4=5$ and $4/x=5$, but that doesn't feel quite right to me...

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    $\begingroup$ Because addition and multiplication are commutative, while exponentiation is not. You are considering functions of two variables and asking for the inverse in one of those variables. When the function is symmetric in the two variables, the two inverses coincide. $\endgroup$ Commented May 23, 2023 at 12:27
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    $\begingroup$ @MichaelBächtold: I'd upvote that for brevity if you made it an answer. $\endgroup$ Commented May 23, 2023 at 14:28
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    $\begingroup$ I don't think roots are an inverse of exponentiation. They are a form of (or extension of) exponentiation. Saying that roots are an inverse of exponentiation is kind of like saying multiplying by 0.1 is the inverse of multiplying by 10. $\endgroup$
    – DKNguyen
    Commented May 24, 2023 at 1:34
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    $\begingroup$ @DKNguyen There is nothing wrong or unusual about inverses being operations of the same type. For example, the inverse of a linear transformation is a linear transformation; the inverse of a rotation is a rotation. $\endgroup$ Commented May 24, 2023 at 11:41
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    $\begingroup$ @DKNguyen: "... is kind of like saying multiplying by 0.1 is the inverse of multiplying by 10." -- Which it is. And I would say so any day. And regarding your last comment, well, the (1-by-1) matrix $0.1$ is the inverse of the (1-by-1) matrix $10$. $\endgroup$ Commented May 24, 2023 at 14:50

5 Answers 5

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The concept of an inverse operation itself is a bit tricky. Often we consider arithmetic operations to be binary operations: $\DeclareMathOperator{\add}{add}$ $\DeclareMathOperator{\subtract}{subtract}$ $$\add(a,b)=a+b,$$ $$\subtract(a,b)=a-b.$$

This is the view we take when we make statements like “addition is commutative,” which is the same as saying $\add(a,b)=\add(b,a)$. However, in this view, addition and subtraction are not inverses of each other. Subtraction does not take the output of addition and return the inputs.

To make sense of the statement that subtraction is the inverse operation of addition, we need to view arithmetic operations as unary operations or single-variable functions: $$\add_a\!(x)=x+a,$$ $$\subtract_a\!(x)=x-a.$$ In this view, it’s true that $\add_a\!$ and $\subtract_a\!$ are inverses. They undo each other. This is why I prefer to tell students that “subtracting $a$ undoes adding $a$.”

However, there’s a problem. We’ve defined $\subtract_a\!(x)$ to mean “subtract $a$ from $x$.” But what about the operation “from $a$ subtract $x$?” From the binary-operation perspective, this is just $\subtract(a,x)$ instead of $\subtract(x,a)$. But from the unary-operation perspective, this is a completely different kind of operation: $$\operatorname{subtractfrom}_a\!(x)=a-x,$$ and it is actually it’s own inverse. Suppose we have a bottle of water that has volume $a$ and we spill an unknown volume $x$. How can we recover the information of how much we spilled? We can take the remaining volume and subtract it from the original volume, $a-(a-x)=x$.

This demonstrates that a non-commutative binary operation has two types of associated unary operations, and they have different inverses. The same is true for division. “Divide $x$ by $a$” and “divide $a$ by $x$” are different functions of $x$ with different inverses. The same is also true for exponentiation, and language for distinguishing the two kinds of functions already exists.

The functions $$\operatorname{power}_a\!(x)=x^a$$ are commonly known as power functions. Their inverses are other power functions, $(x^a)^{1/a}=x$. When $a$ is a positive integer, $x^{1/a}=\sqrt[a]{x}$ is also called a root function.

The functions $$\exp_a\!(x)=a^x$$ are commonly known as exponential functions. Their inverses are logarithmic functions, $\log_a\!(a^x)=x$.

I have to explain this frequently to students in single-variable calculus who are learning derivative rules. Power functions and exponential functions both involve exponentiation, but they are fundamentally different functions with different derivatives: $$\frac{\mathrm d}{\mathrm d x}x^a=ax^{a-1},$$ $$\frac{\mathrm d}{\mathrm d x}a^x=a^x\ln a.$$ The mnemonic I give them is that “$a^x\!$ is an exponential function because the variable is in the exponent.”

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    $\begingroup$ Beautiful, thorough, and educational answer! $\endgroup$ Commented May 24, 2023 at 7:16
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    $\begingroup$ subtact_a(x) vs. subtractfrom_a(x) makes me think of ARM assembly language, which has both a sub instruction and a reverse-subtract instruction, so you can do sub r0, r1, #5 to do r0 = r1 - 5 or rsb r0, r1, #5 to do r0 = 5 - r1. (Where r0 and r1 are registers, and #5 is an immediate constant that becomes part of the machine code for the instruction). Fun fact: it makes sense that ARM has rsb but most ISAs don't, because ARM's final operand can also be a shifted-register, so it's not just immediates constant with asymmetry. "Free" shifts are one of ARM's fun features. $\endgroup$ Commented May 24, 2023 at 14:20
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    $\begingroup$ Good answer. Suggestion: The key part of the answer are the two paragraphs starting with "The functions", so I suggest moving that up to the top and placing the rest as further elaboration underneath. $\endgroup$
    – user18187
    Commented May 25, 2023 at 2:39
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    $\begingroup$ @user24096: Nah, the key part of the answer is the exploration of the relationship between binary operations and unary operations that fix one argument of the binary operation, and how that relates to the concept of an inverse. The section about power and exponential functions is about how those functions fit into the context established by the rest of the answer, and it needs to come after that context is established. $\endgroup$ Commented May 25, 2023 at 11:08
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    $\begingroup$ As a prof. math'n, I think we do a great disservice by teaching add/sub/mult/div are 4 diff. operations and that add & sub and mult & div are inverses. Not only are they non-invertible opns, but there are only 2 opns: add & mult. Add & sub (and mult & div) are the exact same opn. $0$ is the add identity bc for all $x$, you get $x$ back: $x+0=x$ (and $1$ is mult id. bc $x\cdot 1=x$). Then, $a$ has the add inverse $-a$ bc it takes you back to the add id.: $a+-a=0$ (and if $a\ne 0$ its mult inv. is $a^{-1}$ bc $aa^{-1}=1$). Now it's easy to see why you can't div by $0$. $0y=1$. $0^{-1}=y=$? $\endgroup$ Commented May 25, 2023 at 20:09
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Is your variable at the base or the exponent?

An exponential function is a function of the form $f(x)=a^x$ for some constant $a$. In this case, the inverse is indeed given by a logarithm $f^{-1}(x)=\log_a(x)$.

On the other hand, if you place the variable at the base, $g(x)=x^a$, then you get what is better called (to avoid confusion) a power function, whose inverse is given by roots: $g^{-1}(x)=\sqrt[a]{x}$.

This is a nice example to highlight the importance of being precise with the terms being used. "Operation" was not defined properly, nor what an "inverse" of it should be. As a binary operation, "exponentiation" is simply the function $K(x,y)=x^y$, which is not injective and thus does not have any meaningful "inverse".

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    $\begingroup$ Exactly this. When people stop being imprecise, confusion vanishes. $\endgroup$
    – user21820
    Commented May 24, 2023 at 6:01
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(Despite good answers already, I thought the concrete example below could be useful.)

For multiplication and addition, there is exactly one inverse operation, namely division and subtraction.

Yet for subtraction and division, there are two possible inverse operations each. This applies to any non-commutative operation that takes two inputs.

Subtraction: x - y = z

  • To get x back, you compute z + y
  • To get y back, you compute x - z

Division: x / y = z

  • To get x back, you compute z × y
  • To get y back, you compute x / z

Exponentiation: xy = z

  • To get x back, you compute z1/y (y'th root of z)
  • To get y back, you compute logx(z) (base-x logarithm of z)
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    $\begingroup$ (+1) Nice examples and display of them! $\endgroup$ Commented May 24, 2023 at 10:04
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The other answers have shown that it depends on which of the two inputs you fix and which you let vary. As a way to visualize this, here's a plot of the 3D surface $z = x^y$. I've drawn two polynomial curves: $z = x^2$ (blue) and $z = x^3$ (green), and two exponential curves: $z = 2^y$ (red) and $z = 3^y$ (purple).

To visualize the polynomial inverses (roots), rotate the whole picture so that the $z$ axis is horizontal and the $x$ axis is pointing upward. To visualize the exponential inverses (logs), rotate the whole picture so that the $z$ axis is horizontal and the $y$ axis is pointing upward.

A 3D rendering of z = x^y with polynomial and exponential curves illustrated

A sideways look at the polynomial curves and their inverses, with the $y$ axis into/out of the screen:

Head-on view of two polynomial curves and their inverses

A sideways look at the exponential curves and their inverses, with the $x$ axis into/out of the screen:

Head-on view of two exponential curves and their inverses

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In one case base is the variable and in the other case power is the variable. Here you are talking about an inverse of a function and not just an expression. Therefore role of variable is much important and depending on that inverse function must be different.

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