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I was wondering why addition has one inverse (subtraction), multiplication has one inverse (division), but exponentiation has two (radication and logarithm). After a bit of thinking, I thought it might have something to do with the fact that exponentiation isn’t commutative while multiplication and addition ARE, but it still doesn’t entirely answer my question.

Like I said, I know that it has something to do with the fact that exponentiation isn't commutative while addition and multiplication are, but I’d like to know exactly HOW this relates and comparing exponentiation and its inverses to the other two operations and their inverses. Thanks in advance. :)

The_Animator
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7 Answers7

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Addition has two inverses. Subtraction on the left and subtraction on the right...

$\bullet $ Solve $5+x = 12$. Method, subtract $5$ on the left ... $$ 5+x=12 \\ (-5)+(5+x) = -5+12 \\ (-5+5)+x=7 \\ x=7 $$ On the other hand,
$\bullet $ Solve $x+5 = 12$. Subtract $5$ on the right ... $$ x+5=12 \\ (x+5)-5 = 12-5 \\ x+(5-5)=7 \\ x=7 $$ Now, of course, since addition is commutative, we do not need to learn these two methods separately.

But for exponents, $x^y$ is not commutative, and the two inverses are different.
$\bullet $ Solve $2^x = 5$. Use logarithm base $2$... $$ 2^x = 5 \\ \log_2(2^x) = \log_2 5 \\ x = \log_2 5 $$

$\bullet $ Solve $x^2 = 5$. Use radicals ...
$$ x^2 = 5 \\ \sqrt[2]{x^2} = \sqrt[2]{5} \\ x = \sqrt{5} . $$

GEdgar
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    Interesting extension of the first idea: while addition and multiplication are commutative on numbers, multiplication becomes noncommutative for matrices. This makes matrix division "sided", just like in the case of exponentiation for scalars. On paper, we usually write the two divisions as $A^{-1}B$ and $BA^{-1}$, but in computer code (where explicit evaluation of inverses is a bad idea) many matrix-oriented programming languages provide left and right division operators, written maybe A \ B and B / A. – HTNW Dec 18 '23 at 05:17
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    Radicals aren't an "inverse" of exponentiation. It's more exponentiation!!! – Lawnmower Man Dec 18 '23 at 05:55
  • subtraction and division are not commutative, but they do not have two inverses – BlueRaja - Danny Pflughoeft Dec 18 '23 at 06:08
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    Sure they do, at least in the sense that the word "inverse" is being used in this answer (i.e. setting aside the ambiguity surrounding defining an inverse of a binary operation, which is discussed well in another answer). If you have $a - b = c$, then there is one inverse that lets you use $c$ and $b$ to recover $a$, and another inverse that lets you use $c$ and $a$ to recover $b$. (It just so happens that the latter inverse is also a subtraction, so I guess people don't really think of it as an inverse.) – David Z Dec 18 '23 at 09:12
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    @LawnmowerMan Sure, you can think of radicals as exponentiation by the right-hand exponential inverse element. In the same vein, you can think of subtraction as addition by the (left or right-hand) additive inverse element. – Vaelus Dec 18 '23 at 15:17
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If we go back to the definition of function inverses, we recall that for a function $f(x)$, we say that $f^{-1}(x)$ is the inverse of $f(x)$ iff $f^{-1}(f(x)) = x$.

So if we consider the function: $$\operatorname{plus 5}(x) = x + 5$$ then it is obvious that it has an inverse: $$\operatorname{minus 5}(x) = x - 5$$ Hopefully, it is obvious that: $\operatorname{minus 5}(\operatorname{plus 5}(x)) = x$.

If we now consider the function: $$\operatorname{5th power}(x) = 5^x$$

I hope you agree that it has exactly one inverse, namely: $$\log5(x) = \log_5 x$$

and that: $\log5(\operatorname{5th power}(x)) = x$.

On the other hand, if we consider the function: $$\operatorname{5th power}(x) = x^5$$

then it, too, has a single inverse; namely: $$\operatorname{5th root}(x) = x^{1/5}$$ and that: $\operatorname{5th root}(\operatorname{5th power}(x)) = x$.

Now, if you wish to consider both parameters at the same time, we can talk about:

$$\operatorname{plus}(x, y) = x + y$$

but this function doesn't have an inverse, because it must be injective. That is, $\operatorname{plus}(2, 4) = 6$ and $\operatorname{plus}(1, 5) = 6$, so what is $\operatorname{plus}^{-1}(6)$?

So saying that "subtraction is the inverse of addition" is a bit sloppy, as you can see. To even speak of inverses, you have to formulate addition and subtraction in a manner that makes them invertible. The same is true of exponentiation. And once you do this, it becomes clear that there is only one inverse for exponentiation, which is the logarithm. The inverse for raising a value to a fixed power is to again raise it to another fixed power.

More importantly, while the expression "$x^5$" contains an exponent and can be considered an example of "exponentiation", we don't actually call this an exponential function. We call it a "polynomial". That's why recovering $x$ from a polynomial doesn't require a special new function. Whereas, when $x$ is in the exponent, you do have a bona fide "exponential function", and you require a logarithm to recover $x$.

Elements In Space
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Lawnmower Man
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Addition and multiplication are commutative, i.e. $x+y=y+x$ and $x\cdot y=y\cdot x,$ but taking powers, $x^y,$ is not. For example, $2^3=8$ while $3^2=9.$ This is the reason that there are two inverses, one to find $x$ and one to find $y$: if $x^y = a$ then $x=a^{1/y}$ and $y=\log(a)/\log(x).$

md2perpe
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Let's use the extended hyperoperator sequence as a point-of-departure. Now, generally, we can say that there are two orientations for hyperoperators, $\uparrow$ and $\downarrow$, and these orientations are inverses of each other. Set $\uparrow^0$ = the successor operation and $\downarrow^0$ = the predecessor operation (we will run into some complications with this in just a bit, but bear with me). Now, since we're using the version of the hyperoperator sequence with indexes like $\frac{1}{2}$ or $\pi$ or whatever, we have that there are some $\uparrow$-hyperoperations with negative indices. Assume (at least for convenience) that $\downarrow$-hyperoperations with negative indices double-negate to $\uparrow$-hyperoperations with positive indices, e.g.:

$$a \downarrow^{-1} b = a \uparrow^1 b$$

This gives us that every $\uparrow$-hyperoperation has, at most, only two inverses: $\uparrow^{-x}$ and $\downarrow^{x}$.

Now, since $0$ doesn't (usually) have an inverse, the successor operation has only one inverse proper. Note that since the successor operation is (again, usually) taken to be unary rather than binary, it doesn't enter into the unfolding of further hyperoperations via the formula $a \uparrow^n a = a \uparrow^{n+1} 2$. So addition becomes another "base case," or then we set $a \uparrow^{-1} b = a \downarrow^1 b$.± We could justify this, then, by appeal to commutativity, here, and the same will apply to multiplication with division as $a \uparrow^{-2} b = a \downarrow^2 b$.

So this answer isn't meant to displace GEdgar's, incidentally (I think that answer is fit for acceptance even if it doesn't go beyond reasoning from commutativity: that seems to be "all there is to it," ultimately).

±While I remember: first, with addition featuring in the expression of the unfolding of the hyperoperator sequence with indices $n$, that should be an aspect of why addition can be justifiably set to having only one "real" inverse (as a base case). Another aspect of this, then, is that the negative-number operation by itself is unary (though subtraction otherwise speaking is binary), and considering the role of the negative-number operation in the expression of hyperoperation inversion, here (and with subtraction able to reproduced as the addition of a negative number to some other addend).

Kristian Berry
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Maybe a shorter, more elementary answer: In the equation $a=b^c$ there are three numbers. The equation is currently solved for $a$. If you want to solve for $b$, then we have $\sqrt[c]{a} = b$. If you want to solve for $c$, then we have $\log_b a = c.$

There are two "inverses" because $b$ and $c$ play different roles.

B. Goddard
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    While I do know the answer now, if I didn't, this wouldn't have entirely answered my question. I understand your logic, however, I believe this answer could be improved by explaining why this applies solely to exponentation and not multiplication or addition. – The_Animator Dec 19 '23 at 16:49
  • The answer applies to any equation that has 3 variables in it. You can solve for any of the three variables. In the case of addition, two of the solutions look quite a bit alike. – B. Goddard Dec 19 '23 at 19:38
  • Addition and multiplication are commutative. If you have $a = b+c$ you also have $a = c+b$ so the solution for $b$ and the solution for $c$ can be obtained in the same way, via subtraction. Exponentiation is not commutative because $a^b$ is not in general equal to $b^a$. – MJD Dec 19 '23 at 20:30
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$x^y = z$ for a given $z$.

Solve for a given $x$ or solve for a given $y$.

That's 2 ways.

mick
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A perspective I'm surprised not to see here already is that if we consider the binary operator $$a*b = a^b$$ defined on $\mathbb{R}$ say, this is not a very nicely behaved operator. It isn't associative or commutative and doesn't have a left identity.

Addition and multiplication on $\mathbb{R}$, on the other hand, each give you an abelian group, a nice, well understood structure with a lot of theory that results from that structure.

The lack of algebraic structure coming from exponentiation is reflected in there being both "left inverse" and "right inverse" operations (in scare quotes because those descriptions are imprecise and need a bit of care to be formalised precisely).

Joseph Cooper
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