# Why roots aren't the inverse of exponentiation but logarithms?

I think it's easy to see it when we look at the inverse of the function "$$f(x) = a^x$$" but I wonder if there's other way to look at it besides just analyzing the function. I was taught my whole life that if I want to get rid of a root I should just do the inverse and ''exponentiate that with its index'' and now I just learn that roots aren't the inverse of exponentiation and are only its fractional exponent and that the real inverse of expo is getting it logarithm.

• So, the inverse of the function $f(x) = x^y$ is the $y$th root (for fixed $y$). But the inverse of $f(x) = y^x$ is the $y$-base logarithm. These are inverses of different functions. Commented May 27 at 0:31
• If you have $f(a, x)=a^{x}$: given $x$ you can use roots to find $a$, given $a$ you can use logarithms to find $x$ Commented May 27 at 1:01
• You can write ”f(x) = a^x” or ’’f(x) = {a^x}’’ to fix the formatting Commented May 27 at 1:51
• Commented May 27 at 14:33

Because the exponential operator is not commutative in its arguments, it actually has two inverses - a left (commutative) inverse and a right (commutative) inverse (see comments below re: naming).

The root function is the left inverse - it undoes the action of the operation on its left argument, returning the original base.

The logarithm is the right inverse - it undoes the action of the operation on its right argument, returning the original exponent.

• I agree with this comment mostly, except that unless I'm mistaken, associativity is the thing that matters to ensure the left and right inverses are the same, not commutativity. Commented May 27 at 0:44
• Different definitions of "left inverse" and "right inverse" exist for associativity and commutativity. In this case, it's ordering of operands (not operators) that matters. I've yet to encounter a naming scheme that differentiates them. Commented May 27 at 0:49
• @NoName well, in associative operations, if an element $a$ has a left inverse $a^{-1}_l$ and a right inverse $a^{-1}_r$, you can prove the two are the same by associativity: $a^{-1}_l = a^{-1}_l(aa^{-1}r) = (a^{-1}_la)a^{-1}_r = a^{-1}_r$. This is why multiplication in rings, if invertible, has a unique inverse, even tho multiplication in non-commutative. My comment was to say that actual property that makes the left and right inverses non-distinct is not commutativity: it's associativity. And indeed $(x^y)^z \neq x^{(y^{z})}$, so taking powers has different left and right inverses. Commented May 27 at 13:47
• @NoName well, left and right don't even really exist in commutative operations, right? So technically, yes. Commutativity does also need to be broken. Commented May 27 at 14:35
• Wow! That makes lot more sense to me right now, thank you! And i for sure will go back to look at commutative/associativity again haha, couldn't even imagine this relating to c/a Commented May 28 at 1:49

An inverse function is one in which you take an output of your original function and get back the corresponding input. In many classes, this intuition is conveyed as solving for the input, $$x$$, in terms of the output, $$f(x)$$ or $$y$$.

This is entirely dependent on the variable of question. In the case of your variable being raised to some power (i.e., $$f(x) = x^a$$), the “undoing” or inverse of this operation would be raising it to the “fractional exponent” as you call it. However, in your example of $$f(x)=a^x$$, the variable $$x$$ is in the exponent.

If we want to isolate the variable in the exponent, we use logarithms to take a variable out of the exponent. Thus, $$f(x) = a^x \implies f^{-1}(y) = \log_{a}(y)$$

• Thank you so much for this! Seeing that the way we choose to place our variable in this specific problem changes our whole domain and codomain so that our inverse can either be a root or a log made it really easier to see! Commented May 28 at 1:56

First let's talk about this symbol in real number set.

The symbol $$a^x$$ can have several meanings:

• As the exponent: it is a function of $$a$$ and $$x$$ is an interger parameter. In this scenario, $$a^x$$ is understood as the result of multiplying $$a$$ by $$x$$ times if $$x\geq 0$$ i.e. $$a^x = \prod_{i = 1}^{x}a$$, or $$\frac{1}{a^{-x}}$$ when $$x < 0$$. Notable usage includes polynomials, the Taylor expansions,... This function only has an inverse for odd $$x$$ because for even $$x$$ we have $$a^x = (-a)^x$$. The inverse is known as the $$x$$-th root. For even $$x$$ one may choose the positive solution depending on the domain of $$x$$, known as the principal root. Notable usage includes Pythagorean theorems, length, norms, many physical quantities...
• As the power function: it is a function of $$x$$ and $$a$$ is a positive real parameter, known as the base. In this scenario, $$a^x$$ is either understood as either the rational power $$a^{m/n} = \root n \of{a^m}$$ where $$x = m/n$$ is a rational number, and $$\root n \of{x}$$ stands for the $$n$$-th principle root of $$x$$; or the limit $$\displaystyle \lim_{r\rightarrow x} a^r$$ where $$r$$ is rational. Notable usage includes the exponential function $$e^x$$, $$p$$-norm,... This function is reversible for all positive $$a$$ because $$\forall x \neq 0 ,\forall a>0, a^x \neq 1$$ therefore $$\forall x,y, \forall a > 0, x\neq y \Rightarrow a^x \neq a^y$$. The inverse is called the logarithm i.e. $$x = \log_a y$$ if $$a > 0$$ and $$a^x = y$$.
• As the exponentiation: in this case it is considered as a function of both of $$a$$ and $$x$$ and we try to combine the two functions above. However the domain of this function is complicated and can lead to disagreement, for example, on whether $$0^0$$ is defined. This function does not have an inverse.

In the complex number field things are more complicated. Generally for a number $$c$$ and an integer $$n$$ there always exist $$n$$ distinct complex numbers $$z$$ such that $$c = z^n$$. Both the exponent and the exponentation are not bijective in any case because $$x^n = (\zeta x)^n$$ where $$\zeta$$ is an $$n$$-th primitive root of unity and $$a^x = a^{x+i2\pi/\ln a}$$ where $$\ln a$$ is the real natural logarithm of $$a$$. (using the extension of $$e^x$$ to $$x\in \mathbb{C}$$)

It really boils down to what function you are talking about.

If we are talking about (like you said) $$f(x) = a^x$$ then the inverse function, $$f^{-1}(x)$$, is equal to $$\log_a(x)$$. But, if we are talking $$g(x) = x^a$$, then the inverse of that, $$g^{-1}(x)$$, is equal to $$x^{1/a} = \sqrt[a]{x}$$. In other words, taking a root is absolutely an inverse function too.

Part of the point here is that $$x^y$$, when you think of it as a function of two variables, behaves very differently than $$xy$$. The product is commutative and associative, so it's not too sloppy to talk about "the inverse of multiplication". But the two functions that I called $$f(x)$$ and $$g(x)$$ above have very different properties.

And because of that, they have different names. $$\exp(x) = e^x$$ is called the exponential function, and its inverse is the natural logarithm. And exponentials in general are functions where the variable is in the exponent.

Meanwhile, when we talk about raising the variable to a fixed power, we call that a power function. Power functions aren't always invertible, but, when they are, the inverse is a root, just like you were thinking.

So the key mathematical part is that $$x^y$$ is different from $$y^x$$. That means that you get different functions when you plug a constant into one variable or the other, and that means that there is a terminological difference that can be confusing.