# What's the inverse operation of exponents?

You know, like addition is the inverse operation of subtraction, vice versa, multiplication is the inverse of division, vice versa , square is the inverse of square root, vice versa.

What's the inverse operation of exponents (exponents: 3^5)

These functions are the logarithms, and they are fundamentally important. For $$a = b^c$$ (where $$b > 0$$) we write: $$c = \log_b a,$$ which we can take to be the definition of $$\log_b$$. We read the operation as "logarithm, base $$b$$," or "base $$b$$ logarithm".

In particular, we have $$\log_a (a^b) = b \qquad\text{and}\qquad a^{\log_a b} = b.$$ Of special interest is the natural logarithm, denoted by $$\ln$$ or $$\log$$, the logarithm of base $$e$$. (NB that sometimes $$\log$$ can also denote base $$10$$, or base $$2$$, depending on context.)

Logarithmic identities correspond to exponential identities. From example, from the definition we can conclude that $$\log_b (pq) = \log_b p + \log_b q$$ (for $$p, q > 0$$), which corresponds to the identity $$b^{pq} = b^p b^q$$.

Perhaps counterintuitively, sometimes it is convenient to define the natural logarithm first and then define the exponential function $$x \mapsto e^x$$ to be its inverse, which leads to the slightly antiquated name antilog for an exponential function $$x \mapsto b^x$$.

Edit Some of the other answers here pointed out quite rightly that one can also ask about the inverse of functions where the variable is in the base, i.e., functions $$x \mapsto x^a$$, and inverses of these functions$$^*$$ (at least when $$a > 0$$) are just $$x \mapsto x^{1/a}$$, which we often write as $$x \mapsto \sqrt[a]{x}$$. These functions are called power functions (note that the inverse of a power function is again a power function), and we reserve the name exponential function for functions $$x \mapsto b^x$$ where the variable is in the exponent, i.e., those to which the logarithms are inverses.

$$^*$$For some $$a$$ (in particular, even integers), we need to restrict the map $$x \mapsto x^a$$ to $$[0, \infty)$$ in order to take an inverse.

• The other answers were good, but your answer explained it best to me. – warspyking Oct 3 '14 at 12:02
• What about the example in the post? What is the inverse function then? 3^(1/5) or 'base 3 logarithm of 243'? How can I know just by seeing the function which one is the variable here? – Kawrno Oct 24 '19 at 13:05
• That's really the point of the edit: Is the question asking about the inverse of $x \mapsto 3^x$ or the inverse of $x \mapsto x^5$? From just the expression $3^5$ alone there's no way to tell, much like asking about a function whose evaluation yields the expression $1 + 2$ does not indicate whether the question is about the function $x \mapsto x + 2$ or $x \mapsto 1 + x$. As you can see from the question, I initially understood OP to be asking about $x \mapsto 3^x$, since this is an exponential function, and OP asked about "inverse operation of exponents". – Travis Willse Oct 24 '19 at 21:21

Addition and multiplication are commutative, so there is just one inverse function.

Exponents are not commutative; $2^8 \not= 8^2$. So we need two different inverse functions.

Given $b^e = r$, we have the "$n$th root" operation, $b = \sqrt[e] r$. It turns out that this can actually be written as an exponent itself: $\sqrt[e] r = r^{1/e}$.

Again, given $b^e = r$, we have $e = \log_b r$, the "base-$b$ logarithm of $r$".

• Does division have just one inverse function? 8/4 ≠ 4/8 after all. – Dan W Jul 31 '19 at 20:59
• 8/4 and 4/8 have a simple numerical relationship. (One is 2/1, the other is 1/2.) Whereas 2^8 and 8^2 have no particularly simple relationship (256 and 64, respectively.) Admittedly I didn't explain that very well... – MathematicalOrchid Aug 5 '19 at 8:02

There are two inverse operations of exponentiation.

## Logarithm

$$\log _{b} a$$

It's read "base-$b$ logarithm of $a$". And it means "the exponent which $b$ must be raised to, so that the result is $a$".

## Root

$$\sqrt[b] a$$

It's read "$b$-th root of $a$". And it means "the number which, when raised to $b$, produces $a$".

It depends on what you see as the function and what the variable in $3^5$.

Generalising your "square is the inverse of square root" leads to reciprocal exponents being the inverse of exponents, so $3^5 = 243$ corresponds to $3 = 243^{1/5}$.

Alternatively $3^5 = 243$ corresponds to $5 =\log _{3} 243 = \frac{\log _{10} 243}{\log _{10} 3}= \frac{\log _{e} 243}{\log _{e} 3}$ using logarithms.

Logarithms: $$10^x = 100 \iff x=\log _{10} 100 = 2$$

If you take $x=3^5$, to "get the 5 back" you do $log_3(x)$ and, to "get the 3 back", you do $\sqrt{x}$.
The interesting thing here is that there are 2 ways to reverse the operation, while other operations had just one: If you take $x=2+7$, to "get the 2 back" you did $x-7$ and to "get the 7 back", $x-2$. This happens because 2+7 = 7+2. The sum is "symmetrical" (the right term is commutative). If you want to "get the 2 back" from $x=2+7$, just subtract 7. If you want to "get the 2 back" from $y=7+2$, just subtract 7 again (because, after all, $x=y$).
But $x=3^5$ is not the same as $y=5^3$. So you cant expect to use the same operation to "get the 3 back from x" and "get the 3 back from y"