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)
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6 Answers
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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 $r$".
answered Oct 3, 2014 at 13:34
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1$\begingroup$ Does division have just one inverse function? 8/4 ≠ 4/8 after all. $\endgroup$– Dan WCommented Jul 31, 2019 at 20:59
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$\begingroup$ 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... $\endgroup$ Commented Aug 5, 2019 at 8:02
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$\begingroup$ Good stuff, you can't express "nth root" in Excel but you can express x^(1/n) just fine. This gets doubly interesting if the exponent is e.g. 1.26 $\endgroup$ Commented Sep 14, 2021 at 18:21
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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 $b^{p + q} = b^p b^q$.
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.
answered Oct 3, 2014 at 11:41
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3$\begingroup$ The other answers were good, but your answer explained it best to me. $\endgroup$ Commented Oct 3, 2014 at 12:02
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$\begingroup$ 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? $\endgroup$– KawrnoCommented Oct 24, 2019 at 13:05
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$\begingroup$ 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". $\endgroup$ Commented Oct 24, 2019 at 21:21
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$\begingroup$ x = pow(base, exponent); base = root(x, exponent); exponent = logn(base, x); 🤔 I find it easier to remember these relationships by using letters representative of their parts, so 'b' for base (just like you do) but then 'e' for exponent rather than 'c'. (granted, there is also e the constant :/ - alas, only so many letters) $\endgroup$ Commented Jan 13, 2023 at 6:10
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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 $a$".
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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.
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Logarithms: $$10^x = 100 \iff x=\log _{10} 100 = 2$$
Read more about them here
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Depends on what you want to "get back".
If you take $x=3^5$, to "get the 5 back" you do $log_3(x)$ and, to "get the 3 back", you do $\sqrt[5]{x}$.
(btw, if you don't know what a log is yet, do not worry. It is just the operation that gets the exponent back =P)
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"
