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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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These functions are the logarithms, and they turn out to be 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 $\log$, which has base $e$. (NB that sometimes $\log$ can also denote base $10$, or base $2$, depending on context.)

Logarithm 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 logarithms first and then define exponents as inverses of logarithms, 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.

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    $\begingroup$ The other answers were good, but your answer explained it best to me. $\endgroup$ – warspyking Oct 3 '14 at 12:02
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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 $r$".

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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"

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