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Sorry for the trivial question.

If I have the expression $\log(5)$, and the base is $10$, what operation is being performed on the number $5$, in words?

For example, I know that exponents work (say $5^3$) by taking a number and multiplying it by itself the number of time the exponent is equal to.

Can anyone give me a similarly simple definition for logarithms?

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And, how do you understand something like $5^{1/2}$? How do you multiply $5$ by itself half a time? – Arturo Magidin Feb 29 '12 at 19:25
Logarithms are to exponents what division is to multiplication. How do you divide 5 by 12, in words? – Rahul Feb 29 '12 at 19:44
I too am curious about this, when you bang it into a calculator how is the answer approximated, there has to be some kind of operation going on right? – seeker Oct 28 '13 at 18:08
up vote 2 down vote accepted

There is no "operation" (like $+$ or $-$ or $\dots$), the symbolic expression $\log_{10}(5)$ is per definition the one and only number $r \in \mathbb R$ for which $10^r=5$.

For this to make sense, you have to know the following fact: If $a>0$ is any positive number, then there exists a unique $b\in \mathbb R$ such that $10^b=a$. This number depends on $a$, so we write $b=:\log_{10}(a)$.

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It is asking you to find the number, $x$, such that $10^x=5$. There is no simple operation to obtain such an $x$, as with exponents.

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So what is the operation? That's what I'm looking for. – Calvin Froedge Feb 29 '12 at 19:31
The operation is the logarithm base $10$. There is really no simpler way to define it. – Robert Israel Feb 29 '12 at 19:42
I read a history of logarithms and it spoke of a british guy who created a massive table of logarithmic values. When a computer / calculator calculates a logarithm, it has to undergo some process. What is it? – Calvin Froedge Feb 29 '12 at 21:56
Ah, you mean computing a rational approximation, see – Blah Feb 29 '12 at 22:54

We have:
$log_{a}y = x \iff y=a^x$

So in your example, $a=10$ and $y=5$, so $log_{10}5=x \iff 5=10^x$

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