2
$\begingroup$

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?

$\endgroup$
  • 5
    $\begingroup$ And, how do you understand something like $5^{1/2}$? How do you multiply $5$ by itself half a time? $\endgroup$ – Arturo Magidin Feb 29 '12 at 19:25
  • 2
    $\begingroup$ Logarithms are to exponents what division is to multiplication. How do you divide 5 by 12, in words? $\endgroup$ – Rahul Feb 29 '12 at 19:44
  • $\begingroup$ 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? $\endgroup$ – seeker Oct 28 '13 at 18:08
2
$\begingroup$

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 $b=:\log_{10}(a)$.

$\endgroup$
2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ So what is the operation? That's what I'm looking for. $\endgroup$ – Calvin Froedge Feb 29 '12 at 19:31
  • 1
    $\begingroup$ The operation is the logarithm base $10$. There is really no simpler way to define it. $\endgroup$ – Robert Israel Feb 29 '12 at 19:42
  • $\begingroup$ 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? $\endgroup$ – Calvin Froedge Feb 29 '12 at 21:56
  • 1
    $\begingroup$ Ah, you mean computing a rational approximation, see en.wikipedia.org/wiki/Logarithm#Calculation $\endgroup$ – Blah Feb 29 '12 at 22:54
0
$\begingroup$

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

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.