1
$\begingroup$

I'm unclear on why ${{{\log_bn=x}}}$ is the inverse of $b^x=n$, other than the fact someone told me that they were.

$\endgroup$
2
  • 2
    $\begingroup$ What do mean? What else does $\log_bn$ mean? $\endgroup$ Sep 30, 2015 at 0:47
  • 1
    $\begingroup$ +1) For a rather informal but very informative way to learn about logarithms and exponentials being their inverses, you may study a pre calculus book. I think it addresses a lot of doubts you may have. A copy of a pre calculus book can be had for very little from an online vendor. (a math book is never a waste, is it??) $\endgroup$
    – imranfat
    Sep 30, 2015 at 1:20

2 Answers 2

4
$\begingroup$

They are defined that way; the log function is defined to be the inverse of the exponential.

$\endgroup$
1
  • 1
    $\begingroup$ Thank you! I thought maybe there was something that would help me understand the relationship besides memorizing it. Appreciate it. $\endgroup$ Sep 30, 2015 at 1:12
1
$\begingroup$

Logs were created so that multiplication could be done with addition. This means that $\log(ab) =\log(a)+\log(b) $.

Therefore, from a table of logs, look up the logs of $a$ and $b$, add them, and then do an inverse lookup to find the value of $ab$.

To make the inverse lookup easier, tables of the inverse function were created, so that, given $\log(c)$, $c$ could be found. If we call this function $alog$, for $arclog$ or inverse $\log$, then $alog(log(x)) = x$ and $\log(alog(x)) = x$.

This means that $alog(\log(a)+\log(b)) =ab $.

It turns out that, if $v$ is the value such that $\log(v) = 1$, then $\log(v^x) =x \log(v) = x $. This means that taking logs and raising $v$ to a power were inverse operations. Explicitly, $v^x = alog(x)$.

Similarly, it is easy to verify that $\log(a^b) =b\log(a), $ so $\log(\log(a^b)) =\log(b\log(a)) =\log(b)+\log(\log(a)) $. Therefore, $a^b =alog(alog(\log(b)+\log(\log(a)))) $.

And they all computed happily ever after.

$\endgroup$
1
  • $\begingroup$ Thank you marty, very good explanation. Cheers. $\endgroup$ Sep 30, 2015 at 23:00

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.