# Why are logarithms and exponentials inverse?

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.

• What do mean? What else does $\log_bn$ mean? Sep 30, 2015 at 0:47
• +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??) Sep 30, 2015 at 1:20

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

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

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.

• Thank you marty, very good explanation. Cheers. Sep 30, 2015 at 23:00