This note explains how the logarithm works, and is supposed to serve as a memory aid for remembering how the logarithm works.
Introduction
The logarithm to the base $b$ of a number $n$ is defined as follows:
\[\newcommand{\Log}{\mathrm{log}} \newcommand{\Cl}[1]{\textcolor{red}{#1}} \newcommand{\Cll}[1]{\textcolor{blue}{#1}} \newcommand{\Clll}[1]{\textcolor{green}{#1}} \newcommand{\lrm}{\: \longleftrightarrow \:} \newcommand{\lr}{$\lrm$} \newcommand{\mr}[1]{\mathrm{#1}} \Log_b n = x \lrm b^x = n\]If we need to know the logarithm $\Log_b$ of $n$, we can say:
 The base $b$ to the power of what ($x$) equals $n$?
Or in other words:
 The logarithm $\Log_b$ of $n$ is the exponent $x$ to which the base $b$ must be raised in order to result in $n$.
Examples
\[\Log_2 32 = 5 \lrm 2^5 = 32\] \[\Log_{10} 1000 = 3 \lrm 10^3 = 1000\] \[\Log_e 70 = \mathrm{ln} 70 = 4.248495 \lrm e^{4.248495} = 70\]Last example: $e$ is Euler’s number $e = 2.71828$
Logarithm vs. Exponentiation
The logarithm is the inverse operation of exponentiation.
Exponentiation
 Input: a base and an exponent
 Output: a number
Examples:
$\mr{base} = 2$ and $\mr{exponent}=5$
 $2^5 = 32$
$\mr{base} = 10$ and $\mr{exponent}=3$
 $10^3 = 1000$
Logarithm
 Input: a base and a number
 Output: an exponent
Examples:
$\mr{base} = 2$ and $\mr{number}=32$
 $\Log_2 32 = 5$
$\mr{base} = 10$ and $\mr{number}=1000$
 $\Log_{10} 1000 = 3$
Convert Between Different Bases
Suppose we know the logarithm $\Log_{b_1}$ of $n$:
\[\Log_{b_1} n = x_1\]Now we want to know the logarithm $\Log_{b_2}$ of $n$. That is, the logarithm of the same number, but to a different base:
\[\Log_{b_2} n = x_2\]We can calculate $x_2$ as follows:
\[\Log_{b_2} n = \frac{\Log_{b_1} n}{\Log_{b_1} b_2} = x_2\]That is:
 The logarithm $\Log_{b_1} n$, that we already know, divided by the logarithm of the new base $b_2$ to the original base $b_1$
Examples

Convert $\Log_2 128 = 7$ from base $2$ to base $10$:
\[\Log_{10} 128 = \frac{\Log_2 128}{\Log_2 10} = \frac{7}{3.321928} = 2.10721\] 
Convert $\Log_e 512 = 6.238325$ from base $e$ to base 2:
\[\Log_{2} 512 = \frac{\Log_e 512}{\Log_e 2} = \frac{6.238325}{0.6931472} = 9\] 
Convert $\Log_e 1000 = 6.907755$ from base $e$ to base 10:
\[\Log_{10} 1000 = \frac{\Log_e 1000}{\Log_e 10} = \frac{6.907755}{2.302585} = 3\]
Note Regarding BigO Notation
Since $\Log_{b_2} n = \frac{\Log_{b_1} n}{\Log_{b_1} b_2}$, the logarithms of different bases differ by the following factor:
\[\frac{1}{\Log_{b_1} b_2}\]This is a constant factor, because it does not contain the input number $n$, which means that it can be drop from the BigO term.
This is the reason that the bases of logarithms don’t matter in the BigO notation.