# Get a Grip on the Logarithm

Daniel Weibel
Created 28 Aug 2017
Last updated 20 Jan 2018

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:

$\newcommand{\Log}{\mathrm{log}} \newcommand{\Cl}{\textcolor{red}{#1}} \newcommand{\Cll}{\textcolor{blue}{#1}} \newcommand{\Clll}{\textcolor{green}{#1}} \newcommand{\lrm}{\: \longleftrightarrow \:} \newcommand{\lr}{\lrm} \newcommand{\mr}{\mathrm{#1}} \Log_b n = x \lrm b^x = n$

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 Big-O 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 Big-O term.

This is the reason that the bases of logarithms don’t matter in the Big-O notation.