Logarithms - An Introduction

Logarithms are a type of functions that are very useful.

Definition

A logarithm is the inverse of an exponent. So:

$x^a = y \iff \log_xy=a$

$\log_xy$ is read as: "Log $y$ base x", the log function basically asks, $x$ raised to which power will give $y$?

Note that $x$ is greater than 0 and cannot be 1 and $y$ needs to be positive

Logarithmic Identities

Identity 1: $\log_a 1 = 0$

The First Identity comes from a property of exponents. Any number raised to the power of 0 is 1 (except for $0^0$, that is undefined) so $\log_a 1 = 0$

Identity 2: $\log_a a = 1$

For all $a$ $a^1=a$ Using the Definition of Logarithms: $\therefore \log_a a = 1 \blacksquare$

Identity 3: $\log_a xy = \log_a x +\log_ay$

Let $\log_a x = u$ and $\log_ay=v$

Then $a^u =x$ and \(a^v = y)

\[xy = a^ua^v = a^{u+v}\]

Taking the Log of both sides:

$\begin{array} {l l } \log_a xy & = \log_a a^{u+v} \\ & = u+v\\ & = \log_ax+\log_ay \blacksquare \end{array}$

**Identity 4** : $\log_a \frac xy = \log_a x - \log_ay$

Let $\log_a x = u$ and $\log_ay=v$

Then $a^u =x$ and \(a^v = y)

\[\frac xy = \frac{a^u}{a^v} = a^{u-v}\]

Taking the Log of both sides:

$\begin{array} { l l } \log_a \frac xy & = \log_a a^{u-v} \\ & = u-v \\ &= \log_ax-\log_ay \blacksquare \end{array}$

**Identity 5:** $\log_a x^n = n\log_ax$

Let $\log_ax=u$

Then according to the definition of Logarithms: $a^u=x$

By Raising both sides to the power of n: $x^n = (a^u)^n = a^{nu}$

Take the log of both sides: $\begin{array} {l l } \log_a x^n & = \log_a a^{nu}\\ &= nu\\ &= n\log_ax \blacksquare \end{array}$

This Identity is very important, it is very helpful in problems where something is raised to the power of a variable, you can just take the log and then "take" the variable down and the log term becomes a constant

Identity 6: $\log_ax=\frac{\log_bx}{\log_ba}$

Let $log_ax=u$

Then $a^u=x$

Taking the log of both sides with a base of a completely new number $b$:

$\log_ba^u=\log_bx$

Using Identity 5 which we proved earlier:

$\begin{array} {l l} u\log_ba & =\log_bx \\ u& =\frac{\log_bx}{\log_ba} \\ \log_ax & =\frac{\log_bx}{\log_ba}\blacksquare \end{array}$

Special Logarithms

If a logarithm doesn't have a base (like $\log$) then it is called the "Common Logarithm", it is a logarithm in base 10

$e$ is a special number that is approximately 2.71828... $\log_e x$ is often written as $\ln x$ This is called the Natural Logarithm

Why do we need logarithms?

When Scientists do experiments their results often do not have a linear relationship but instead have exponential relationships that often aren't recognisable easily, but when the logarithms of the results are taken the results will often show linear relationships.

Example:

A Radioactive source has a half life of 1 year, how many years will it take for the radioactive source to have only one tenth of its radioactive elements remaining?

Let $N_0$ be the number of radioactive elements in the beginning.

$\frac 1{10} N_0 = N_0\left(\frac 12\right)^n$ $\frac 1{10} = \left(\frac 12\right)^n$

Before you learnt logarithms, you wont know how to solve this, as the $n$ is at the top, but since we know logs, we can solve it like this:

$\log \frac 1{10} = \log\left(\frac 12\right)^n$ $-1 = n\log\left(\frac 12\right)$ $n = \frac{-1}{\log\left(\frac 12\right)}$

Using a calculator: $n = \frac{-1}{-0.301}$ $n =3.22$

It will take approximate 3.22 years