# 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

Note by Yan Yau Cheng
5 years, 6 months ago

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Here's another example, taken from my local math team:

Let $a, b$ be positive real numbers such that $\log_a (b^5) = 3.$ Determine the value of $\log_b (a^5).$

- 5 years, 6 months ago

First of all, we have

$\log_a (b^5) = 3$

$5\log_a b =3$

$\log_a b =\frac{3}{5}$

Applying $\log$ on RHS,

$\log _a b = \log_a a^{\frac{3}{5}}$

Canceling $\log$ from both sides

$b = a^{\frac{3}{5}}$

$b^{\frac{5}{3}} = a$

Therefore,

$\log_b a^5 = \log_b (b^{\frac{5}{3}})^5$

$\log_b a^5 = \log_b b^{\frac{25}{3}}$

$\log_b a^5 = \frac{25}{3}\log_b b$

$\log_b a^5 = \frac{25}{3}\blacksquare$

- 5 years, 6 months ago

How about using $\displaystyle \log_b a = \frac {1}{\log_a b}$ ?

- 5 years, 6 months ago

That's a great idea. We have

$\frac{1}{\log_a b^5} = \log_{b^5}{a} = \frac{1}{5}\log_b a = \frac{1}{3}$.

Then,

$\log_b a =\frac{5}{3}$

$\implies \log_b a^5 = 5\log_b a = \frac{25}{3}\blacksquare$

- 5 years, 6 months ago

loga(b^5)=3. Implies a^3=b^5. Implies, a=b^5/3. Therefore a^5=b^25/3. From this we get 25/3.

- 5 years, 6 months ago

I didn't understand the right side of the equation in the radioactive element example. What is (1/2)^n ? Please explain

- 5 years, 6 months ago

Every year the number of elements will decrease by half, n is the number of years

- 5 years, 6 months ago

Thanks! :D

- 5 years, 6 months ago

logarithms, i had seen them in physics from last year, taking log then antilogs, it was confusing the way our school teachers taught us, so i used the concept without using "log" word.

eg. 23.86 x 3.14 = 10^(1.377) x 10^(0.497) = 10^(1.377 + 0.497) = 10^1.874 = 74.81

• above is possible if you got a log table :) but still we can sometimes aprox it.

- 5 years, 6 months ago

Think logarithm in terms of function and expansion, you will get clear idea.

- 5 years, 6 months ago

what is the relation that is used here

- 5 years, 6 months ago

Actually, the base of the logarithm, $x,$ does not have to be greater than $1.$ It may also be from $0$ to $1.$ However, these logarithms are admittedly not as nice, and are hardly necessary: for example, $\log_{1/2} y$ is the same thing as $-\log _2 y.$

- 5 years, 6 months ago

can anyone tell me how to solve a log within a log..log a ^(log b^a)/log b^(log a^b)..

- 5 years, 1 month ago