# 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
4 years, 7 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 $$\log_b (a^5).$$

- 4 years, 7 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,

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

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

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

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

- 4 years, 7 months ago

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

- 4 years, 7 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,

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

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

- 4 years, 7 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.

- 4 years, 7 months ago

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

- 4 years, 1 month 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 $$-\log _2 y.$$

- 4 years, 7 months ago

what is the relation that is used here

- 4 years, 7 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.

- 4 years, 7 months ago

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

- 4 years, 7 months ago

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

- 4 years, 7 months ago

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

- 4 years, 7 months ago

Thanks! :D

- 4 years, 7 months ago