Chapter

Logarithms
4
Syllabus reference: 1.2, 2.2, 2.6, 2.8
Contents: A Logarithms in base 10
B Logarithms in base a
C Laws of logarithms
D Natural logarithms
E Exponential equations using logarithms
F The change of base rule
G Graphs of logarithmic functions
H Growth and decay
110 LOGARITHMS (Chapter 4)

OPENING PROBLEM
In a plentiful springtime, a population of 1000 mice will
double every week.
The population after t weeks is given by the exponential
function P (t) = 1000 £ 2t mice.
Things to think about:
a What does the graph of the population over time look
like?
b How long will it take for the population to reach
20 000 mice?
c Can we write a function for t in terms of P , which
determines the time at which the population P is
reached?
d What does the graph of this function look like?

A LOGARITHMS IN BASE 10
y = f(x)
Consider the exponential function f : x 7! 10x y
or f (x) = 10x . y=x

The graph of y = f(x) is shown alongside, along with

its inverse function f ¡1 .
f!1
Since f is defined by y = 10x , 1
f ¡1 is defined by x = 10y .
finterchanging x and yg 1 x
y is the exponent to which the base 10 is raised in order
to get x.

We write this as y = log10 x and say that y is the logarithm in base 10, of x.
Logarithms are thus defined to be the inverse of exponential functions:

If f (x) = 10x then f ¡1 (x) = log10 x.

WORKING WITH LOGARITHMS

Many positive numbers can be For example: 10 000 = 104
easily written in the form 10x . 1000 = 103
100 = 102
10 = 101
1 = 100
0:1 = 10¡1
0:01 = 10¡2
0:001 = 10¡3
 LOGARITHMS (Chapter 4) 111

p p 1
Numbers like 10, 10 10 and p
5
can also be written in the form 10x as follows:
10
p 1
2 p 1 1
¡5
10 = 10 10 10 = 101 £ 100:5 p
5
= 10
10
= 100:5 = 101:5 = 10¡0:2
In fact, all positive numbers can be written in the form 10x by using logarithms in base 10.

The logarithm in base 10 of a positive number is the exponent when the number is written
as a power of 10.

For example:
² Since 1000 = 103 , we write log10 1000 = 3 or log 1000 = 3.
² Since 0:01 = 10¡2 , we write log10 (0:01) = ¡2
If no base is indicated we
or log(0:01) = ¡2: assume it means base 10.
log a means log10 a.
a = 10 log a for any a > 0.

Notice that a must be positive since 10x > 0 for all x 2 R .

Notice also that log 1000 = log 103 = 3
and log 0:01 = log 10¡2 = ¡2.

We hence conclude that log 10 x = x for any x 2 R .

Example 1 SelfSelf Tutor

p
a Without using a calculator, find: i log 100 ii log( 4 10).
b Check your answers using technology.
p 1
1
a i log 100 = log 102 = 2 ii log( 4 10) = log(10 4 ) = 4
b i Casio fx-CG20 ii TI-84 Plus

GRAPHICS
CALCUL ATOR
INSTRUCTIONS

p
log 100 = 2 log( 4 10) = 0:25

EXERCISE 4A
1 Without using a calculator, find:
a log 10 000 b log 0:001 c log 10 d log 1
µ ¶
p p 1 ¡ p ¢
e log 10 f log( 3 10) g log p 4
h log 10 10
10
µ ¶
p
3 100 ¡ p ¢ ¡ p ¢
i log 100 j log p k log 10 £ 3 10 l log 1000 10
10

Check your answers using your calculator.

 LOGARITHMS (Chapter 4) 115

THEORY OF KNOWLEDGE
Acharya Virasena was an 8th century Indian mathematician. Among other areas, he worked with the
concept of ardhaccheda, which is how many times a number of the form 2n can be divided by 2.
The result is the integer n, and is the logarithm of the number 2n in base 2.
In 1544, the German Michael Stifel published Arithmetica Integra which contains a table expressing
many other integers as powers of 2. To do this, he had created an early version of a logarithmic
table.
In the late 16th century, astronomers spent a large part of their
working lives doing the complex and tedious calculations of
spherical trigonometry needed to understand the movement of
celestial bodies. In 1614, the Scottish mathematician John Napier
formally proposed the idea of a logarithm, and algebraic methods
for dealing with them. It was said that Napier effectively doubled
the life of an astronomer by reducing the time required to do
calculations.
Just six years later, Joost Bürgi from Switzerland published
a geometric approach for logarithms developed completely
independently from John Napier. John Napier

1 Can anyone claim to have invented logarithms?

2 Can we consider the process of mathematical discovery an evolution of ideas?

Many areas of mathematics have been developed over centuries as several mathematicians have
worked in a particular area, or taken the knowledge from one area and applied it to another field.
Sometimes the process is held up because a method for solving a particular class of problem has not
yet been found. In other cases, pure mathematicians have published research papers on seemingly
useless mathematical ideas, which have then become vital in applications much later.
In Everybody Counts: A report to the nation on the future of Mathematical Education by the National
Academy of Sciences (National Academy Press, 1989), there is an excellent section on the Nature
of Mathematics. It includes:
“Even the most esoteric and abstract parts of mathematics - number theory and logic, for
example - are now used routinely in applications (for example, in computer science and
cryptography). Fifty years ago, the leading British mathematician G.H. Hardy could boast
that number theory was the most pure and least useful part of mathematics. Today, Hardy’s
mathematics is studied as an essential prerequisite to many applications, including control of
automated systems, data transmission from remote satellites, protection of financial records,
and efficient algorithms for computation.”

3 Should we only study the mathematics we need to enter our chosen profession?
4 Why should we explore mathematics for its own sake, rather than to address the needs of
science?
116 LOGARITHMS (Chapter 4)

C LAWS OF LOGARITHMS

INVESTIGATION DISCOVERING THE LAWS OF LOGARITHMS

What to do:
1 Use your calculator to find:
a log 2 + log 3 b log 3 + log 7 c log 4 + log 20
d log 6 e log 21 f log 80
From your answers, suggest a possible simplification for log a + log b.
2 Use your calculator to find:
a log 6 ¡ log 2 b log 12 ¡ log 3 c log 3 ¡ log 5
d log 3 e log 4 f log(0:6)
From your answers, suggest a possible simplification for log a ¡ log b:
3 Use your calculator to find:
a 3 log 2 b 2 log 5 c ¡4 log 3
d log(23 ) e log(52 ) f log(3¡4 )
From your answers, suggest a possible simplification for n log a.

From the Investigation, you should have discovered the three important laws of logarithms:

If A and B are both ² log A + log B = log(AB)

positive then: µ ¶
A
² log A ¡ log B = log
B
² n log A = log (An )

More generally, in any base c where c 6= 1, c > 0, we have these laws of logarithms:

If A and B are both ² logc A + logc B = logc (AB)

positive then: µ ¶
A
² logc A ¡ logc B = logc
B
² n logc A = logc (An )

Proof:

² clogc A + logc B ² clogc A ¡ logc B

= clogc A £ clogc B =
clogc A
=A£B clogc B
A
= clogc (AB) =
B ¡ ¢
) logc A + logc B = logc (AB) logc
A

=c B
³ ´
A
) logc A ¡ logc B = logc
B