Levelling-Up Basic Mathematics

Logarithms
Asante Jude

The aim of this document is to provide a short,

self assessment programme for students who
wish to acquire a basic competence in the use
of logarithms.
Copyright judeiquee@gmail.com
Last Revision Date: January 16, 2001 Version 1.00
Table of Contents
1. Logarithms
2. Rules of Logarithms
3. Logarithm of a Product
4. Logarithm of a Quotient
5. Logarithm of a Power
6. Use of the Rules of Logarithms
7. Quiz on Logarithms
 3
8. Change of Bases
Solutions to Quizzes
Solutions to Problems
Section 1: Logarithms

1. Logarithms (Introduction)
Let a and N be positive real numbers and let N = an. Then n is called the
logarithm of N to the base a. We write this as

n = loga N.
Examples 1

(a) Since 16 = 24, then 4 = log2 16.

(b) Since 81 = 34, then 4 = log3 81.

(c) Since 3 = , then 1/2 = log9 3.

(d) Since 3−1 = 1/3, then −1 = log3(1/3).

 5
Section 1: Logarithms

Exercise
Use the definition of logarithm given on the previous page to
determine the value of x in each of the following.

1. x = log3 27
2. x = log5 125
3. x = log2(1/4)
4. 2 = logx(16)
 6
5. 3 = log2 x
Section 2: Rules of Logarithms

2. Rules of Logarithms
Let a,M,N be positive real numbers and k be any number. Then the
following important rules apply to logarithms.

(a) (b) (c) (d)

 7

Section 3: Logarithm of a Product

3. Logarithm of a Product
1. ←− Proof that loga MN = loga M + loga N.
 8
Examples 2
(a) log6 4 + log6 9 = log6(4 × 9) = log6 36.

If x = log6 36, then 6x = 36 = 62.

log
Thus log6 4 + 6 9 = 2.
.
Now 20 = 5 so log .

Quiz. To which of the following numbers does the

expression log3 15 + log3 0 · 6 simplify?

4 3 2 1

(a) (b) (c) (d)

 9
Section 4: Logarithm of a Quotient

4. Logarithm of a Quotient
1. ←− Proof that

Examples 3
.
If x = log2 8 then 2x = 8 = 23, so x = 3.

(b) If log3 5 = 1.465 then we

can find log3 0 6.
Since 3/5 = 0 · 6, then log.

Now log3 3 = 1, so that log3 0 · 6 = 1 − 1 · 465 = −0 · 465

(a) (b) (c) (d)
 10
Quiz. To which of the following numbers does the
expression log simplify?
0 1 2 4
Section 5: Logarithm of a Power

5. Logarithm of a Power
1. ←− Proof that Examples 4

(a) Find log10 (1/10000). We have 10000 = 104, so 1/10000 = 1/104 =

10−4.
Thus log , where
we have used rule 4 to write log10 10 = 1.

(b) Find log36 6. We have 6 = .

(a) (b) (c) (d)

 11

Thus log .

Quiz. If log3 5 = 1·465, which of the following numbers is log3 0·04?

-2.930 -1.465 -3.465 2.930

Section 6: Use of the Rules of Logarithms

6. Use of the Rules of Logarithms

In this section we look at some applications of the rules of logarithms.

Examples 5

(a) log4 1 = 0.
 12
(b) log10 10 = 1.

(a) (b) (c) (d)

 13
Section 6: Use of the Rules of Logarithms

Exercise
Use the rules of logarithms to simplify each of the
following.

1.

2.

3.

4.
Section 8: Change of Bases 14
5.
Section 7: Quiz on Logarithms

7. Quiz on Logarithms
In each of the following, find x.
Begin Quiz
1. logx 1024 = 2
(a) 23 (b) 24 (c) 22 (d) 25

2.

3. logc(10 + x) − logc x = logc 5)

 15
(a) 2.5 (b) 4.5 (c) 5.5 (d) 7.5

End Quiz
8. Change of Bases
There is one other rule for logarithms which is extremely useful in
practice. This relates logarithms in one base to logarithms in a
different base. Most calculators will have, as standard, a facility for
finding logarithms to the base 10 and also for logarithms to base e
(natural logarithms). What happens if a logarithm to a different base,
for example 2, is required? The following is the rule that is needed.

loga c = loga b × logb c

Section 8: Change of Bases 16
1. ←− Proof of the above rule
Section 8: Change of Bases 17

The most frequently used form of the rule is obtained by rearranging

the rule on the previous page. We have

loga c = loga b × logb c so log .

Examples 6

(a) Using a calculator we find that log 10 3 = 0 · 47712 and log 10 7 = 0 ·

84510. Using the above rule,

.
(b) We can do the same calculation using instead logs to base e.
Using a calculator, loge 3 = 1 · 09861 and loge 7 = 1 · 94591. Thus
Section 8: Change of Bases 18

.
The calculations have all been done to five decimal places, which
explains the slight difference in answers.
(c) Given only that log10 5 = 0 · 69897 we can still find log 2 5, as

follows. First we have 2 = 10/5 so

Then
Solutions to Quizzes 19

.
Solutions to Quizzes
Solution to Quiz:

Using rule 1 we have

log3 15 + log3 0 · 6 = log3(15 × 0 · 6) = log3 9

But 9 = 32 so

log3 15 + log3 0 · 6 = log3 32 = 2.

Section 8: Change of Bases 20

End Quiz
Solutions to Quizzes 21

Solution to Quiz:

Using rule 2 we have

Now we have 12 .
Thus log .
If x = log2 2 , then 2 = 2 , so x = 4. End Quiz Solution to Quiz:
4 x 4
Solutions to Problems 22

Note that
0 · 04 = 4/100 = 1/25 = 1/52 = 5−2.

Thus
.
·−
Since log3 5 = 1 · 465, we have

log3 0 · 05 = −2 × 1 · 465 = −2.930.

End Quiz

Solutions to Problems
Problem 1.
Solutions to Quizzes 23

Since
x = log3 27
then, by the definition of a logarithm, we have 3x
= 27.
But 27 = 33, so we have
3x = 27 = 33,
giving
Solutions to Problems 24

x = 3.
Solutions to Problems 25

Problem 2.

Since x = log25 5 then, by the definition of a

logarithm,
25x = 5.
Now
,
so that
,
Solutions to Problems 26

From this we see that x = 1/2.

Problem 3.

Since x = log2(1/4), then, by the definition of a

logarithm,
2x = 1/4 = 1/(22) = 2−2.
Thus x = −2.
Solutions to Problems 27

Problem 4.

Since 2 = logx(16) then, by the definition of

logarithm,
x2 = 16 = 42.
Thus
x = 4.
Problem 5.
Solutions to Problems 28

Since 3 = log2 x, by the definition of logarithm, we

must have
23 = x.
Thus x = 8.
Solutions to Problems 29

Problem 1.

Let m = loga M and n = loga N, so, by definition, M = am and

N = an. Then
MN = am × an = am+n ,

where we have used the appropriate rule for exponents. From this,
using the definition of a logarithm, we have
m + n = loga(MN).

But m+n = loga M +loga N, and the above equation may be written
Solutions to Problems 30

loga M + loga N = loga(MN), which is what we

wanted to prove. Problem 1.

As before, let m = loga M and n = loga N. Then M = am and

N = an. Now we have

,
where we have used the appropriate rule for indices. By the definition
of a logarithm, we have

.
From this we are able to deduce that
Solutions to Problems 31

Problem 1.

Let m = loga M, so M = am. Then

Mk = (am)k = amk = akm,

where we have used the appropriate rule for indices. From this we
have, by the definition of a logarithm,
.
But m = loga M, so the last equation can be written
Solutions to Problems 32

,
which is the result we wanted.
Problem 1. First of all, by rule 3, we have 3log3 2 = log3 (23) =

log3 8. Thus the expression becomes

.
Using rule 1, the first expression in the [ ] brackets becomes

.
Solutions to Problems 33

The expression then simplifies to

log3 4 − log3 4 = 0.

Problem 2. First we

use rule 3:

and
.
Thus
,
Solutions to Problems 34

where we have used rule 1 to obtain the right hand side. Thus

and, using rule 2, this simplifies to

Problem 3.

Dealing first with the expression in brackets, we have

,
where we have used, in succession, rules 3 and 2. Now
Solutions to Problems 35

so that, finally, we have

Problem 4.

Dealing first with the expression in brackets we have

,
where we have used rule 3 first, and then rule 1. Now, using rule 3 on
the first term, followed by rule 2, we obtain
Solutions to Problems 36

5log3 6 − (2log3 4 + log3 18) = log

= log

= log

= log
= 3log
since log3 3 = 1.

Problem 5.
Solutions to Problems 37

The first thing we note is that √3 can be written as 3 . We first

simplify some of the terms. They are

.
Putting all of this together:
Solutions to Problems 38

Problem 1.

Let x = loga b and y = logb c. Then, by the definition of logarithms,

ax = b and by = c.

This means that

c = by = (ax)y = axy,
with the last equality following from the laws of indices. Since c = axy, by
the definition of logarithms this means that

loga c = xy = loga b × logb c.