# Laws of Logarithms

Further Mathematics SS 1 FIRST TERM

WEEK TWO

## SS1 FURTHER MATHS FIRST TERM

LOGARITHMS

CONTENTS:

1. Laws of Logarithms
2. Change of Base of Logarithms.
3. Use of Tables (greater than one and less than one).
4. Logarithmic Equations.

SUB TOPIC: LAWS OF LOGARITHMS

In the last topic indices, we learnt that p = ax , e. g 1000 = 103 where 3 is called the index. We can express the same in logarithms form. Log a p = x or log 10 1000 = 3

What then is logarithms?

The logarithms of a number p to base a, where a is a positive number not equal to 1 is the index to which a must be raised to give p. This shows clearly that indices and logarithms are the same.

Log28 = 3 because 8 = 23

Log 39 = 2 because 9 = 32

Laws of Logarithms

1. Loga(pq) = Logap + Logaq = Multiplication rule

e.g. if Log3(6×5) = Log36 + Log35

1. Logaa = 1 e.g Log1010 = 1
2. Loga(x/y) = Logax – Logay = Division Rule
3. Loga(x)n = nLogax
4. Loga1 = 0
5. Loga(1/x) = Logax-1 = -1Logax
6. If Logby = 1 then y = b
7. If Logba = 1/Logab
8. Logb = 1/nLogbx
9. Logby. Logyb = 1 for b and y positive and not equal to 1
10. Log by = Log by / Log ab

Examples:

1. Simplify Log39 + Log321 – Log37

Solution:

Log39 + Log321 – Log37

= Log3(9×21÷7)

= Log3(9×21/7)

= Log327

= Log333 = 3 Log 33

But Log33 = 1

Therefore 3 x 1 =3

Log39 + Log321 – Log37 = 3

1. Solve completely for x in the equation 4Logx5 = Log5x

Solution:

4Logx5 = Log5x

4Logx5 = 4/Log5x since Logyx = 1/Logxy

therefore 4/Log5x = Log5x = 4 = (Log5x)2

take square root of both sides

x = ±2

Therefore Log5x = ±2

Hence, Log5x = 2 or Log5x = -2

x = 52 or 5-2

x = 25 or 1/25

1. Solve the equation Log4(x2 + 6x + 11) = 1/2

Solution:

Log4(x2 + 6x + 11) = 1/2

x2 + 6x + 11 = 41/2 = 2

x2 + 6x + 11 – 2 = 0

x2 + 6x + 9= 0

(x +3)(x + 3) = 0

x = -3 twice

CLASS ACTIVITIES:

1. Simplify the following

(a) Log327 + 2Log 39 (b) Log x x9 (c) Log5

1. Solve the following

(i) Log10(x2+4) = 2 + Log10x – Log1020 (ii) Log22n – 2Log8n = 4

SUB TOPIC CHANGE OF BASE OF LOGARITHMS

To change the base of logarithms, we follow the procedure below

Let Logax= y then x = ay

Logbx = yLogba

yLogba = Logbx

y = Logbx/Logba

Therefore Logax = Logbx/Logba

Example:

Change the base of the following logarithms to base 10

1. Log381 (b) Log5125 (c) Logcx = d

Solution

1. Log381 = x

Log381 = Log1081/ Log103

= Log334/ Log103

= 4Log103/ Log103 = 3

Therefore Log381 = 4

1. Log5125 = Log3125/ Log105

Log5125 = Log1053/ Log105

Log5125 = 3Log5/ Log5

Therefore Log5125 = 3

1. Logcx = d

d = Log10x/ Log10c

therefore Logcx = Log10x/ Log10c

CLASS ACTIVITICES:

1. Show that

SUB TOPIC: USE OF TABLES (GREATER THAN ONE AND LESS THAN ONE)

Generally logarithms with base 10 are universal. This is why we have the table of logarithms in base ten. The logarithm of any number has two parts. These are the characteristics and mantissa. The characteristic is the integer part.

Consider the Log10530, Log1053 and Log105.3.

The mantissas of all of them are the same. The difference is in the characteristics.

530 = 5.3 x 102 , 53 = 5.3 x 101 whereas 5.3 = 5.3 x 100.

No Log

530 2.7243

53 1.7243

5.3 0.7243

This is useful to evaluate problems of multiplication and division

To check from table of logarithms, you will need to follow the examples

Example:

1. use tables to evaluate the following
2. 65.43 x 1453 (b) 86.31 x 0.6218 (c) 0.07304 ÷ 0.8931

Solution:

No Standard form Log

65.43 6.543 x 101 1.815

1453 1.453 x 103 3.1623

4.9781

-we look up 65 under difference 3

8 1 5 6 Add the difference

• 2

8 1 5 8

14 under five difference 3

16.14 + 9 = 1623

–For multiplication we add

–Look up the antilog table

0.97 under 8 difference 1

9506 + 2 = 9508

The characteristic is used to determine the decimal point location.

Antilog of 0.9781 = 9.508, Hence 9.508 x 104 = 95080

65.43 x 1453 = 95080

1. 86.31 x 0.6218

No Standard form Log

86.31 8.631 x 101 1.9361

0.6218 6.218 x 10-1 1.7937

1.7298

-note: 1 is called bar 1.

Mantissa is always positive but characteristic can be negative or positive; we put the negative sign on it. When there is no sign then, it is positive.

9360 + 1 = 9361

7931 + 6 = 7937

From Antilog table

.7298 = 5358 + 10 = 5368

1. x 101 = 53.6886.31 x 0.6218 = 53 .68

1. 0.07304 ÷ 0.8931

No Standard form Log

0.07304 7.304 x 10-2 2.8635

0.8931 8.931 x 10-1 1.9509

2.9126 subtract for division

8633 + 2 = 8635

9509 + 0 = 9509

Antilog of .9126 = 8166 + 11 = 8177

8.177 x 10-2 = 0.08177

CLASS ACTIVITIES:

Use tables to solve the following

1. 23.82 x 142.8
2. 0.03167 x 102.8 x 0.325
3. 14.87 ÷ 2.314
4. (12.31)2
5. (33.28) ÷ 4.689

SUB TOPIC: LOGARITHMIC EQUATIONS

EXAMPLE 1: Solve for m in the equation

Solution

Equating powers

-4(2m+3) =

• 4(2m + 3) = 5m+1

-12 – 1 = 5m + 8m

-13 = -13m

-1 = m

EXAMPLE 2: Given that , solve for x and y respectively.

Solution

Applying the laws of logarithm

We can now equate terms

8 = yx – x ………………………………………………….. (i)

Also,

Applying the laws of logarithm,

We can now equate terms

4 = yx + x ………………………………………………………(ii)

Solving (i) and (ii) simultaneously, we eliminate yx.

8 = yx – x

-(4 = yx + x)

4 = -2x

X = = – 2

Put x = – 2 into (2)

4 = yx + x will become

4 = y(- 2) + (- 2)

4 = – 2y – 2

4 + 2 = – 2y

6 = – 2y

y =

x, y means – 2, – 3.

CLASS ACTIVITIES:

1. Find n if
2. If

PRACTICE EXERCISE

Objective test:

Choose the correct answer from the alternatives

1. Evaluate Log0.258 (a) ½ (b)2/3 (c) -2/3 (d) -3/2
2. If 2Log42 = x + 1, find the value of x. (a) -2 (b)-1 (c) 0 (d) 1
3. Given that Log3(x-y) = 1 and Log3(2x +y) = 2, find the value of x. (a) 1 (b)2 (c) 3 (d) 4

Essay Questions:

Solve for x giving your answer correct to 3 S. F.

1. 2Log10x + 3Log105 = 2
2. Using logarithm table, evaluate

÷ correct to 3 s.f

1. If , find x and y respectively.
2. Given that
3. Using logarithm table to evaluate

ASSIGNMENT

1. Without using tables, simplify
2. If m and n are positive real numbers such that

Find: (i) a relation between m and n which does not involve logarithms; (ii) the value of n if m = 16. (leave your answer as in surd in the simplest form).

1. If
2. Using logarithms table, evaluate giving your answer to three significant figures.
3. Use tables to evaluate 13.81×142.8

KEY WORDS

• LOGARITHM
• ANTILOGARITHM
• CHARACTERICTICS
• MANTISSA
• DIFFERENCE
• BAR