ENGINEERING MATHEMATICS 2

CONTENTS

1.0 INDICES PAGES
1.1 Laws of Indices
Exercise 1.1 1
1.2 Simplify index expressions by using laws of 3
indices 5
Exercise 1.2 6
Exercise 1.3
7
2.0 LOGARITHM 9
2.1 Law of Logarithms
Exercise 2.1 11
Exercise 2.2 12
Exercise 2.3 14
Exercise 2.4 15
Exercise 2.5 16
2.2 Simplify logarithm expressions by using laws 18
of logarithms 19
Exercise 2.6 22
Exercise 2.7
Exercise 2.8 23
28
PAST YEAR QUESTION 34

DBM 20023 JUNE 2019 SESSION 40
41
DBM 2013 JUNE 2019 SESSION 42
43
DBM 2013 DECEMBER 2018 SESSION

DBM 2013 JUNE 2018 SESSION



INDICES AND
LOGARITHM

Upon completion of this topic, students
should be able to:

CLO 1 : Use algebra and calculus
knowledge to describe the relationship
between various physical phenomena.



DBM20023 : ENGINEERING MATHEMATICS 2
INDICES AND LOGARITHM

1.0 INDICES

Indices are a convenient tool in mathematics to compactly denote the process of
taking a power or a root of a number. Taking a power is simply a case of repeated
multiplication of a number with itself while taking a root is just equivalent to taking
a fractional power of the number. Therefore, it is important to clearly understand
the concept as well as the laws of indices to be able to apply them later in
important applications.

We will first understand the formal notation for writing a number with an index,
followed by the laws governing it.

Index (or Exponent Numbers)
Notation Index (or Exponent Numbers)
Notation
Laws of Indices

In the expression The index or exponent is 2
The base is 3

1

DBM20023 : ENGINEERING MATHEMATICS 2
INDICES AND LOGARITHM

If we can write a number in the following form :

The number y then is said to be equal to the number a raised to the power n. a is

known as the base (the number which is to be multiplied by itself successively)

and n is known as the power or index to which a is raised, or simply the exponent
of a.

It can also be equivalently interpreted as :

n can be any real number. a can be any real number for n ∈ (Integer) and is
restricted to being a positive real number for fractional values of n.

2

DBM20023 : ENGINEERING MATHEMATICS 2
INDICES AND LOGARITHM

1.1 Laws of Indices

For real numbers , and valid bases , , the following basic laws hold,

LAWS OF INDICES

1 × = + When expressions Example :

with the same base are
multiplied, the indices a. 6 3 × 6 2 = 6 3 + 2 = 6 5

are added.

b. −2 × 5 = −2 + 5 = 3

= − When expressions Example :

2 with the same base are

Or divided, the indices are a. 3 ÷ −3 = 3 − (−3) = 6
subtracted.

÷ = − b. 25 = 25−3 = 22
23

3 0 = 1 If any integer or Example :

unknown is power of 0,

the answer is 1 a. 7 0 = 1

b. 0 = 1

4 i. ( ) = × Note that and have Example :
ii. ( ) = a. �22�3 = 2 2 × 3 = 2 6
iii. � � been multiplied to
yield the new index

b. (4 ) 2 = 4 2 2 = 16 2

c. �2� 2 = 22 = 4
32 9
8

3

DBM20023 : ENGINEERING MATHEMATICS 2
INDICES AND LOGARITHM

5 − = 1 , ≠ 0 A negative power Example :
. means “one over” so
everything is sent to −3 1
the bottom of a a. = 3
fraction.
b. �3 −1� 2 = 3 −2 = 1
32

6 A fractional power Example :

. = � √ � 3
3 1
= √ means a root. The a. = �234 5 �
(243) 5

bottom of the fraction = 33
tell you which root to = 27
take and the top tells

you which power. 33

b. �16� 4 = �4�16�

81 81

�23� 3 8
27
= =

Wrong Method Correct Method Wrong Method Correct Method

32 × 33 = (3 × 3)2+3 32 × 33 = 32+3 34 ÷ 32 = �33�4−2 34 ÷ 32 = 34−2
= 95 = 35 = 12 = 32

Wrong Method Correct Method Wrong Method Correct Method

�33�2 = 332 = 39 �33�2 = 33×2 = 36 30 = 3 30 = 0 30 = 1

4

DBM20023 : ENGINEERING MATHEMATICS 2
INDICES AND LOGARITHM

Solve the following 2. × × × × 3. 3 2 × 3 −5 × 3 4
1. × × × ×

∶ 5 ∶ 3 2 ∶ 3 1 3

4. 2 8 ÷ 2 4 5. 7 × 6 ÷ 5 6. 3� 3� 5

∶ 2 4 ∶ 8 ∶ 3 15

7. 2 2 × 3 4 −2 1 1
2 2 5
8. � 4 −2 2�2 9. √4 2

∶ 6 6 −1 ∶ 2 ∶ 4
�

5

DBM20023 : ENGINEERING MATHEMATICS 2
INDICES AND LOGARITHM

1.2 Simplify index expressions by using laws of
indices.

Simplify each of the following:

1. 2 2 + 2 2. 23 +1 × 4 −3
= 23 +1 × 22( −3)
4 −2+ = 23 +1 × 22 −6
= 23 +1+2 −6
= 1 2 + 2 − (−2 + ) = 25 −5
2

= 1 2 + 2 + 2 −
2

= 1 + 4
2

+ 4
=2

3. 3 −1 × 27 ÷ 9 4. 7 +1 − 7 −2
= 3 −1 × 33 ÷ 32 = 7 �7 − 7 −2�
= 3 −1+3 −2 = 7 �34492�
= 34 −3 = 7 −2(342)

6

DBM20023 : ENGINEERING MATHEMATICS 2
INDICES AND LOGARITHM

Simplify index expression by using laws of indices.

1. 53 × 252 −1 ÷ 54 +1 2. −3 − 421 +1 + 3�2 +1�

83

∶ 1 ∶ 17�2 −3�

3. +1 4. 2 −1 + 32 +1 − 2�32 �

2 −1× 3−5 92

∶ 4 −1 ∶ 4�32 −1�
7

1 DBM20023 : ENGINEERING MATHEMATICS 2
2 2 INDICES AND LOGARITHM
5.
3 6. 27� 3−2� − 921 −1 + 2�3 +1�

2

∶ ∶ 6(3 )

Solve the following equations.

1. 3 = 3 +1 − 6 2. 2 +1 = 1
8 −2
3 = 3 ∙ 31 − 6 → 3 =
2 +1 = 1
= 3 − 6 23( −2)

− 3 = −6 2 +1 = 1
23 −6
−2 = −6
2 +1 = 2−(3 −6)
−6
= −2 2 +1 = 2−3 +6

= 3 + 1 = −3 + 6

∴ 3 = + 3 = 6 − 1
4 = 5
3 = 3

3 = 31 ∴ = 1 5
4
=

8

DBM20023 : ENGINEERING MATHEMATICS 2
INDICES AND LOGARITHM

Solve each of the following equations.

1. 3(25 ) = 75 2. 2 −5 = 1
2
16

∶ = 1 ∶ = 4

3. 3 +2 = 3 + 72 4. 3 +2 + 3 +4 = 90

∶ = 2 ∶ = 0
9

5. 2 +2 = 2 +1 + 16 DBM20023 : ENGINEERING MATHEMATICS 2
INDICES AND LOGARITHM

6. 7 − 350 = −7 +2

∶ = 3 ∶ = 1

10

DBM20023 : ENGINEERING MATHEMATICS 2
INDICES AND LOGARITHM

2.0 LOGARITHM

Logarithms were introduced by John Napier in the early 17th century as a
means to simplify calculations. They were rapidly adopted by navigators,
scientists, engineers, and others to perform computations more easily, using
slide rules and logarithm tables. Tedious multi-digit multiplication steps can be
replaced by table look-ups and simpler addition because of the fact—important
in its own right.

In mathematics, the logarithm is the inverse function to exponentiation. That
means the logarithm of a given number x is the exponent to which another
fixed number, the base b, must be raised, to produce that number x.

Expressed mathematically, is the logarithm of to the base if ,

in which case one writes

For example :

11

DBM20023 : ENGINEERING MATHEMATICS 2
INDICES AND LOGARITHM

Therefore, is the logarithm of to base , or

2.1 Law of Logarithms

There are a number of rules known as the laws of logarithms. These allow
expressions involving logarithms to be rewritten in a variety of different ways.
The laws apply to logarithms of any base but the same base must be used
throughout a calculation.

LAWS OF LOGARITHMS

1. (Identity Rule) The logarithm of Example :
loga = 1
a number that is

equal to its base is a. log3 3 = 1

just 1.

b. logp = 1

2. (Zero Rule) The logarithm of 1 Example :
loga 1 = 0 with > 1 equals

zero. a. log5 1 = 0

b. logp 1 = 0

12

DBM20023 : ENGINEERING MATHEMATICS 2
INDICES AND LOGARITHM

3. (Change Base Rule) Pick a new base Example :
and the formula
says it is equal to a. log38 = 10 8
the log of the 10 3
loga = number in the
new base divided
by the log of the b. log29 = 4 9
old base in the 4 2
new base.

4. (Product Rule) The logarithm of Example :

loga =loga +loga a product of two a. log2 =log2 +log2
or more b. logn 3 =logn 3 +logn
numbers is equal
to the sum of the

logarithms of the

numbers or

factors of the

product.

5. (Quotient Rule) The logarithm of Example :

a fraction is

loga =loga −loga equal to the a. log232 =log22 − log23

difference

between the b. log2 3 = log2 3 − log2

logarithm of the

numerator and

the logarithm of

the denominator.

13

DBM20023 : ENGINEERING MATHEMATICS 2
INDICES AND LOGARITHM

6. (Power Rule) The logarithm of Example :
loga = loga an exponential a. log7 7 3 = 3 log7
number is the b. log3 2 = 2 log3
exponent times
the logarithm of
the base.

7. = ⇐⇒ loga = Exponential form Example :

convert to

logarithmic a. 32 = 2 5 ⇐⇒log232 = 5
equation.

b. 10−2 = 0.0 ⇐⇒log20.01 = −2

Write in logarithm form. d. 100 = 10 2
c. 5 2 = 25 f. 3 3 = 27
e. = h. 42 = 7 2
g. 6 1 = 6

14

DBM20023 : ENGINEERING MATHEMATICS 2
INDICES AND LOGARITHM

i. 10 −3 = 0.001 1

3 j. 64 2 = 8

k. �1 � = 1 l. 3 −3 = 1
27
28

Write in Indices form. 2. log2 2 = 1
1. log4 = 3 4. log4 64 = 3
3. logx = 24 6. 2 =log7 49
5. log8 64 = 2
8. log7 343 = 3
7. 4 =log7 2401
9. log 0.001 = −3 10. log4096 8 = 1
4

15

DBM20023 : ENGINEERING MATHEMATICS 2
INDICES AND LOGARITHM

Solve log264 Method : Change to 10 Method : Change base
log264 =log22 6
Method : Indices log2 64 = 10 64
log2 64 = 10 2 = 6log22
= 6 (1)
2 = 64 1.806 =6
2 = 2 6 = 0.301
∴ = 6
=6

Exercise 2.3 : Solve the following 2. log5 125
1. log2 8

∶ 3 ∶ 3
∶ −4
3. log4 256 4. log5 0.0016

∶ 4
16

DBM20023 : ENGINEERING MATHEMATICS 2
INDICES AND LOGARITHM

5. log4 1 6. log3 9 3
64

∶ −3 ∶ 6
∶ 0
7. log2 0.015625 8. logx 1

∶ −6

9. log3 27 2 10. log2 1
4

∶ 6 ∶ −2

11. logm� 1 12. 5 loga 5√

∶ − 1 ∶ 1
2 17

DBM20023 : ENGINEERING MATHEMATICS 2
INDICES AND LOGARITHM

Find the value of . 2. log3 = 1
1. logx 8 = 3

∶ = 2 ∶ = 3

3. logx 64 = 2 4. 2 logx 3 = 1

∶ = 8 ∶ = 9

5. log2 2 = 16 6. 2 log2x 4 = 2

∶ = 256 ∶ = 2

18

DBM20023 : ENGINEERING MATHEMATICS 2
INDICES AND LOGARITHM

Express in term of logn and logn ,

a. logn� 2 b. logn�

�

= logn +logn 2 −logn 1

= logn + 2 logn −logn = logn � � 2

= 1 logn � �
2
= logn +logn
= 1 [logn −( logn + logn )]
2

= 1 [logn − logn − logn ]
2

= 1 [−logn ]
2

= − 12logn

Express the following expression by using laws of logarithms.

1. log 2 3 2. log 1
2 102 2

∶ 2 log + 3 log ∶ −log 102 − 2log
2 19

DBM20023 : ENGINEERING MATHEMATICS 2
INDICES AND LOGARITHM

3. logb 3 2 4. logn 3
2

∶ 3 logb + 2 logb ∶ 3logn − 2 logn
5. logn � 3
6. − 2 loga 3
3 2

∶ 1 logn + 3 logn − 1 logn ∶ −2 loga − 2 loga + 4 loga
2 2 2 3 3

20

DBM20023 : ENGINEERING MATHEMATICS 2
INDICES AND LOGARITHM

7. log� 4 8. log 3
2 3√

∶ 1 log − log − 2 log ∶ 3 log + log − 1 log
2 3

21

DBM20023 : ENGINEERING MATHEMATICS 2
INDICES AND LOGARITHM

2.2 Simplify logarithm expressions by using laws of
logarithms.

Simplify the following expression into a single logarithm,

a. loga + 1 loga b. 4logx4 + 2logx1
3 = logx44 +logx12
= logx16 +logx1
= loga + loga ( )31 = logx16(1)
= logx16
= loga ( ) 1
3

= loga 1 1
3 3

= loga 4 1 or loga3√ 4
3 3

22

DBM20023 : ENGINEERING MATHEMATICS 2
INDICES AND LOGARITHM

Simplify logarithm expressions by using laws of logarithms.

1. 2 log −log 2. 2 log + 3 log − log

∶ log 2 ∶log 2 3
3. log − log − 2 log 3

4. log9 − 1 + 3 log9
2

∶ log9 ∶ log9 3
3

23

DBM20023 : ENGINEERING MATHEMATICS 2
INDICES AND LOGARITHM

5. 2 log2 + 1 log2 3 6. 4 loga + 2 − 3 loga
3 3

∶ log2 ∶ loga 4


7. 1 log10 + 2 log10 − log10 2 8. logx − 1 logx − 3 logx
2 2

∶ log10 √ ∶logx 21√

24

DBM20023 : ENGINEERING MATHEMATICS 2
INDICES AND LOGARITHM

9. log4 9 − log4 36 10. log4 2 − log4 32

∶ −1 ∶ -2
11. 4 loga 2 − 3 loga 4 + loga 6 12. 2 log4 2 − log4 3 + log4 12

∶ loga 3 ∶ 2
2 25

DBM20023 : ENGINEERING MATHEMATICS 2
INDICES AND LOGARITHM

13. 2 log5 10 − 3 log5 2 + log5 32 14. 2 log2 5 − log2 100 + 3 log2 4

∶ log5 400 ∶ 4

Given log4 6 = 0.7737 and log4 8 = 1.1606, Determine the value of followings
without using calculator.

a. log4 36 b. log4 0.75 c. log4 48

= log4 6 2 = log4 6 = log4 6(8)
= 2log4 6 8 = log4 6 +log4 8

= 2 (0.7737) = log4 6 −log4 8 = 0.7737 + 1.1606
= 1.5474 = 1.9343
= 0.7737 − 1.1606

= −0.3869

26

DBM20023 : ENGINEERING MATHEMATICS 2
INDICES AND LOGARITHM

Given log2 3 = 0.1213 and log2 5 = 1.7172, Determine the value of followings without
using calculator.

a. log3 15 b. log2725 c. log4125

= 2 15 = 2 25 = 2 125
2 3 2 27 2 4

= 2 3(5) = 2 5 2 = 2 5 3
23 2 3 3 2 2 2

= 2 3+ 2 5 2 2 5 3 2 5
2 3 3 2 3 2 2 2
= =

0.1213 + 1.7172 2 (1.7172) 3 (1.7172)
= 0.1213 = 3 (0.1213) = 2 (1)

= 15.1566 3.4344 5.1516
= 0.3639 =2

= 9.4378 = 2.5758

27

DBM20023 : ENGINEERING MATHEMATICS 2
INDICES AND LOGARITHM

1. Given loga 3 = 0.477 and loga 5 = 0.699, calculate the value of

a. loga 15 b. loga 0.6

∶ 1.176 ∶ −0.222

c. loga 3√5 d. 45
9

∶ 0.8265 ∶ 0.846

28

DBM20023 : ENGINEERING MATHEMATICS 2
INDICES AND LOGARITHM

2. Given log 2 = 0.3010 and log 3 = 0.4771, calculate the value of

a. log 36 b. log 1.5

∶ 1.5562 ∶ 0.1761

3. Given log3 = and log3 = , 4. Given log5 2 = and log5 7 = ,
find log3 2 2 in terms of and find log5 (2.8)2 in terms of and
. .

∶ 2 + 2 ∶ 2 + 2 − 2
29

DBM20023 : ENGINEERING MATHEMATICS 2
INDICES AND LOGARITHM

5. Given log5 = and log5 = , find the following in terms of and .

a. log5 � 2 � b. log5 �25 � �

∶ 2 − ∶ 2 + + 1
2

Solve the following equations.

1. 9 ∙ 3 −1 = 81 2. 25 = 15625
32 ∙ 3 −1 = 34 52 = 56
32 + − 1 = 34 2 = 6
= 3
33 − 1 = 34

3 − 1 = 4

3 = 5

= 5
3

30

DBM20023 : ENGINEERING MATHEMATICS 2
INDICES AND LOGARITHM

3. 3 − 1 = 1 4. 5 + 1 = 5 − 1 + 24
27 − 2 5 + 1 − 5 − 1 = 24
5 �5 − 5 −1� = 24
3 −1 = 1
33( − 2)

3 −1 = 1 5 �254� = 24
33 −
6

3 − 1 = 3 −(3 − 6) 5 5
24
3 − 1 = 3 −3 + 6 = 24 ×

− 1 = −3 + 6 5 = 5 → 5 = 51

+ 3 = 6 + 1 = 1

4 = 7

= 7
4

5. 4 + 1 = 9 6. 45 −43 = 5

3

log 4 + 1 = log 9 −43 5
× 45
= 3

( + 1)log 4 = log 9

+ 1 = 9 11
4 43 = 27

+ 1 = 1.5850 43 = 27

= 1.5850 − 1 = �4√27�3

= 0.5850 = 81

31

DBM20023 : ENGINEERING MATHEMATICS 2
INDICES AND LOGARITHM

7. log3 3 + log9 = 5 8. log4 ( − 3) = 2 log4 5 − log4 10
2 log4 ( − 3) = log4 52 − log4 10
log4 ( − 3) = log4 �2105�
log3 3 + 3 = 5 − 3 = 2.5
3 9 2 = 2.5 + 3
= 5.5
log3 3 + 3 = 5
3 32 2

log3 3 + 3 = 5
2 3 3 2

log3 3 + 3 = 5
2(1) 2

log3 3 + 1 log3 = 5
2 2

log3 3 + log3 1 = 5
2 2

log3�3 × 21 � = 5
2

log3�3 1+21 � = 5
2

3 3 = 5
2
32

5
32
3 = 3
2

3 = 325−1
2

3 = 3
2
32

= 3

32

DBM20023 : ENGINEERING MATHEMATICS 2
INDICES AND LOGARITHM

9. log2 − log2 (1 − ) = 4 10. log4 ( + 4) − log4 (3 + 1) = 1
2

Log2 1 − = 4 log4 ( + 4) = 1
(3 + 1) 2

1 − = 24 ( + 4) 1
3 + 1 = 42
= 16(1 − )
( + 4)
= 16 − 16 3 + 1 = 2

+ 16 = 16 ( + 4) = 2(3 + 1)

17 = 16 + 4 = 6 + 2

= 16 6 − = 4 − 2
17
= 2

= 2
5

33

DBM20023 : ENGINEERING MATHEMATICS 2
INDICES AND LOGARITHM

Solve each of the following equations.

1. 3 +3 − 3 +1 = 72 2. 5 −1 − 5 −2 = 100

∶ = 1 ∶ = 4

3. 5 = 12 4. 73 +2 = 31

∶ = 1.544 ∶ = −1.619

34

DBM20023 : ENGINEERING MATHEMATICS 2
INDICES AND LOGARITHM

5. 3 + 4 6. 26 − 810 − 2 = 0

3 + 2 × 4 −

∶ 2 − 3 ∶ = 5
2

7. 4 × 33 = 9( +1) 8. 2 + 8(2− ) = 9

∶ = 0.7381 ∶ = 0 ; = 3
35

9. 42 + 1 = 25 DBM20023 : ENGINEERING MATHEMATICS 2
INDICES AND LOGARITHM

10. 62 + 1 = 17

∶ = 0.661 ∶ 0.2906

11. 3 = 5 + 2 12. 3 + 1 = 108 − 3

∶ = −6.3 ∶ = 3
36

DBM20023 : ENGINEERING MATHEMATICS 2
INDICES AND LOGARITHM

13. log3 ( − 6) = 4 − log3 3 14. logx 8 + logx 4 = 5

∶ = 9 ; = −3 ∶ = 2
15. 2 3 ( − 2) = 1
16. 3 2 = 9
2

∶ = 7 ∶ = 4
3 37

DBM20023 : ENGINEERING MATHEMATICS 2
INDICES AND LOGARITHM

17. 3 logx 3 + 4 logx 4 −logx 12 = 1 18. Log2 ( + 2) + log4 ( + 2) = 3
2

∶ = 576 ∶ = 0
19. 3 log 2 + log(4 − 1) =log(7 − 8 ) 20. 2 log 2 + log 3 =log 96

∶ = 3 : = 2
8

38

DBM20023 : ENGINEERING MATHEMATICS 2
INDICES AND LOGARITHM

21. log ( − 5) = 2 + log ( − 1) 22. log2 �5√ � + log4 (16 ) = 16

∶ = 0.9596 ∶ = 3276.8
24. 2 logp 10 − 2 logp �52� = 4 − logp 2
23. 1 log3 + log9 4 = 0
2

∶ = 1 ∶ = 4
2 39

DBM20023 : ENGINEERING MATHEMATICS 2
INDICES AND LOGARITHM

[DBM 20023] JUNE 2019 SESSION
QUESTION 1

CLO 1 (a)
C3 Express each of the following expressions in the simplest form.

i. 3 + 4 ans : a- 3+4n [3 marks]
3 + 2 × 4 − ans : -2 [4 marks]

ii. log5 1
25

CLO 2 (b) Solve the following questions using a suitable method.
C3

i. 32 − 1 × 273 − = 1 ans : x = 8 [4 marks]

ii. logx 8 +logx 4 = 5 ans : x = 2 [4 marks]

40

DBM20023 : ENGINEERING MATHEMATICS 2
INDICES AND LOGARITHM

[DBM 2013] JUNE 2019 SESSION
QUESTION 1

CLO 1 (a) Simplify the following expressions.
C2

i. 4 ÷ �256 − 1 × 128� ans : 2-7x + 1 [5 marks]

ii. log9 36 − log9 1 − log9 4 ans : 1 [5 marks]
9

CLO 1 (b) Solve the following equations.
C3

i. 22 − 5 = 643 ans : − 5 or -0.3125 [4 marks]
16

ii. 5 − 3 = 3 ans : 9.452 [6 marks]

iii. Given that logm 3 = and logm 5 = , write [5 marks]
logm 45 in terms of and .
ans : 2p + q

41

DBM20023 : ENGINEERING MATHEMATICS 2
INDICES AND LOGARITHM

[DBM 2013] DECEMBER 2018 SESSION
QUESTION 1

CLO 1 (a) Simplify the following expressions
C2

i. log4 2 + log4 14 − log4 4 ans : log4 5 [3 marks]
7 3 15 [3 marks]
[4 marks]
ii. 32 − 1 × 3 − 3 ans : 1
33 − 2 9

iii. 25 × 5 + 1 ans : 512m - 5
1252−3

CLO 1 (b) Solve the following equations
C3

i. 42 �8 + 3� = 32(4 − ) ans : x = 9 [4 marks]
8 [5 marks]
[6 marks]
ii. 32 + 1 = 5 ans : x = 0.232

iii. log3 (2x + 5) = 1 + log3 (2x – 3)

ans : x = 7
2

42

DBM20023 : ENGINEERING MATHEMATICS 2
INDICES AND LOGARITHM

[DBM 2013] JUNE 2018 SESSION
QUESTION 1

CLO 1 (a) Express each of the following expressions in the simplest form.
C2

i. 24 × 25 ans : 2-3n [2 marks]
46

ii. 1 logx 16 − logx 4 − 3 logx 2 ans : logx 1 [4 marks]
2 8

iii. 75 +2 1 [4 marks]
33− × 152 +1 × 52 9
ans : x =

CLO 1 (b) Solve the following equations using the suitable method.
C3

i. log3 (3 + 1) − log3 ( − 7) = 4 [5 marks]
ans : x = 7.359

ii. 2 + 8(2− ) = 9 ans : x = 1 , x = -9 [5 marks]

iii. Given that log5 3 = 0.6826 and log5 7 = 1.2091, [5 marks]
Calculate the value of log3 7 + log5 √7 − log59.

ans : 2.8243

43