1.0 INDICES AND LOGARITHMS

CONTENTS 1.0

2.0

PAGES

INDICES 1.1 Laws of Indices Exercise 1.1 1.2 Simplify index expressions by using laws of indices Exercise 1.2 Exercise 1.3

1 3 5 6

LOGARITHM 2.1 Law of Logarithms Exercise 2.1 Exercise 2.2 Exercise 2.3 Exercise 2.4 Exercise 2.5 2.2 Simplify logarithm expressions by using laws of logarithms Exercise 2.6 Exercise 2.7 Exercise 2.8

11 12 14 15 16 18 19 22

PAST YEAR QUESTION DBM 20023 JUNE 2019 SESSION DBM 2013 JUNE 2019 SESSION DBM 2013 DECEMBER 2018 SESSION DBM 2013 JUNE 2018 SESSION

7 9

23 28 34

40 41 42 43

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

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

2

can be any real number for

DBM20023 : ENGINEERING MATHEMATICS 2 INDICES AND LOGARITHM

1.1 Laws of Indices For real numbers 𝑚𝑚, 𝑛𝑛 and valid bases 𝑎𝑎, 𝑏𝑏, the following basic laws hold, LAWS OF INDICES

1

2

𝑎𝑎 𝑚𝑚 × 𝑎𝑎 𝑛𝑛 = 𝑎𝑎 𝑚𝑚 + 𝑛𝑛 When expressions Example : with the same base are multiplied, the indices a. 6 3 × 6 2 = 6 3 + 2 = 6 5 are added. 𝑎𝑎 𝑚𝑚 = 𝑎𝑎 𝑚𝑚 − 𝑛𝑛 𝑎𝑎 𝑛𝑛

Or

𝑎𝑎 𝑚𝑚 ÷ 𝑎𝑎 𝑛𝑛 = 𝑎𝑎 𝑚𝑚 − 𝑛𝑛

3 𝑎𝑎 0 = 1

4 i. (𝑎𝑎 𝑚𝑚 )𝑛𝑛 = 𝑎𝑎 𝑚𝑚 × 𝑛𝑛 ii. (𝑎𝑎𝑎𝑎)𝑛𝑛 = 𝑎𝑎 𝑛𝑛 𝑏𝑏 𝑛𝑛 𝑎𝑎 𝑛𝑛

iii. �𝑏𝑏�

b. 𝑑𝑑 −2 × 𝑑𝑑 5 = 𝑑𝑑

−2 + 5

= 𝑑𝑑 3

When expressions Example : with the same base are divided, the indices are a. 𝑞𝑞 3 ÷ 𝑞𝑞 −3 = 𝑞𝑞 3 − (−3) = 𝑞𝑞 6 subtracted. b.

25

2

3

= 25−3 = 22

If any integer or Example : unknown is power of 0, the answer is 1 a. 7 0 = 1 b. 𝑐𝑐 0 = 1

Note that 𝑚𝑚 and 𝑛𝑛 have Example : been multiplied to 3 yield the new index 𝑚𝑚𝑚𝑚 a. �22 � = 2 2 × 3 = 2 6

b. (4𝑏𝑏) 2 = 4 2 𝑏𝑏 2 = 16𝑏𝑏 2 2 2

c. �8� =

22 32

=

4 9

3

DBM20023 : ENGINEERING MATHEMATICS 2 INDICES AND LOGARITHM

5 𝑎𝑎 −𝑛𝑛 = .

6 .

𝑚𝑚

𝑎𝑎 𝑛𝑛

𝑛𝑛

1

𝑎𝑎 𝑛𝑛

= � √ 𝑎𝑎 � =

𝑛𝑛

, 𝑎𝑎 ≠ 0

𝑚𝑚

√𝑎𝑎 𝑚𝑚

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

A fractional power means a root. The bottom of the fraction tell you which root to take and the top tells you which power.

Example :

3 5

a. (243) = �234

3

b.

Wrong Method

32 × 33 = (3 × 3)2+3 = 95 Wrong Method

2

2

�33 � = 33 = 39 4

Correct Method

32 × 33 = 32+3 = 35 Correct Method

2

�33 � = 33×2 = 36

= 33 = 27

16 4 �81�

4

16

1 5

3

�

3

= � �81�

2 3 8 =� � = 3 27

Wrong Method

Correct Method

3 4−2 34 ÷ 32 = � � 3 = 12

34 ÷ 32 = 34−2 = 32

30 = 3 𝑜𝑜𝑜𝑜 30 = 0

30 = 1

Wrong Method

Correct Method

DBM20023 : ENGINEERING MATHEMATICS 2 INDICES AND LOGARITHM

Solve the following 1. 𝑝𝑝 × 𝑝𝑝 × 𝑝𝑝 × 𝑝𝑝 × 𝑝𝑝

4. 2 8 ÷ 2 4

𝑎𝑎𝑎𝑎𝑎𝑎 ∶ 𝑝𝑝 5

𝑎𝑎𝑎𝑎𝑎𝑎 ∶ 2 4

7. 2𝑎𝑎 2 𝑏𝑏 × 3𝑎𝑎 4 𝑏𝑏 −2 𝑐𝑐

𝑎𝑎𝑎𝑎𝑎𝑎 ∶ 6𝑎𝑎 6 𝑏𝑏 −1 𝑐𝑐

2. 𝑚𝑚 × 𝑚𝑚 × 𝑚𝑚 × 𝑛𝑛 × 𝑛𝑛

𝑎𝑎𝑎𝑎𝑎𝑎 ∶ 𝑚𝑚 3 𝑛𝑛 2

5. 𝑧𝑧 7 × 𝑧𝑧 6 ÷ 𝑧𝑧 5

1

8.

4 2 2 �𝑏𝑏 𝑐𝑐 −2 𝑑𝑑 �

𝑎𝑎𝑎𝑎𝑎𝑎 ∶

𝑎𝑎𝑎𝑎𝑎𝑎 ∶ 𝑧𝑧 8

𝑏𝑏 2 𝑑𝑑 𝑐𝑐

3. 3 2 × 3 −5 × 3 4

6. 3�𝑥𝑥 3 �

1

9.

5

2𝑥𝑥𝑥𝑥 2 𝑧𝑧 5

𝑎𝑎𝑎𝑎𝑎𝑎 ∶ 3 1 𝑜𝑜𝑜𝑜 3

𝑎𝑎𝑎𝑎𝑎𝑎 ∶ 3𝑥𝑥 15

√4𝑥𝑥 2 𝑦𝑦𝑦𝑦

𝑎𝑎𝑎𝑎𝑎𝑎 ∶

𝑧𝑧 4

𝑥𝑥�𝑦𝑦 5

DBM20023 : ENGINEERING MATHEMATICS 2 INDICES AND LOGARITHM

1.2 Simplify index expressions by using laws of indices.

Simplify each of the following: 1.

2𝑎𝑎 2𝑚𝑚 + 2

4𝑎𝑎 −2+𝑚𝑚

1 = 𝑎𝑎 2𝑚𝑚 + 2 − (−2 + 𝑚𝑚) 2 =

1 2𝑚𝑚 + 2 + 2 − 𝑚𝑚 𝑎𝑎 2 = =

1 𝑚𝑚 + 4 𝑎𝑎 2

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

𝑎𝑎𝑚𝑚 + 4 2

3. 3𝑛𝑛−1 × 27𝑛𝑛 ÷ 9

= 3𝑛𝑛−1 × 33𝑛𝑛 ÷ 32 = 3𝑛𝑛−1+3𝑛𝑛−2 = 34𝑛𝑛−3

6

2. 23𝑛𝑛+1 × 4𝑛𝑛−3

4. 7𝑛𝑛+1 − 7𝑛𝑛−2

= 7𝑛𝑛 �7 − 7𝑛𝑛−2 � 342 = 7𝑛𝑛 � � 49

= 7𝑛𝑛−2 (342)

DBM20023 : ENGINEERING MATHEMATICS 2 INDICES AND LOGARITHM

Simplify index expression by using laws of indices.

1. 53 × 252𝑛𝑛−1 ÷ 54𝑛𝑛+1

3.

𝑎𝑎𝑛𝑛+1 𝑎𝑎2𝑛𝑛−1 ×𝑎𝑎3−5𝑛𝑛

2. 8

𝑎𝑎𝑎𝑎𝑎𝑎 ∶ 1

𝑎𝑎𝑎𝑎𝑎𝑎 ∶ 𝑎𝑎4𝑛𝑛−1

4. 9

𝑛𝑛−3 3

2𝑥𝑥−1 2

1

− 42𝑛𝑛+1 + 3�2𝑛𝑛+1 �

+ 32𝑥𝑥+1 − 2�32𝑥𝑥 �

𝑎𝑎𝑎𝑎𝑎𝑎 ∶ 17�2𝑛𝑛−3 �

𝑎𝑎𝑎𝑎𝑎𝑎 ∶ 4�32𝑥𝑥−1 � 7

DBM20023 : ENGINEERING MATHEMATICS 2 INDICES AND LOGARITHM

1

5.

6. 27�

𝑥𝑥2 𝑥𝑥2 3

𝑥𝑥2

𝑚𝑚−2 � 3

1

− 92𝑚𝑚−1 + 2�3𝑚𝑚+1 �

𝑎𝑎𝑎𝑎𝑎𝑎 ∶ 𝑥𝑥

𝑎𝑎𝑎𝑎𝑎𝑎 ∶ 6(3𝑚𝑚 )

Solve the following equations.

1. 3𝑥𝑥 = 3𝑥𝑥+1 − 6

3𝑥𝑥 = 3𝑥𝑥 ∙ 31 − 6 𝑞𝑞 = 3𝑞𝑞 − 6

→ 𝑖𝑖𝑖𝑖 3𝑥𝑥 = 𝑞𝑞

𝑞𝑞 − 3𝑞𝑞 = −6

∴ 3 = 𝑞𝑞 3 =3 8

1

2

3(𝑛𝑛−2)

1 23𝑛𝑛−6

𝑛𝑛 + 1 = −3𝑛𝑛 + 6

𝑥𝑥

1

8

2𝑛𝑛+1 = 2−3𝑛𝑛+6

𝑞𝑞 = 3 𝑥𝑥

1 𝑛𝑛−2

2𝑛𝑛+1 = 2−(3𝑛𝑛−6)

−6 𝑞𝑞 = −2

3 =3

2𝑛𝑛+1 = 2𝑛𝑛+1 =

−2𝑞𝑞 = −6

𝑥𝑥

2. 2𝑛𝑛+1 =

∴ 𝑥𝑥 = 1

𝑛𝑛 + 3𝑛𝑛 = 6 − 1 4𝑛𝑛 = 5 𝑛𝑛 =

5 4

DBM20023 : ENGINEERING MATHEMATICS 2 INDICES AND LOGARITHM

Solve each of the following equations.

1. 3(25𝑥𝑥 ) = 75

3. 3𝑥𝑥+2 = 3𝑥𝑥 + 72

−5

1

2. 2𝑠𝑠 2 = 16

𝑎𝑎𝑎𝑎𝑎𝑎 ∶ 𝑥𝑥 = 1

𝑎𝑎𝑎𝑎𝑎𝑎 ∶ 𝑥𝑥 = 2

4. 3𝑥𝑥+2 + 3𝑥𝑥+4 = 90

𝑎𝑎𝑎𝑎𝑎𝑎 ∶ 𝑠𝑠 = 4

𝑎𝑎𝑎𝑎𝑎𝑎 ∶ 𝑥𝑥 = 0 9

DBM20023 : ENGINEERING MATHEMATICS 2 INDICES AND LOGARITHM

5. 2𝑥𝑥+2 = 2𝑥𝑥+1 + 16

6. 7𝑚𝑚 − 350 = −7𝑚𝑚+2

𝑎𝑎𝑎𝑎𝑎𝑎 ∶ 𝑥𝑥 = 3

10

𝑎𝑎𝑎𝑎𝑎𝑎 ∶ 𝑚𝑚 = 1

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,

2.1

is the logarithm of

to base

, or

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) loga 𝑎𝑎 = 1

2. (Zero Rule) loga 1 = 0

12

The logarithm of Example : a number that is equal to its base is a. log3 3 = 1 just 1. b. logp 𝑝𝑝 = 1 The logarithm of 1 Example : with 𝑎𝑎 > 1 equals zero. a. log5 1 = 0 b. logp 1 = 0

DBM20023 : ENGINEERING MATHEMATICS 2 INDICES AND LOGARITHM

3. (Change Base Rule) loga 𝑏𝑏 =

𝑙𝑙𝑙𝑙𝑙𝑙𝑐𝑐 𝑏𝑏

𝑙𝑙𝑙𝑙𝑙𝑙𝑐𝑐 𝑎𝑎

4. (Product Rule) loga 𝑀𝑀𝑀𝑀 =loga 𝑀𝑀 +loga 𝑁𝑁

5. (Quotient Rule) 𝑀𝑀

loga 𝑁𝑁 =loga 𝑀𝑀 −loga 𝑁𝑁

Pick a new base Example : and the formula a. log38 = says it is equal to the log of the b. log29 = number in the new base divided by the log of the old base in the new base.

𝑙𝑙𝑙𝑙𝑙𝑙10 8 𝑙𝑙𝑙𝑙𝑙𝑙10 3

𝑙𝑙𝑙𝑙𝑙𝑙4 9 𝑙𝑙𝑙𝑙𝑙𝑙4 2

The logarithm of Example : a product of two or more a. log2 𝑥𝑥𝑥𝑥 =log2 𝑥𝑥 +log2 𝑦𝑦 numbers is equal to the sum of the b. logn 3𝑚𝑚 =logn 3 +logn 𝑚𝑚 logarithms of the numbers or factors of the product.

The logarithm of Example : a fraction is 2 equal to the a. log23 =log22 − log23 difference between the 3 b. log2 𝑝𝑝 = log2 3 − log2 𝑝𝑝 logarithm of the numerator and the logarithm of the denominator.

13

DBM20023 : ENGINEERING MATHEMATICS 2 INDICES AND LOGARITHM

6. (Power Rule) loga 𝑁𝑁 𝑝𝑝 = 𝑃𝑃 loga 𝑁𝑁

7. 𝑁𝑁 = 𝑎𝑎 𝑥𝑥 ⇐⇒ loga 𝑁𝑁 = 𝑥𝑥

The logarithm of Example : an exponential number is the a. log7 7 3 = 3 log7 𝑁𝑁 exponent times the logarithm of b. log3 𝑃𝑃2 = 2 log3 𝑃𝑃 the base.

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.

14

c. 5 2 = 25

d. 100 = 10 2

e. 𝑋𝑋 𝑛𝑛 = 𝑝𝑝

f. 3 3 = 27

g. 6 1 = 6

h. 42 = 7 2

DBM20023 : ENGINEERING MATHEMATICS 2 INDICES AND LOGARITHM

i.

10 −3 = 0.001 1

3

1

k. � 2 � = 8

1

j.

64 2 = 8

l.

3 −3 = 27

1

Write in Indices form.

1. log4 𝑃𝑃 = 3

2. log2 2 = 1

3. logx 𝑌𝑌 = 24

4. log4 64 = 3

5. log8 64 = 2

6. 2 =log7 49

7. 4 =log7 2401

8. log7 343 = 3

9. log 0.001 = −3

10. log4096 8 = 4

1

15

DBM20023 : ENGINEERING MATHEMATICS 2 INDICES AND LOGARITHM

Solve log264 Method : Indices log2 64 = 𝑚𝑚

2 𝑚𝑚 = 64 2

𝑚𝑚

= 2

6

Method : Change to 𝑙𝑙𝑙𝑙𝑙𝑙 10

log2 64 =

=

𝑙𝑙𝑙𝑙𝑙𝑙 10 64 𝑙𝑙𝑙𝑙𝑙𝑙 10 2

Method : Change base log264 =log22 6

= 6log22

1.806 0.301

= 6 (1)

=6

∴ 𝑚𝑚 = 6

=6

Exercise 2.3 : Solve the following 1. log2 8

3. log4 256

2. log5 125

𝑎𝑎𝑎𝑎𝑎𝑎 ∶ 3

𝑎𝑎𝑎𝑎𝑎𝑎 ∶ 4 16

4. log5 0.0016

𝑎𝑎𝑎𝑎𝑎𝑎 ∶ 3

𝑎𝑎𝑎𝑎𝑎𝑎 ∶ −4

DBM20023 : ENGINEERING MATHEMATICS 2 INDICES AND LOGARITHM

5. log4

1

6. log3 9

64

7. log2 0.015625

𝑎𝑎𝑎𝑎𝑎𝑎 ∶ −3

9. log3 27 2

𝑎𝑎𝑎𝑎𝑎𝑎 ∶ −6

11. logm�𝑚𝑚1

𝑎𝑎𝑎𝑎𝑎𝑎 ∶ 6

𝑎𝑎𝑎𝑎𝑎𝑎 ∶ −

1 2

3

𝑎𝑎𝑎𝑎𝑎𝑎 ∶ 6

8. logx 1

𝑎𝑎𝑎𝑎𝑎𝑎 ∶ 0

1

10. log2 4

5

12. 5 loga √𝑎𝑎

𝑎𝑎𝑎𝑎𝑎𝑎 ∶ −2

𝑎𝑎𝑎𝑎𝑎𝑎 ∶ 1 17

DBM20023 : ENGINEERING MATHEMATICS 2 INDICES AND LOGARITHM

Find the value of 𝒙𝒙.

1. logx 8 = 3

3. logx 64 = 2

5. log2 𝑋𝑋 2 = 16

2. log3 𝑥𝑥 = 1

𝑎𝑎𝑎𝑎𝑎𝑎 ∶ 𝑥𝑥 = 2

𝑎𝑎𝑎𝑎𝑎𝑎 ∶ 𝑥𝑥 = 8

𝑎𝑎𝑎𝑎𝑎𝑎 ∶ 𝑋𝑋 = 256 18

4. 2 logx 3 = 1

6. 2 log2x 4 = 2

𝑎𝑎𝑎𝑎𝑎𝑎 ∶ 𝑥𝑥 = 3

𝑎𝑎𝑎𝑎𝑎𝑎 ∶ 𝑥𝑥 = 9

𝑎𝑎𝑎𝑎𝑎𝑎 ∶ 𝑥𝑥 = 2

DBM20023 : ENGINEERING MATHEMATICS 2 INDICES AND LOGARITHM

Express in term of logn 𝑋𝑋 and logn 𝑌𝑌, 𝑋𝑋𝑌𝑌 2 � 𝑌𝑌

a. logn�

𝑋𝑋

b. logn�𝑋𝑋𝑋𝑋

= logn 𝑋𝑋 +logn 𝑌𝑌 2 −logn 𝑌𝑌

𝑋𝑋

= logn �𝑋𝑋𝑋𝑋�

= logn 𝑋𝑋 + 2 logn 𝑌𝑌 −logn 𝑌𝑌

=

= logn 𝑋𝑋 +logn 𝑌𝑌

= = =

1

2 1 2 1 2 1 2

1 2

𝑋𝑋

logn �𝑋𝑋𝑋𝑋�

[logn 𝑋𝑋 −( logn 𝑋𝑋 + logn 𝑌𝑌 )] [logn 𝑋𝑋 − logn 𝑋𝑋 − logn 𝑌𝑌] [−logn 𝑌𝑌] 1

= − 2logn 𝑌𝑌

Express the following expression by using laws of logarithms.

1

3

2. log 102𝑏𝑏2

1. log 𝑎𝑎 2 𝑏𝑏 2

3

𝑎𝑎𝑎𝑎𝑎𝑎 ∶ 2 log 𝑎𝑎 + 2 log 𝑏𝑏

𝑎𝑎𝑎𝑎𝑎𝑎 ∶ −log 102 − 2log 𝑏𝑏 19

DBM20023 : ENGINEERING MATHEMATICS 2 INDICES AND LOGARITHM

3. logb 𝑥𝑥 3 𝑦𝑦 2

𝑎𝑎𝑏𝑏 5. logn � 𝑐𝑐

1

3

𝑦𝑦 3

4. logn 𝑥𝑥 2

𝑎𝑎𝑎𝑎𝑎𝑎 ∶ 3 logb 𝑥𝑥 + 2 logb 𝑦𝑦

3

1

𝑎𝑎𝑎𝑎𝑎𝑎 ∶ 2 logn 𝑎𝑎 + 2 logn 𝑏𝑏 − 2 logn 𝑐𝑐 20

2

6. − 3 loga

𝑥𝑥 3 𝑧𝑧 𝑝𝑝2

𝑎𝑎𝑎𝑎𝑎𝑎 ∶ 3logn 𝑦𝑦 − 2 logn 𝑥𝑥

2

4

𝑎𝑎𝑎𝑎𝑎𝑎 ∶ −2 loga 𝑥𝑥 − 3 loga 𝑧𝑧 + 3 loga 𝑝𝑝

DBM20023 : ENGINEERING MATHEMATICS 2 INDICES AND LOGARITHM

𝑥𝑥

8. log

7. log�𝑎𝑎 2 𝑦𝑦 4

1

𝑎𝑎𝑎𝑎𝑎𝑎 ∶ 2 log 𝑥𝑥 − log 𝑎𝑎 − 2 log 𝑦𝑦

𝑥𝑥 3 𝑦𝑦 3

√𝑧𝑧

1

𝑎𝑎𝑎𝑎𝑎𝑎 ∶ 3 log 𝑥𝑥 + log 𝑦𝑦 − 3 log 𝑧𝑧

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, 1 3

a. loga 𝑋𝑋 + loga 𝑋𝑋𝑋𝑋

b. 4logx4 + 2logx1

= logx44 +logx12

1

= loga 𝑋𝑋 + loga (𝑋𝑋𝑋𝑋)3

= logx16 +logx1

1

= loga 𝑋𝑋(𝑋𝑋𝑋𝑋) 3 1 3

= loga 𝑋𝑋𝑋𝑋 𝑌𝑌 4 3

= loga 𝑋𝑋 𝑌𝑌

22

1 3

= logx16(1)

1 3

or

3

loga √𝑋𝑋 4 𝑌𝑌

= logx16

DBM20023 : ENGINEERING MATHEMATICS 2 INDICES AND LOGARITHM

Simplify logarithm expressions by using laws of logarithms.

1. 2 log 𝑋𝑋 −log 𝑌𝑌

2. 2 log 𝑋𝑋 + 3 log 𝑌𝑌 − log 𝑍𝑍

𝑋𝑋 2

𝑎𝑎𝑎𝑎𝑎𝑎 ∶ log 𝑌𝑌

3. 𝑎𝑎 log 𝑋𝑋 − 𝑏𝑏 log 𝑌𝑌 − 2 log 3

𝑋𝑋 𝑎𝑎

𝑎𝑎𝑎𝑎𝑎𝑎 ∶ log9𝑌𝑌 𝑏𝑏

𝑋𝑋 2 𝑌𝑌 3

1

4. log9 𝑋𝑋 − 2 + 3 log9 𝑌𝑌

𝑎𝑎𝑎𝑎𝑎𝑎 ∶log

𝑎𝑎𝑎𝑎𝑎𝑎 ∶ log9

𝑍𝑍

𝑋𝑋𝑌𝑌 3 3

23

DBM20023 : ENGINEERING MATHEMATICS 2 INDICES AND LOGARITHM

5.

7.

2 3

1 2

1

6. 4 loga 𝑥𝑥 + 2 − 3 loga 𝑎𝑎

log2 𝑋𝑋 + 3 log2 𝑋𝑋𝑌𝑌 3

𝑎𝑎𝑎𝑎𝑎𝑎 ∶ log2 𝑋𝑋𝑋𝑋

log10 𝑎𝑎 + 2 log10 𝑎𝑎 − log10 𝑎𝑎2

𝑎𝑎𝑎𝑎𝑎𝑎 ∶ log10 √𝑎𝑎 24

1

𝑥𝑥 4

𝑎𝑎𝑎𝑎𝑎𝑎 ∶ loga

𝑎𝑎

8. logx 𝑎𝑎 − 2 logx 𝑏𝑏 − 3 logx 𝑎𝑎

1

𝑎𝑎𝑎𝑎𝑎𝑎 ∶logx𝑎𝑎2

√𝑏𝑏

DBM20023 : ENGINEERING MATHEMATICS 2 INDICES AND LOGARITHM

9. log4 9 − log4 36

10. log4 2 − log4 32

𝑎𝑎𝑎𝑎𝑎𝑎 ∶ −1

11. 4 loga 2 − 3 loga 4 + loga 6

3

𝑎𝑎𝑎𝑎𝑎𝑎 ∶ loga 2

12. 2 log4 2 − log4 3 + log4 12

𝑎𝑎𝑎𝑎𝑎𝑎 ∶ -2

𝑎𝑎𝑎𝑎𝑎𝑎 ∶ 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

𝑎𝑎𝑎𝑎𝑎𝑎 ∶ 4

𝑎𝑎𝑎𝑎𝑎𝑎 ∶ log5 400

Given log4 6 = 0.7737 and log4 8 = 1.1606, Determine the value of followings without using calculator. a. log4 36

= log4 6 2 = 2log4 6

= 2 (0.7737) = 1.5474

26

b. log4 0.75 = log4

6 8

= log4 6 −log4 8

= 0.7737 − 1.1606 = −0.3869

c. log4 48

= log4 6(8)

= log4 6 +log4 8

= 0.7737 + 1.1606 = 1.9343

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

= =

= =

𝑙𝑙𝑙𝑙𝑙𝑙 2 15 𝑙𝑙𝑙𝑙𝑙𝑙 2 3

𝑙𝑙𝑙𝑙𝑙𝑙 2 3(5) 𝑙𝑙𝑙𝑙𝑙𝑙 2 3

𝑙𝑙𝑙𝑙𝑙𝑙 2 3 + 𝑙𝑙𝑙𝑙𝑙𝑙 2 5 𝑙𝑙𝑙𝑙𝑙𝑙 2 3 0.1213 + 1.7172 0.1213 = 15.1566

c. log4125

= = = =

𝑙𝑙𝑙𝑙𝑙𝑙 2 25 𝑙𝑙𝑙𝑙𝑙𝑙 2 27 𝑙𝑙𝑙𝑙𝑙𝑙 2 5 2

𝑙𝑙𝑙𝑙𝑙𝑙 2 3 3

2 𝑙𝑙𝑙𝑙𝑙𝑙 2 5 3 𝑙𝑙𝑙𝑙𝑙𝑙 2 3

2 (1.7172) 3 (0.1213)

=

3.4344 0.3639

= 9.4378

𝑙𝑙𝑜𝑜𝑜𝑜 2 125 𝑙𝑙𝑙𝑙𝑙𝑙 2 4

= = = =

𝑙𝑙𝑙𝑙𝑙𝑙 2 5 3

𝑙𝑙𝑙𝑙𝑙𝑙 2 2 2

3 𝑙𝑙𝑙𝑙𝑙𝑙 2 5 2 𝑙𝑙𝑙𝑙𝑙𝑙 2 2

3 (1.7172) 2 (1)

=

5.1516 2

= 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

c. loga 3√5

28

b. loga 0.6

𝑎𝑎𝑎𝑎𝑎𝑎 ∶ 1.176

𝑎𝑎𝑎𝑎𝑎𝑎 ∶ 0.8265

d.

𝑙𝑙𝑙𝑙𝑙𝑙𝑎𝑎 45

𝑙𝑙𝑙𝑙𝑙𝑙𝑎𝑎 9𝑎𝑎

𝑎𝑎𝑎𝑎𝑎𝑎 ∶ −0.222

𝑎𝑎𝑎𝑎𝑎𝑎 ∶ 0.846

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

3.

b. log 1.5

𝑎𝑎𝑎𝑎𝑎𝑎 ∶ 1.5562

𝑎𝑎𝑎𝑎𝑎𝑎 ∶ 0.1761

Given log3 𝑥𝑥 = 𝑚𝑚 and log3 𝑦𝑦 = 𝑛𝑛, find log3 𝑥𝑥 2 𝑦𝑦 2 in terms of 𝑚𝑚 and 𝑛𝑛.

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

𝑎𝑎𝑎𝑎𝑎𝑎 ∶ 2𝑚𝑚 + 2𝑛𝑛

𝑎𝑎𝑎𝑎𝑎𝑎 ∶ 2𝑝𝑝 + 2𝑞𝑞 − 2 29

DBM20023 : ENGINEERING MATHEMATICS 2 INDICES AND LOGARITHM

5. Given log5 𝑥𝑥 = 𝑎𝑎 and log5 𝑦𝑦 = 𝑏𝑏, find the following in terms of 𝑎𝑎 and 𝑏𝑏. 𝑥𝑥 2

b. log5 �25𝑥𝑥�𝑦𝑦�

a. log5 � 𝑦𝑦 �

1 𝑎𝑎𝑎𝑎𝑎𝑎 ∶ 2 + 𝑎𝑎 + 𝑏𝑏 2

𝑎𝑎𝑎𝑎𝑎𝑎 ∶ 2𝑎𝑎 − 𝑏𝑏

Solve the following equations.

1. 9𝑥𝑥 ∙ 3𝑥𝑥−1 = 81

32𝑥𝑥 ∙ 3𝑥𝑥−1 = 34 2𝑥𝑥 + 𝑥𝑥 − 1

3

3𝑥𝑥 − 1

3

4

=3

= 34

3𝑥𝑥 − 1 = 4 3𝑥𝑥 = 5 𝑥𝑥 =

30

5 3

2. 25𝑥𝑥 = 15625

52𝑥𝑥 = 56 2𝑥𝑥 = 6

𝑥𝑥 = 3

DBM20023 : ENGINEERING MATHEMATICS 2 INDICES AND LOGARITHM

3. 3𝑛𝑛 − 1 =

1 27

𝑛𝑛 − 2

𝑛𝑛 − 1

3

=

3𝑛𝑛 − 1 =

1 33(𝑛𝑛 − 2) 1

3𝑛𝑛 − 6

3

3𝑛𝑛 − 1 = 3 −(3𝑛𝑛 − 6)

4. 5𝑥𝑥 + 1 = 5𝑥𝑥 − 1 + 24

5𝑥𝑥 + 1 − 5𝑥𝑥 − 1 = 24

5𝑥𝑥 �5 − 5 −1 � = 24 5𝑥𝑥 �

24 � = 24 5

3𝑛𝑛 − 1 = 3 −3𝑛𝑛 + 6

5𝑥𝑥 = 5

𝑛𝑛 − 1 = −3𝑛𝑛 + 6

𝑥𝑥 = 1

𝑛𝑛 + 3𝑛𝑛 = 6 + 1 4𝑛𝑛 = 7 𝑛𝑛 =

log 4𝑥𝑥 + 1 = log 9

(𝑥𝑥 + 1)log 4 = log 9 𝑙𝑙𝑙𝑙𝑙𝑙 9 𝑥𝑥 + 1 = 𝑙𝑙𝑙𝑙𝑙𝑙 4

𝑥𝑥 = 1.5850 − 1 𝑥𝑥 = 0.5850

5 24 → 5 = 51

7 4

5. 4𝑥𝑥 + 1 = 9

𝑥𝑥 + 1 = 1.5850

5𝑥𝑥 = 24 ×

3

5

6. 45𝑝𝑝−4 = 3

3

𝑝𝑝−4 = 1

3

𝑝𝑝4

5 3 × 45 =

1 27

3

𝑝𝑝4 = 27 4

𝑝𝑝 = � √27�

3

𝑝𝑝 = 81

31

DBM20023 : ENGINEERING MATHEMATICS 2 INDICES AND LOGARITHM

5

8. log4 (𝑥𝑥 − 3) = 2 log4 5 − log4 10

7. log3 3𝑥𝑥 + log9 𝑥𝑥 = 2 𝑙𝑙𝑙𝑙𝑙𝑙3 𝑥𝑥

log3 3𝑥𝑥 + 𝑙𝑙𝑙𝑙𝑙𝑙

log3 3𝑥𝑥 +

9

3

𝑙𝑙𝑙𝑙𝑙𝑙3 𝑥𝑥

𝑙𝑙𝑜𝑜𝑜𝑜3 32 𝑙𝑙𝑙𝑙𝑙𝑙3 𝑥𝑥

log3 3𝑥𝑥 + 2 𝑙𝑙𝑙𝑙𝑙𝑙 log3 3𝑥𝑥 +

3

3

𝑙𝑙𝑜𝑜𝑜𝑜3 𝑥𝑥 2(1)

1

log4 (𝑥𝑥 − 3) = log4 52 − log4 10

5

=2

25

log4 (𝑥𝑥 − 3) = log4 � 10 �

5

=2

𝑥𝑥 − 3 = 2.5

5

=2

𝑥𝑥 = 2.5 + 3

5

=2

𝑥𝑥 = 5.5

5

log3 3𝑥𝑥 + 2 log3 𝑥𝑥 = 2 1

5

log3 3𝑥𝑥 + log3 𝑥𝑥 2 = 2 1

5

log3�3𝑥𝑥 × 𝑥𝑥 2 � = 2 1

5

log3�3𝑥𝑥 1+2 � = 2 3

5

3𝑥𝑥 2 = 32 5

3 𝑥𝑥 2 3

32 = 3

3

3

5

𝑥𝑥 2 = 32−1 𝑥𝑥 2 = 32 𝑥𝑥 = 3

32

DBM20023 : ENGINEERING MATHEMATICS 2 INDICES AND LOGARITHM

1

9. log2 𝑥𝑥 − log2 (1 − 𝑥𝑥) = 4 𝑥𝑥

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

Log2 1 −𝑥𝑥 = 4

𝑥𝑥 = 24 1 − 𝑥𝑥

𝑥𝑥 = 16(1 − 𝑥𝑥) 𝑥𝑥 = 16 − 16𝑥𝑥

𝑥𝑥 + 16𝑥𝑥 = 16 17𝑥𝑥 = 16 𝑥𝑥 =

16 17

(𝑥𝑥 + 4)

1

log4 (3𝑥𝑥 + 1) = 2

1 (𝑥𝑥 + 4) = 42 3𝑥𝑥 + 1 (𝑥𝑥 + 4) =2 3𝑥𝑥 + 1

(𝑥𝑥 + 4) = 2(3𝑥𝑥 + 1) 𝑥𝑥 + 4 = 6𝑥𝑥 + 2

6𝑥𝑥 − 𝑥𝑥 = 4 − 2 𝑥𝑥 = 2 𝑥𝑥 =

2 5

33

DBM20023 : ENGINEERING MATHEMATICS 2 INDICES AND LOGARITHM

Solve each of the following equations.

1. 3𝒙𝒙+3 − 3𝒙𝒙+1 = 72

3. 5𝑥𝑥 = 12

34

2. 5𝑥𝑥−1 − 5𝑥𝑥−2 = 100

𝑎𝑎𝑎𝑎𝑎𝑎 ∶ 𝑥𝑥 = 1

𝑎𝑎𝑎𝑎𝑎𝑎 ∶ 𝑥𝑥 = 1.544

4. 73𝑥𝑥+2 = 31𝑥𝑥

𝑎𝑎𝑎𝑎𝑎𝑎 ∶ 𝑥𝑥 = 4

𝑎𝑎𝑎𝑎𝑎𝑎 ∶ 𝑥𝑥 = −1.619

DBM20023 : ENGINEERING MATHEMATICS 2 INDICES AND LOGARITHM

5.

6. 26𝑥𝑥 − 810 − 2𝑥𝑥 = 0

𝑎𝑎 3 + 4𝑛𝑛

𝑎𝑎3𝑛𝑛 + 2 × 𝑎𝑎4 − 𝑛𝑛

7. 4 × 33𝑥𝑥 = 9

(𝑥𝑥+1)

𝑎𝑎𝑎𝑎𝑎𝑎 ∶ 𝑎𝑎2𝑛𝑛 − 3

𝑎𝑎𝑎𝑎𝑎𝑎 ∶ 𝑥𝑥 = 0.7381

8. 2𝑥𝑥 + 8(2−𝑥𝑥 ) = 9

𝑎𝑎𝑎𝑎𝑎𝑎 ∶ 𝑥𝑥 =

5 2

𝑎𝑎𝑎𝑎𝑎𝑎 ∶ 𝑥𝑥 = 0 ; 𝑥𝑥 = 3 35

DBM20023 : ENGINEERING MATHEMATICS 2 INDICES AND LOGARITHM

9. 42𝑥𝑥 + 1 = 25

11. 3𝑥𝑥 = 5𝑥𝑥 + 2

36

10. 62𝑥𝑥 + 1 = 17

𝑎𝑎𝑎𝑎𝑎𝑎 ∶ 𝑥𝑥 = 0.661

𝑎𝑎𝑎𝑎𝑎𝑎 ∶ 𝑥𝑥 = −6.3

12. 3𝑥𝑥 + 1 = 108 − 3𝑥𝑥

𝑎𝑎𝑎𝑎𝑎𝑎 ∶ 0.2906

𝑎𝑎𝑎𝑎𝑎𝑎 ∶ 𝑥𝑥 = 3

DBM20023 : ENGINEERING MATHEMATICS 2 INDICES AND LOGARITHM

13. log3 (𝑥𝑥 − 6) = 4 − log3 3𝑥𝑥

15. 2 𝑙𝑙𝑙𝑙𝑙𝑙3

(𝑥𝑥 − 2)

1

14. logx 8 + logx 4 = 5

𝑎𝑎𝑎𝑎𝑎𝑎 ∶ 𝑥𝑥 = 9 ; 𝑥𝑥 = −3

=2

𝑎𝑎𝑎𝑎𝑎𝑎 ∶ 𝑥𝑥 =

7 3

16. 3 𝑙𝑙𝑙𝑙𝑙𝑙2 𝑥𝑥 = 9

𝑎𝑎𝑎𝑎𝑎𝑎 ∶ 𝑥𝑥 = 2

𝑎𝑎𝑎𝑎𝑎𝑎 ∶ 𝑥𝑥 = 4 37

DBM20023 : ENGINEERING MATHEMATICS 2 INDICES AND LOGARITHM

3

17. 3 logx 3 + 4 logx 4 −logx 12 = 1

18. Log2 (𝑥𝑥 + 2) + log4 (𝑥𝑥 + 2) = 2

𝑎𝑎𝑎𝑎𝑎𝑎 ∶ 𝑥𝑥 = 576

19. 3 log 2 + log(4𝑥𝑥 − 1) =log(7 − 8𝑥𝑥)

𝑎𝑎𝑎𝑎𝑎𝑎 ∶ 𝑥𝑥 = 38

3 8

𝑎𝑎𝑎𝑎𝑎𝑎 ∶ 𝑥𝑥 = 0

20. 2 log 2𝑥𝑥 + log 3𝑥𝑥 =log 96

𝑎𝑎𝑎𝑎𝑎𝑎: 𝑥𝑥 = 2

DBM20023 : ENGINEERING MATHEMATICS 2 INDICES AND LOGARITHM

21. log (𝑥𝑥 − 5) = 2 + log (𝑥𝑥 − 1)

23.

1 2

22. log2 �5√𝑥𝑥� + log4 (16𝑥𝑥) = 16

𝑎𝑎𝑎𝑎𝑎𝑎 ∶ 𝑥𝑥 = 0.9596

log3 𝑥𝑥 + log9 4𝑥𝑥 = 0

𝑎𝑎𝑎𝑎𝑎𝑎 ∶ 𝑥𝑥 =

1 2

𝑎𝑎𝑎𝑎𝑎𝑎 ∶ 𝑥𝑥 = 3276.8 5

24. 2 logp 10 − 2 logp �2� = 4 − logp 𝑝𝑝2

𝑎𝑎𝑎𝑎𝑎𝑎 ∶ 𝑝𝑝 = 4 39

DBM20023 : ENGINEERING MATHEMATICS 2 INDICES AND LOGARITHM

[DBM 20023]

JUNE 2019 SESSION

QUESTION 1 CLO 1 C3

(a)

Express each of the following expressions in the simplest form. i.

ii.

CLO 2 (b) C3

1

log5 25

ans : a- 3+4n

ans : -2

[3 marks]

[4 marks]

Solve the following questions using a suitable method. i. ii.

40

𝑎𝑎3 + 4𝑛𝑛 𝑎𝑎3𝑛𝑛 + 2 × 𝑎𝑎4 − 𝑛𝑛

32𝑥𝑥 − 1 × 273 − 𝑥𝑥 = 1

logx 8 +logx 4 = 5

ans : x = 8

[4 marks]

ans : x = 2

[4 marks]

DBM20023 : ENGINEERING MATHEMATICS 2 INDICES AND LOGARITHM

[DBM 2013]

JUNE 2019 SESSION

QUESTION 1 CLO 1 C2

(a)

Simplify the following expressions. i. ii.

CLO 1 C3

(b)

4𝑥𝑥 ÷ �256𝑥𝑥 − 1 × 128�

ans : 2-7x + 1

[5 marks]

log9 36 − log9 9 − log9 4

ans : 1

[5 marks]

1

Solve the following equations. i. ii. iii.

22𝑥𝑥 − 5 = 643𝑥𝑥 5𝑥𝑥 − 3 = 3𝑥𝑥

5

ans : − 16 or -0.3125

ans : 9.452

Given that logm 3 = 𝑝𝑝 and logm 5 = 𝑞𝑞, write logm 45 in terms of 𝑝𝑝 and 𝑞𝑞. ans : 2p + q

[4 marks] [6 marks]

[5 marks]

41

DBM20023 : ENGINEERING MATHEMATICS 2 INDICES AND LOGARITHM

[DBM 2013]

DECEMBER 2018 SESSION

QUESTION 1 CLO 1 C2

(a)

Simplify the following expressions 2

14

i.

log4 7 + log4

ii.

32𝑛𝑛 − 1 × 3𝑛𝑛 − 3

3

4

− log4 15

ans : log4 5

ans :

33𝑛𝑛 − 2

iii.

25𝑚𝑚 × 5𝑚𝑚 + 1 125

CLO 1 C3

(b)

[3 marks]

1 9

ans : 512m - 5

2−3𝑚𝑚

[4 marks]

Solve the following equations i. ii. iii.

42𝑥𝑥 �8𝑥𝑥 + 3 � = 32 32𝑥𝑥 + 1 = 5

(4 − 𝑥𝑥)

ans : x = 8

[4 marks]

ans : x = 0.232

[5 marks]

9

log3 (2x + 5) = 1 + log3 (2x – 3)

[6 marks] 7

ans : x = 2

42

[3 marks]

DBM20023 : ENGINEERING MATHEMATICS 2 INDICES AND LOGARITHM

[DBM 2013]

JUNE 2018 SESSION

QUESTION 1 CLO 1 C2

(a)

Express each of the following expressions in the simplest form. i.

24𝑛𝑛 × 25𝑛𝑛

ans : 2-3n

1

ans : logx 8

[2 marks]

46𝑛𝑛

ii.

iii.

logx 16 − logx 4 − 3 logx 2 2 75𝑝𝑝+2

33−𝑝𝑝 × 152𝑝𝑝+1 × 52 CLO 1 C3

(b)

1

1

ans : x = 9

[4 marks]

[4 marks]

Solve the following equations using the suitable method. i. ii. iii.

log3 (3𝑥𝑥 + 1) − log3 (𝑥𝑥 − 7) = 4 ans : x = 7.359 2𝑥𝑥 + 8(2−𝑥𝑥 ) = 9

ans : x = 1 , x = -9

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

[5 marks] [5 marks]

[5 marks]

ans : 2.8243

43