DUM10122 WORKBOOK

DUM 10122: Engineering Mathematics 1

5. Solve the right angled triangle Determine ∡ B from.
a)

6.8 cm B

5.8 cm Determine ∡ x, ∡ y, m
b)

m 5.3 cm
y
x
6.5 cm

c) Determine side r, p and ∡ θ
r
p

180
9.5 cm

13

DUM 10122: Engineering Mathematics 1

After completing the unit, students should be able to:
1. Sketch the graph of trigonometric functions.

UNIT LEARNING
OUTCOMES

3.3 GRAPHING TRIGONOMETRIC FUNCTIONS

Graph y = a sin θ for 0  θ < 2π

Ө 00 900 1800 2700 3600
(in degree)  2
0 1  0 32  0
Ө 2 -a
(in radian)
0a
sin

Based on table, the graph y = sin x ; for 0o  x  360o
y
a

0 1   32 2π θ
-a 2

• The shape of the graph of y =
sin x from x = 00 to x = 3600 is
repeated for each complete
cycle.

• The function y = sin x is
periodic with the period of
3600.

• The maximum and minimum
values of the function y = sin x
are 1 and -1 respectively. This
value is also called as
amplitude.

14

DUM 10122: Engineering Mathematics 1

Graph y = a cos θ for 0  θ < 2π

Ө 00 900 1800 2700 3600
(in degree) 2
0 1   32  a
Ө 2
(in radian)
a 0 -a 0
Cos

Based on table, the graph y = cos x ; for 0o  x  360o

ay

0 2Π
-a

▪ The shape of the graph of y = cos x
from x = 0 o to x = 360 o is repeated
for each complete cycle.

▪ The function y = cos x is periodic
with a period of 360o.

▪ The maximum and minimum value
of the function y = cos x are 1 & -1
respectively.

15

DUM 10122: Engineering Mathematics 1

Graph y = a tan θ for 0  θ < 2π

Ө 00 900 1800 2700 3600
(in degree)  32  2
0 1  0 0
Ө 2 
(in radian)

Tan 0 

Based on table, Sketch the graph y = tan x ; for 0o  x  360o

y

0

▪ The shape of the graph of y = tan x from x = 00
to x = 1800 is repeated for each complete cycle.

▪ The function y = tan x is periodic with a period
of 1800.

▪ The function y = tan x does not have any
maximum or minimum values.

16

DUM 10122: Engineering Mathematics 1

Sketch the graph of the trigonometric equation given

a) b) 0  Ө  2π 3 7 2
a) y = sin 2Ө 1  2 4
2
1
Ө 0 4

sin 2Ө

y
1

π 2π
-1 

b) y = cos 2Ө 0Өπ
Ө 0 1

cos 2Ө 2
y

π 2π

17

DUM 10122: Engineering Mathematics 1

c) y = 3 sin 2Ө 0°  Ө  360°
90° 180°
Ө 0° 270° 360°
3 sin 2Ө

y

d) y = a2) cobs)2x 0°  x  180° 180°
X0 90°
2 cos 2x

y

18

DUM 10122: Engineering Mathematics 1
TUTORIAL 3.3
1. Sketch each of the following trigonometric functions in 0 °  x  3600
(a) y = 2 sin x

(b) y = 1 cos x
2

(c) y = 3 cos 2x

19

DUM 10122: Engineering Mathematics 1
2. Sketch each of the following trigonometric functions in 0 °  x  2π
(a) y = 4 sin 4x

(b) y = 2 cos 3x

(c) y = 1 sin 3x
2

20

DUM 10122: Engineering Mathematics 1

UNIT LEARNING After completing the unit, students should be able to:
OUTCOMES
1. Identify the sign of trigonometric ratios for all quadrant.
2. Solve trigonometric equations.

3.4 TRIGONOMETRIC EQUATIONS

THE QUADRANT
The Cartesian plane can be divided into 4 quadrants as shown below:

Quadrants II Quadrants I

Quadrants III Quadrants IV

ANGLES IN QUADRANT y

In the first quadrant, the angles,  is in 0o ≤  ≤ 90o. Commonly called acute angle. x
This angle is also known is reference angle,
y
=α
α
In the second quadrant, the angles,  is in 90o ≤  ≤ 180o commonly called obtuse x
angle.
y
 = 180o – α 

In the third quadrant, the angle,  is in 180o ≤  ≤ 270o commonly called reflex αx
angle.

 = 180o + α

In the fourth quadrant, the angle,  is in 270o ≤  ≤ 360o commonly called reflex y
angle.

 = 360o – α x
α


21

DUM 10122: Engineering Mathematics 1

TRIGONOMETRIC RATIO FOR ALL QUADRANTS
y

SECOND QUADRANT FIRST QUADRANT
in Positive ll Positive

 = 180o – α  =α x
 = 180o + α   = 360o – α

an Positive os Positive
THIRD QUADRANT FOURTH QUADRANT

The sign of the trigonometric ratios can be determine in the following figure:
QUADRANT I

● sin θ = y  cosec θ = r
r y
r y
cos θ = x  sec θ = r
 r x
x
tan θ = y  cot θ = x
x y

QUADRANT II

● sin θ = y  cosec θ = r
yr r  y

 cos θ = − x sec θ = − r
–x r x

tan θ = − y cot θ = − x
x y

22

DUM 10122: Engineering Mathematics 1

QUADRANT III

–y –x sin θ = − y  cosec θ = − r
●  r  y

r cos θ = − x sec θ = − r
r x

tan θ = y cot θ = x
x y

QUADRANT IV

x –y sin θ = − y  cosec θ = − r
 ● r  y

r cos θ = x sec θ = r
r x

tan θ = − y cot θ = − x
x y

THE VALUE OF TRIGONOMETRIC FUNCTION FOR PARTICULAR ANGLES (SPECIAL
ANGLES): 30, 45 60

30 30 45
22 1

60 60 45

11 1
Figure 3.14

The values of the functions sin θ, cos θ and tan θ for the particular angles 30, 45 and 60 are easily
obtained without calculator as shown below :

Angle, θ sin θ cos θ tan θ
30 1
45 1 3 3
60 2 2 1
1 1
2 2 3

3 1
2 2

23

DUM 10122: Engineering Mathematics 1

TRIGONOMETRIC EQUATIONS
Trigonometric equation is an equation that involved one or more terms of trigonometric function.
Equation given is a process for getting value that satisfying the equation.

The steps to solve single trigonometric equations are as follows:
1. Determine the quadrants of the angle and should be in based on the given trigonometric
equation.
2. Find the acute angle using a scientific calculator,  for angle θ.
3. Determine the range of values of the required angles, for example the range of values of
angles 2θ @ 3θ.
4. Determine the values of angles in those quadrants (where is located).

Example :

Find the solutions in 0 ≤ θ ≤ 360 of the following equations:

a) tan θ = 0.5

 = 26.57 26.57

θ = 26.57, 180 + 26.57 26.57
= 26.57, 206.57
tan θ is positive, so θ are in first
and third quadrant

b) cos θ = -0.6428

= 130, 230
24

DUM 10122: Engineering Mathematics 1

c) tan 2θ = 1.732

θ = 30, 120, 210, 300

d) sin θ = 0.7071
2

θ = 90, 270

e) cos (θ - 25) = 0.9848

θ = 35 ,375
25

DUM 10122: Engineering Mathematics 1

TUTORIAL 3.4

1. Find the solutions in 0 ≤ θ ≤ 360 of the following equations.

a) cos θ = -0.7760 b) cosec θ = -2

= 140.9, 219.1 = 210, 330

c) tan 2 θ = 3 d) sin  = 0.6428

2

= 30, 120,210, 300 = 80, 280
e) 5 sin θ = tan θ
f) sec θ cot θ = 5

= 0, 78.14,180, 282 = 11.54, 168.46
26

DUM 10122: Engineering Mathematics 1

2. Solve each of the following trigonometric equations for 0  x  3600

a) sin x = 0.4233 b) cos x = 0.3412

= 25.04, 154.96 = 70.05, 289.95

c) tan x = 1.8849 d) sec x = -2.345

= 62.05, 242.05 = 115.25, 244.75
e) cos x = -0.7324 f) tan x = -2.2755

= 137, 223 = 113.72, 293.72
27

DUM 10122: Engineering Mathematics 1

3. Find the angles between 0° and 360° that satisfy each of the following trigonometric

equations

a) sin 2 θ = 0.5327 b) cos 3 θ = -0.5473

= 16.1, 73.91,196.1, 253.91 = 41.06,118.41, 161.06,
198.94,281.06, 318.94
c) tan 2θ = -2.4325
d) sin  = 0.4453
2

= 56.18, 146.18,236.18, 326.18 = 52.88, 307.12

e) cos 1  = 0.4775 f) tan 1  = -2.7458
3 2

= 184.44, 895.56 = 220, 580
28

g) 2 sin θ = 0.7443 DUM 10122: Engineering Mathematics 1
h) sin 2θ = -0.7569

= 21.85, 158.15 = 114.6, 155.4,294.5, 335.4
i) 3 tan θ = 3.4533 j) 2 tan 3θ = -0.4357

= 49.02, 229.02 = 55.9,115.9, 175.9, 235.9,295.9,
355.9

29

DUM 10122: Engineering Mathematics 1

UNIT LEARNING After completing the unit, students should be able to:
OUTCOMES
1. Solve oblique triangle using Sine Rule and Cosine Rule
2. Calculate the area of a given oblique triangle by using formula.

3.5 SOLUTIONS OF TRIANGLES

OBLIQUE TRIANGLE
An oblique triangle is a triangle that does not contain a right angle (90o).

It could be an acute triangle (all three angles of the triangle are less than right angles) or it
could be an obtuse triangle (one of the three angles is greater than a right angle).

ARBITRARY TRIANGLE
An arbitrary triangle is a triangle with no special characteristic as shown below:
B
ca

AbC

The angles at the vertices A, B and C will be denoted by A, B and C.
The sides will be denoted by a, b and c.

THE SUM OF ANGLES IN ANY TRIANGLE IS 180°.

30

SINE RULE AND COSINE RULE C DUM 10122: Engineering Mathematics 1
b
a
A c
Sine Rule B
Cosine Rule
a=b=c
sinA sinB sinC a2 = b2 + c2 - 2bc cos A
b2 = a2 + c2 - 2ac cos B
Sine Rule can be used when the following c2 = a2 + b2 - 2ab cos C
values are given:
a. two angles and any one side. Cosine Rule can be used when the following
b. two sides and 1 non-included values are given:
a. two sides and one included angle is
angle (angle opposite one of two
sides given) given
b. three sides are given.
TIPS (A – Angles, S – Side)
TIPS (A – Angles, S – Side)

31

DUM 10122: Engineering Mathematics 1

QUICK CHECK !

1. Circle which of the following situations we can use the Sine rule to calculate a missing side or angle (Hint: look for
2 sets of corresponding sides and angles)

2. Circle which of the following situations we can use the Cosine rule to calculate a missing side or angle (Hint: look
for 2 sets of corresponding sides and angles)

32

DUM 10122: Engineering Mathematics 1

Figure below shows the triangle ABC. Given that ∡ ABC = 680, ∡ ACB = 330,

AC = 7.9 cm. Determine the length of AB.

A Solution:
7.9
330 = 680

7.9 cm AB = 7.9 33
680

680 330 C = 7.9 0.5446
B : =
0.9272
Sine Rule
= 4.64 cm

Figure below shows the triangle PQR. Given that ∡ PRQ = 1180, ∡ PQR = 210, PQ
= 9.7 cm. Find the length of PR.

P

9.7 cm

1180 Q
R

PR = 3.93cm

JKL is an oblique triangle, Calculate the length of JK
K

l 127.50 j

250 L

J
26 cm

JK= 15.13 cm
33

DUM 10122: Engineering Mathematics 1

R

5.8 cm 61.50

S T
7.1 cm

Figure shows a triangle RST. Given that ∡R = 61.50, RS = 5.8 cm, ST = 7.1 cm. Find

a) ∡T

b) The length of RT

Solution:
a)

∡ T= 45.880
b)

RT = 7.709 cm
34

6.4 cm B DUM 10122: Engineering Mathematics 1
4.7 cm
Figure show the triangle ABC. Given
A C that AB = 6.4 cm, BC = 4.7 cm and
AC = 7.8 cm. Find

1. ∡A
2. ∡C

7.8 cm

∡ A= 37.020, ∡C = 55.070
35

AREA OF TRIANGLE DUM 10122: Engineering Mathematics 1
The area of a triangle is given by:
Area = 1 ab sin C
B 2

ca = 1 bc sin A
2
AC
= 1 ac sin B
2

In Figure below shows Δ PQR. Such that PQ = 13 cm, QR = 10 cm and
∡Q = 57.80. Find the area of the triangle Δ PQR.

Q Solution:

13 cm 57.80 10 cm The area of the triangle PQR
PR = 1 x 13 x 10 sin 57.80
2
= 55 cm2

36

DUM 10122: Engineering Mathematics 1

A Figure shows Δ ABC such that
AC = 14 cm, BC = 11 cm ∡C =
14 cm 1250. Find
a) AB
1250 b) ∡B
C 11 cm B c) The area of the triangle ABC

AB= 22.22 cm, ∡B = 31.10, 63.1 cm2
37

DUM 10122: Engineering Mathematics 1

TUTORIAL 3.5

1. Solve the folllowing oblique triangles of ABC :

ANGLE ( o ) SIDES ( cm )
Abc
No. A B C
119
I 126 27 15 21

Ii 43 375
58
Iii 15 72 228 304

Iv 125 32

V 46.3

2. Figure shows the triangle ABC. Given that ∡AB = 810, ∡B = 850, BC = 51.3 cm. Find
AC

A C
810 51.3 cm

850

B

3. Figure shows the triangle PQR. Given that ∡Q = 1240, QR = 16cm, PR = 20cm. Find
(a) ∡P
(b) Length PQ

P

20 cm R
1240
Q 16 cm

38

DUM 10122: Engineering Mathematics 1

4. Figure shows the triangle MNP. Given that ∡P= 240, PN = 36 cm, MN = 19.9 cm. Find:

(a) ∡M M
(b) ∡N 19.9 cm
(c) Length MP

240 P
N

36 cm

39

DUM 10122: Engineering Mathematics 1

5. Solve the following oblique triangles of ABC: SIDES ( cm )
ANGLE ( 0 ) Abc
186 179
No. A B C 11.3 15.6 12.8
I 129 128 152
Ii 1.95 1.46
Iii 27.3 77.3 81.4
Iv 51.4
V 35.2

6. Figure shows the triangle ABC. Given that ∡A = 700, AB = 25 cm, AC = 17.2 cm. Find:

(a) BC A 25 cm B
(b) ∡B 700

17.2 cm

C

7. Figure shows the triangle PQR. Given that ∡Q = 1070, PQ = 37 cm, QR = 43 cm. Find:

(a) PR P
(b) ∡P

(c) The area of the triangle PQR 37 cm

107° R
Q

40

DUM 10122: Engineering Mathematics 1

8. Figure shows the triangle JKL. Given that JK = 14 cm, KL = 6.9 cm, JL = 9.7 cm. Find:

a) ∡L

b) ∡K J 14 cm K
c) The area of the triangle JKL

9.7 cm 6.9 cm

L

9. PQR and PRS are two triangles. Calculate
a) Length of PR
b) ∠PSR
S

9 cm

P 40o R

8.2 cm

85o 6 cm
Q

41

DUM 10122: Engineering Mathematics 1

10. Two cars drive from from point B to point A and C respectively. Determine the area that cover
the distance by the two cars.

B

39 km 58 km

C 87 km A

11. PQR and PST are two triangles. Given ∠PQR = 101o, ∠PTS = 140o, ∠QPS = 34o, QR
= 5 cm, PT = 8cm and ST = 11 cm. Calculate

Q

101o 5 cm

P 34o R S
11 cm
8 cm 140o
T

c) ∠QRP
d) Length of PR
e) Length of RS

42

43

Unit 4 DUM 10122: Engineering Mathematics 1

INDICES & LOGARITHMS

INTRODUCTION

Indices is another word for powers - the singular of indices is index. They OVERVIEW

are useful (not only in mathematics, but also in physics, astronomy and

other sciences) because they enable us to write large and small numbers very concisely.

A knowledge of powers, or indices as they are often called, is essential for an

understanding of most algebraic processes. In this section, you will learn about powers

and rules for manipulating them through a number of worked examples. They also present

useful properties for manipulating using what are called the law of indices. Furthermore,

logarithms and indices are closely related, and in order to understand logarithms a good

knowledge of indices is required. In this unit, we learn laws of indices and laws of

logarithms. Then, solves indicial equations and logarithmic equations.

CONTENT CONTENT

4.1. Indices And Laws Of Indices
4.2. Logarithms & Laws Of Logarithms
4.3. Conversion Between Indicial Form And Logarithmic Form
4.4. Solutions Of In Indicial Equations
4.5. Solutions Of Logarithmic Equations

1

DUM 10122: Engineering Mathematics 1

UNIT LEARNING After completing the unit, students should be able to:
OUTCOMES
1. Evaluate indices with and without using calculator
2. Use Laws of Indices to :

▪ simplify indicial expressions
▪ evaluate indicial expressions

4.1 INDICES AND LAWS OF INDICES

4.1.1 Definition
If a is a real number and n is a positive integer, then
an = a x a x a x a … x a
n times
The number a is called the base and n is called the index, and an is read as a to
the power of n.

i. 34 = 3 x 3 x 3 x 3 = 81
4 times

ii. 3x = 3 x 3 x … x 3
x times

iii. e2 = e x e = 7.389 (3 d.p) where: e = 2.718281828..(irrational number)
2 times

2

DUM 10122: Engineering Mathematics 1

Express in the form of index with the base being the smallest positive integer

a) 81 = 92 b) 8 = 23
= (32)2 = 34

c) 32 = d) 125 =

e) 216 = f) 49 =

g) 1 = h) 1 =
243 27

i) 125 = j) 16 =
27 25

k) -27 = l) -512 =

Evaluate each of the following indices without using calculator.

a) 102 b) 42 + 33
Solution :
Solution : 42 + 33 = (4 x 4) + (3 x 3 x 3)
102 = 10 x 10 = 100 = 16 + 9 = 25

3

DUM 10122: Engineering Mathematics 1

Evaluate the following without using calculator.
a) (-3)4
b) (-5)3 + (-2)2

Evaluate each of the following indices using calculator.

a) (0.2)4

Solution :

PRESS 1.6 x 10-03

0.2 ^ 4 =

= 1.6 x 10-3 @ 0.0016

1 PRESS ( 1/2 )

b) 642 64 ^ = 8

=8

c) e3

= 20.086

4

DUM 10122: Engineering Mathematics 1

Evaluate the following using calculator.

a) 12-5 e) −3 =

16 4

b) (0.3)-4 f) (-6)0 =

1 ( )1

c) 156253 = g) 24 2

d)  2 0 +  8 1

 3 =
 3  125 

Answers : a) 4.02 x 10-6, b) 123.457, c) 25, d) 1.4, e) 0.125, f) 1, g) 4
4.1.2 Laws of Indices

LAWS TO REMEMBER

5. a0 = 1
1. a−m = 1
2. 6. am

1

3. 7. a m = m a

4. m

8. a n = (n a )m

Simplify the following. b) 18a7b2
6a2b2
a) 2n x 22n
= 2n+2n = 3a7 – 2 b2 – 2
= 23n
= 3a5b0
= 3a5

5

DUM 10122: Engineering Mathematics 1

c) (m2n)4 d) 23n x 4n – 1 x 8n
= (m2) 4 n4 = 23n x (22)n – 1 x (23)n
= m8n4 = 23n x 22n – 2 x 23n
= 23n + 2n – 2 + 3n
Simplify the following :
a) 3x x 34x = = 28n – 2

Comman errors on Laws of Indices

am + an ≠ am+n
am - an ≠ am-n

b) m5 x (m2n3)4 =

d)  3  2  t5 =
 t4 
c) (2a2)3 x 5ab2 =

( )e) 2mn4 3 = f) 3xy5x4 y2 =
4mn9
h) 12 p4q5 =
g) k 4m2  km = 6p2

( )i) 2a 2 2  6a3 = ( )j) a 2 4  a3
4a 5 ( )a4 3

6

DUM 10122: Engineering Mathematics 1

Answers : a) 35x, b) m13n12 ,c) 40a7b2, d) 9 , e) 2m2n3, f) 15x5y3, g) k3m, h) 2p2q5, i) 6a2, j) a-1,
t 13

Simplify and evaluate the following.

a) 23 x 22 2 −3
= 25
= 32 b) 27 3 9 2

= 2  (3 2 ) − 3
2
(33 ) 3

= 32 x 3-3
= 3-1

=1

3

Simplify and evaluate the following :

a) 43 x 42 = b) (-3)5 x (-3)-2 =

c)  2 4   2 0 = d) (0.25)− 1  (0.25)32 =
5 5 2

 3  − 1
2 3
e) 22 × (4)3 = f) =
9

Answers : a) 1024, b) -27, c) 16 , d) 16, e) 256, f) 1
625 3

7

TUTORIAL 4.1 DUM 10122: Engineering Mathematics 1
1. Simplify or evaluate the following :
a) 32  33n  34 = b) 5n  52n 52n =

52 d) 34 ÷ 32 =
c) 5−1

e) 51− 2n  5n + 2  5n −3 34n + 5
f) 32n 3n − 2

g) (h-2 )3  2h 4 (h) 4m  8

8

34n−1  27 DUM 10122: Engineering Mathematics 1
i) 9 n+1
j) (16)2n 22n (8)3

(k) 82n  2n+1 4−n = k5
(l) k2m k8−m

(m) 92x  27 x −3  35 − x = n) (4a2b)3 x 2a3 b2

9

DUM 10122: Engineering Mathematics 1

UNIT LEARNING After completing the unit, students should be able to:
OUTCOMES
1. Evaluate logarithms using calculator.
2. Use Laws of Logarithms

• simplify logarithmic expressions
• evaluate logarithmic expressions

4.2 LOGARITHMS & LAWS OF LOGARITHMS

4.2.1 Logarithms “the logarithm of 100 to the
The statement base 10 is 2”

is abbreviated to log10 100 = 2

A. Common Logarithms
The logarithm to the base 10 or log10 can be written as lg and is referred as the common
logarithm, or log10 x = lg x.

B. Natural Logarithms
The logarithm to the base e or loge can be written as ln and is referred as the natural or naperian
logarithm, or loge x = ln x.

10

DUM 10122: Engineering Mathematics 1

Evaluate each of the following logarithms using the calculator, leave you answer to 4 decimal places.

a) log10 8 0.90308998
7
Solution :

PRESS

log 8 =
=

= 0.9031 (4 d.p.)

b) lg 0.2
= -0.6990 (4 d.p.)

c) ln 0.25 =

d) ln 6 =
5

a) lg 5 Evaluate the following logarithms using calculator.
b) log10 12

c) ln 1 d) lg 2.3 + lg 1.3
2 f) ln e

e) log 18 - log 4 + log 54

11

4.2.2 Laws of Logarithms DUM 10122: Engineering Mathematics 1
LAWS TO REMEMBER

1. 5. loga1= 0
2. 6. logaa = 1
3. 7. logaan = n

4.

Express the following in terms of loga x, loga y and/or loga z.

a) loga xy 2
z3

Solution :

loga xy 2 = loga x + loga y2 − loga z3
z3

= loga x + 2 loga y − 3loga z

b) loga x3
y4

Solution : 1
a) loga
x3 = loga  x3  2
y4 y4

=  3  loga  x  = loga x − loga y
loga   y
x2 
 y2

3 loga xn= n loga x

= loga x 2 −loga y2

= 3 loga x − 2loga y
2

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DUM 10122: Engineering Mathematics 1

Express the following in terms of loga m and loga n.

a) loga m
n

Solution :

b) loga (mn)2

Solution :

Simplify the following using laws of logarithms :

a) 3log2 x – 4log2 y

Solution :
3log2 x – 4log2 y = log2 x3 – log2 y4

= log2  x3 
y4

b) ln 6 + ln y – ln 2

Solution:

ln 6 + ln y – ln 2 = ln 6 y 
2

= ln 3y

Express the following as a single logarithm:

a) 3log3 x + 2log3 y (b) 2ln x + ln y

13

DUM 10122: Engineering Mathematics 1

Evaluate the following:

a) log6 8 + log6 27 ( without using calculator)

Solution :
log6 (8×27) = log6 216
= log6 63
=3

b) log4 7 + log3 8 ( by using calculator)

Solution:

log4 7 + log3 8 = log10 7 + log10 8
log10 4 log10 3

= 1.4037 + 1.8928
= 3.2965

a) Evaluate log8 16 + log8 4 without using the calculator.
b) Evaluate log5 6 by using the calculator.

loga xy  loga x  loga y

loga  x   loga x  loga y
y

14

DUM 10122: Engineering Mathematics 1

Given log2 3 = 1.58 and log2 4 = 2. Find the value of each of the following.

a) log2 3 b. log2 12
4
Solution :
Solution :
log2 12 = log2 (3 x 4)
log2 3 = log2 3 – log2 4
4 = log2 3 + log2 4
= 1.58 – 2 = 1.58 + 2
= 3.58
= -0.42 d. log2 36

c. log2 1.5

Solution : Solution :

1. Given that logp 3 = a and logp 5 = b, express the following in terms
of a and/or b.

i. log3 p ii. log5 3

2. Given that loga x = m and logb x = n, express logx ab in terms of m and n.

15

DUM 10122: Engineering Mathematics 1

TUTORIAL 4.2

1. Express the following in terms of loga x, loga y and loga z

a) 2loga x4 b) loga y
y2 x5

c) loga 3 x4 d) loga 4 1
y2 x

e) loga x4yz5 f) loga xy
z

loga x h) loga x2y4
y2z z
g)

16

DUM 10122: Engineering Mathematics 1

2. Express the following as a single logarithm:

a) loge 25 – loge t b) logp 4 + logp 3

c) 3log2 12 – log2 27 d) 31
2 log3 x − 5 log3 y

e) 2log x + log y – log z f) log a + 3log b -2log c

g) ln x2y3 – ln x + ln y h) 1
ln x − ln x + ln xy

2

17

DUM 10122: Engineering Mathematics 1

3. Evaluate the following without using the calculator.

a) log2 6 – log2 3 b) log4 3 – log4 48

c) log3 81 d) log4 1
e) log5 1 + 2log5 5 64

f) log8 45 – 1 log8 81 + 7log8 2 – log8 10
2

4. Evaluate the following by using the calculator :

a) log3 12 b) 2log3 5

18

c) log2 5 – log3 6 DUM 10122: Engineering Mathematics 1

d) 1 log2 3 + log7 3
2

5. Given log5 2 = 0.43 and log5 3 = 0.68. Find the value of each of the following

a) log5 24 b) log5 60

c) log5 0.3 d) log5 6

6. Given that r = log10 3 , express the following in terms of r

a) log10 90 b) log10 0.3

c) log3 10 log10  1 
 27 
d)

19