INDEX AND LOG ENGINEERING MATH, Lecture notes of Engineering Mathematics

index and log for engineering math semester 2 university level

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2024/2025

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CHAPTER 1: INDEX AND LOGARITHMS
1.1 LAWS OF INDEX
1.1.1 Define Index
Index are shorthand for repeated multiplication of the same thing by itself. The Index of a number
says how many times to use that number in a multiplication. It is written as a small number to the
right and above the base number.
In this example: 82 = 8 × 8 = 64
# In words: 82 could be called "8 to the second power", "8 to the power 2" or simply "8 squared"
Other names for Index are index or power.
1.1.2 Index Law
If a, b, m and n are real numbers, then;
Name Of The Law
The Index Law
i. ( a m ) n = a mn
ii. (ab)n = an b n
a n
a
n
1. Index
iii.
b
n
b
iv. am x an = a m+n
v. am an = a m - n
2.
Zero Index
a0
= 1
; a 0
i. a
n
=
1
; a 0
a n
3.
Negative Index
ii.
a
m
b
n
; a 0 dan b 0
b n
a m
4.
Fraction Index
m
n
a m
a n
pf3
pf4
pf5
pf8

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CHAPTER 1: INDEX AND LOGARITHMS

1.1 LAWS OF INDEX

1.1.1 Define Index

Index are shorthand for repeated multiplication of the same thing by itself. The Index of a number

says how many times to use that number in a multiplication. It is written as a small number to the

right and above the base number.

In this example: 8^2 = 8 × 8 = 64

# In words: 8^2 could be called "8 to the second power", "8 to the power 2" or simply "8 squared"

Other names for Index are index or power.

1.1.2 Index Law

If a, b, m and n are real numbers, then;

Name Of The Law The Index Law

i. ( a m^ ) n^ = a mn

ii. (ab)n^ = an^ b n

 a  n^ a n

1. Index iii.    ; b  0

 b  b n

iv. am^ x an^ = a m+n

v. am^  an^ = a m^ -^ n

2. Zero Index a

0 = 1 ; a  0

i. a – n^ =

; a  0

a n

3. Negative Index

ii.

a  m  b n

; a  0 dan b  0

b  n^ a^

m

4. Fraction Index

m

a n^ ^ n^ a^ m

(a) Positive Index

1. All non-zero numbers can be written in the index form an^ where a ≠ 0 and n is a

positive integer.

2. an^ is read as a to the power of n. its means a is multiplied by itself n times, that is

Where a is the base and n is the Index

Examples 1:

Find the value of

a. 4 3

d.

b. 

3 

c. 7

e.

Solution

Question Solution

a. 4 3 4 ^4 ^4 ^64

b. ^3 ^ ^3 ^3 ^3 ^27 3

c. 7

d.  ^ ^ ^ 

e. 0.3^4 0.30.30.30.3^ ^ 0.

(b) Zero Index

Any non-zero number with zero Index is equal to 1, that is

a 0  1

Where a ≠ 0

Examples 2

Solution

a.

b.

c.

Question

6 ^3

Solution

6 ^1  1  6  1  0.

6 ^3  1  (6  6  6) 

 4.63  10 ^3

 1  0.3  0.3 0.3  37.

(d) Scientific Notation

By using Index, we can reformat numbers. For very large or very small numbers, it is

sometimes simpler to use "scientific notation"

Examples 4 :

Write in scientific notation:

a. 300 b. 7201 c. 0.

Solution

Question Solution

a. 300

2 1

b. 7201

3 2 1

0.0 0 0 2 56 = 2.56  10 ^4

c. 0.

1 2 3 4

(e) Fraction Index

 a 2 is a positive nth root of a, that is

a^1 n  n a

Where a≠ 0

m

 a^ n^ is the mth^ power of^ n^ a^ or the nth^ root of am^ , that

a m^  a    a 

n n m^ n m

Where a≠ 0

Examples 5

Find the value of:

a. 3 2  3 4  3 8 b. 5 2  5 4  (^58)

Solution

Question Solution

a. 3 2  3 4  3 8 1 5 ^1 ^5

3 2  4 ^  8 ^  3

b. 5 ^5 4 ^5 8 5 2  4 ^  8 ^  5

2 4 1 1 ^2 ^1

c. ^ ^  4 1 2

2 4 2 4 ^3 

Examples 6

Simplify the following expressions using the laws of Index.

a. 34 x 3^3 b. 8^5  82

c. (5^8 )^3 d. ( 2^4 ) -^3

e. 8 0

f. 27

Solution

Question The Index Law Solution

c.

d. n  2  8

2 ^ ^2 

2 n

n  (^2 3) n

42 n^ ^2 2  2 n

 2 n  2  2 3 n  2 4 n

 2  n ^2 ^3 n ^4 n

EXERCISE 1 A

1. Find the value of

a. 63

c.  

b. 0.^

2

d. 2

  1. Simplify each of the following below

a. a^5 x a 6

d. 4 n^ x 16 2n^ x 32 – 2n^ x 8 - n

g. z 7 x z 6  z 5

3. Simplify

b. 3 x 3 8 x 3-^4 c. z 2 y^5 x y –^2

e. m^12 ^ m^3 f. 2 8 ^2

h. 25

n 5 – 2n x 125 2n

a. ( x^3 )^5 b. ( 3x^4 )^2 c. ( 2x^2 y^3 z )^5

d. ( 10^3 )^4 e. 3 ( ab^2 )^ –^4 f. ( 2m^2 )^ ^ ( 4n )^3

4. Simplify each of the following below

a. 8

2 2 n ^3 ^4

d.

81  n

b. 32 n ^3  27  9 n ^2

2 n^ ^2 n ^2

e.

2 2 n ^3

n  1

n

c. 3 2  273 ^9 n

b n ^1

f. b 2 n  1  b 3  4 n

Answer

1. a. 216 b. 0.

c.

1 d. 8

2. a. a^11 b. 3^5

d. 2 – 3n^ e. m 9

g. z 8 h. 5 10n

a. x^15 b. 9x 8

d. 10000

e.

81 a

b

a.

1 b. 81

d. 2 5 n  2 e. 32

c. z^2 y 3

f. 2^4

c. 32 x 10 y 15 z 5

m^2

f.

32 n^3

(^11) n  2

c. 3 2

f. b

3 n  1