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Laws of Indices2
2.1 Simplifying Algebraic Expressions Involving Indices2.2 Zero and Negative Integral Indices
2.3 Simple Exponential Equations
Chapter Summary
Mathematics in Workplaces
2.4 Different Numeral Systems
2.5 Inter-conversion between Different NumeralSystems
P. 2
BiologistIn the 1840’s, biologists found that all plants and animals, including humans, are made up of cells.
Cells are created from cell division. Each time a cell division takes place, a parent cell divides into 2 daughter cells. Solving exponential equations like 2n 215 can help biologists determine the growth rate of cells.
Mathematics in Workplaces
P. 3
2.1 Simplifying Algebraic Expressions Involving Indices
A. Law of Index of (am)n
Suppose m and n are positive integers, we have
times
)(n
mmmmnm aaaaa
timesn
mmmma
amn
For any positive integers h and k, we have ah ak ah k.
If m and n are positive integers, then (am)n amn.
P. 4
Simplify each of the following expressions.(a) (q3)x (b) (q3)8 (c) (q2y)5
2.1 Simplifying Algebraic Expressions Involving Indices
A. Law of Index of (am)n
Example 2.1T
Solution:(a) (q3)x q3 x
q3x
(b) (q3)8 q3 8
q24
(c) (q2y)5 q2y 5
q10y
P. 5
)...()...( times times
nn
bbbbaaaa
2.1 Simplifying Algebraic Expressions Involving Indices
B. Law of Index of (ab)n
Suppose n is a positive integer, we have
times
)()()()()(n
n ababababab
anbnGroup the terms of a and b separately.
If n is a positive integer, then (ab)n anbn.
P. 6
2.1 Simplifying Algebraic Expressions Involving Indices
B. Law of Index of (ab)n
Example 2.2TSimplify each of the following expressions.(a) (11u2)2 (b) (3b4)3
Solution:(a) (11u2)2 112u2 2
121u4
(b) (3b4)3 33b4
3 27b12
P. 7
timesn
n
b
a
b
a
b
a
b
a
b
a
times
times
n
n
bbbb
aaaa
n
n
b
a
2.1 Simplifying Algebraic Expressions Involving Indices
n
ba
C. Law of Index of
is undefined.0
1
If n is a positive integer,
then , where b 0.n
nn
b
a
b
a
When a fraction is multiplied by itself n times,
where b 0 and n is any positive integer, we can simplify the expression as follows:
ba
P. 8
2 4
22
4 (b)
d
c
d
c
2.1 Simplifying Algebraic Expressions Involving Indices
n
ba
C. Law of Index of
Example 2.3T
42
n
Simplify each of the following expressions.
(a) , n 0 (b) , d 02
4
d
c
Solution:
8
2
d
c
4
44 22 (a)
nn
4
16
n
P. 9
)()2(
2 (a)
32
535
kh
kh2
4
233
7)7( (b)
v
uuv
82
4393
7
7
v
uuv
2.1 Simplifying Algebraic Expressions Involving Indices
n
ba
C. Law of Index of
Example 2.4T
)()2(
232
535
kh
kh2
4
233
7 )7(
v
uuv
Simplify each of the following expressions.
(a) , h and k 0 (b) , v 0
For any positive odd integer m, (1)m 1.For any positive even integer n, (1)n 1.
Solution:
223252 kh 28hk
22
3535
2
2
h
kh
2 42
2 233 33
7)7(
v
uuv
4389237 uvvu77
P. 10
2.1 Simplifying Algebraic Expressions Involving Indices
n
ba
C. Law of Index of
Example 2.5TSimplify 64y 8x 42y. Solution:
yxy 2236 )2()2()2( yxy 436 222
yxy 4 3 62 yx 232
64y 8x 42y
Change the numbers to the same base before applying the laws of indices, i.e., write64 26, 8 23 and4 22.
P. 11
2.2 Zero and Negative Integral Indices
A. Zero Index
In Book 1A, we learnt that am an am n for m n.
Consider the case when m n: am n a0
However, if we calculate the actual value of the expression 32 32,
32 32 9 9 1
Hence, we define the zero index of any non-zero number as follows:
We can conclude that 30 1.
For example, 32 32 32 2 30.
If a 0, then a0 1.
00 is undefined.
P. 12
B. Negative Integral Indices
Consider am an am n. If m n, then m n is negative.
The expression am n has a negative index.For example, 52 53 52 3 51.
However, 52 53 25 125 . 5
1
We can conclude that 51 .5
1
Hence, we define the negative index of any non-zero number as follows:
2.2 Zero and Negative Integral Indices
If a 0 and n is a positive integer,
then .n
n
aa
10n is undefined.
P. 13
32123 )a( 50
1)7(
1)2()7( )b(
303
2323
)10(
1
5
1)10(5 )c(
125
100
Example 2.6T
B. Negative Integral Indices
2.2 Zero and Negative Integral Indices
Find the values of the following expressions without using a calculator.(a) 30 25 (b) (7)3 (2)0 (c) 53 (10)2
Solution:
32
1
343
1
1
10
5
1 2
3
5
4
P. 14
2.2 Zero and Negative Integral Indices
Summarizing the previous results, we have the following laws of integral indices.
If m and n are integers, then1. am an am n
2. am an am n (where a 0)3. (am)n amn
4. (ab)n anbn
5. (where b 0)
6. a0 1 (where a 0)
7. (where a 0)
nn
aa
1
n
nn
b
a
b
a
B. Negative Integral Indices
P. 15
5 12 2
5122
)()( )a(
uu
uu
4
41
)(
13
)()3( )b(
ss
ss
33s
Example 2.7T
2.2 Zero and Negative Integral Indices
B. Negative Integral Indices
Simplify the following expressions and express the answers with positive indices.(a) (u2)2(u1)5, u 0 (b) (3s1) (s)4, s 0
Solution:
Since it is stated that each answer should be written with positive indices, it is incorrect to express the answer as u1.
)5( 4 u1u
u
1
43s
s
1 43 s
P. 16
43 23
4
0332
)3(
1)1(2
)3(
)3()2( (a)
yz
y
yz
Example 2.8T
2.2 Zero and Negative Integral Indices
B. Negative Integral Indices
5
23
4
)(
qr
p
Simplify the following expressions and express the answers with positive indices.
(a) , y 0 (b) , p, q and r 04
0332
)3(
)3()(2 y
yz
Solution:
463 )3(2 yz 4463 32 yz
64648 zy
Alternative Solution:
)4()4(3 23
4
0332
)3(1 2
)3(
)3()2(
yz
y
yz
64648 zy
P. 17
5
4
235
23
4 )(
)( (b)
p
qr
qr
p
Example 2.8T
2.2 Zero and Negative Integral Indices
B. Negative Integral Indices
5
23
4
)(
qr
p
Simplify the following expressions and express the answers with positive indices.
(a) , y 0 (b) , p, q and r 04
0332
)3(
)3()(2 y
yz
Solution:
1020
15
qp
r
5 4
5 25 3)(
p
qr
20
1015
p
qr
Since it is stated that each answer should be written with positive indices, it is incorrect to express the answer in terms of q 10.
P. 18
The variable x of this equation appears as an index.
Such equations are called exponential equations.
Method of solving exponential equations:First express all numbers in index notation with the same base.
For example, 2x 82x 23
x 3
2.3 Simple Exponential Equations
Consider the equation 2x 8.
Then simplify the expression using laws of integral indices if necessary.For example, (9t)2 81
92t 92
2t 2 t 1
(am)n amn
Express both sides as powers of 2
Equate the indices on both sides
P. 19
2.3 Simple Exponential Equations
Example 2.9T
216
1Simplify the following exponential equations.(a) 103k 1000 (b) 2k 1 (c) 6k
Solution:(a) 103k 1000103k 103
3k 3 k 1
(b) 2k 12k 20
k 0
k 3
(c) 6k
216
1
6k
36
1
6k 63
P. 20
1 2 6
1 6
)7(
17
49
17 )a(
mm
mm
2.3 Simple Exponential Equations
Example 2.10TSimplify the following exponential equations.
(a) (b) 2x 1 5 2x 281
6
49
17
m
m
Solution:
8226
)22(6
mmmmm
2 2 6
7
17
m
m
)2 2( 6 77 mm
(b) 2x 1 5 2x 28 2 2x 5 2x 28 (2 5) 2x 28 7 2x 28 2x 4 2x 22
x 2
Express all numbers inindex notation with thesame base.
Apply the techniques ofsolving linear equationswith one unknown.
P. 21
The most commonly used numeral system today is the denary system.
Numbers in this system are called denary numbers.
The denary system consists of 10 basic numerals: ‘0’, ‘1’, ‘2’, ‘3’, ‘4’, ‘5’, ‘6’, ‘7’, ‘8’ and ‘9’.
Consider the expanded form of 236 with base 10:
236 = 2 102 + 3 101 + 6 100
The numbers 102, 101 and 100 are the place values of the corresponding positions/digits of a number.
The place values of numbers in this system differ by powers of 10.
A. Denary System
2.4 Different Numeral Systems
P. 22
Another commonly used numeral system is the binary system.
Numbers in this system are called binary numbers.
The binary system consists of only 2 numerals: ‘0’ and ‘1’.
For example, the expanded form of 1011(2) is:
1011(2) = 1 23 + 0 22 + 1 21 + 1 20
The numbers 23, 22, 21 and 20 are the place values of the corresponding positions/digits of a number.
The place values of the digits in a binary number differ by powers of 2.
2.4 Different Numeral Systems
B. Binary System
P. 23
Another commonly used numeral system is the hexadecimal system.
Numbers in this system are called hexadecimal numbers.
The hexadecimal system consists of 16 numerals and letters: ‘0’, ‘1’, ‘2’, ‘3’, ‘4’, ‘5’, ‘6’, ‘7’, ‘8’, ‘9’, ‘A’, ‘B’, ‘C’, ‘D’, ‘E’ and ‘F’.The letters A to F represent the values 10(10) to 15(10) respectively.
For example, the expanded form of 13A(16) is:
13A(16) = 1 162 + 3 161 + 10 160
The numbers 162, 161 and 160 are the place values of the corresponding positions/digits of a number.
The place values of the digits in a hexadecimal number differ by powers of 16.
2.4 Different Numeral Systems
C. Hexadecimal System
P. 24
)2(012 101212021 )a(
)10(012 490100109104 )b(
2.4 Different Numeral Systems
C. Hexadecimal System
Example 2.11T(a) Express 1 22 0 21 1 20 as a binary number.(b) Express 4 102 9 101 0 100 as a denary number.Solution:
P. 25
It can be done by summing up all the terms in the expanded form.
A. Convert Binary/Hexadecimal Numbers into Denary Numbers
2.5 Inter-conversion between Different Numeral Systems
We can make use of the expanded form to convert binary/hexadecimal numbers into denary numbers.
P. 26
012)2( 212121111 (a)
0123)2( 212020211001 (b)
2.5 Inter-conversion between Different Numeral Systems
A. Convert Binary/Hexadecimal Numbers into Denary Numbers
Example 2.12TConvert the following binary numbers into denary numbers.(a) 111(2)
(b) 1001(2)
Solution:
124
18
)10(7
)10(9
P. 27
01)16( 16616666 )a(
012)16( 1612162161C12 )(b
2.5 Inter-conversion between Different Numeral Systems
A. Convert Binary/Hexadecimal Numbers into Denary Numbers
Example 2.13TConvert the following hexadecimal numbers into denary numbers.(a) 66(16)
(b) 12C(16)
Solution:
696 102
1232256 300
P. 28
2.5 Inter-conversion between Different Numeral Systems
B. Convert Denary Numbers into Binary/Hexadecimal Numbers
It can be done by considering all the remainders in the short division.
We make use of division to convert denary numbers into binary/hexadecimal numbers.
P. 29
2.5 Inter-conversion between Different Numeral Systems
B. Convert Denary Numbers into Binary/Hexadecimal Numbers
Example 2.14TConvert the denary number 33(10) into a binary number.
Solution:2 33
16 … 1
8 … 0
2
2
2 4 … 0
2 2 … 0
1 … 0
33(10) 100001
(2)
P. 30
2.5 Inter-conversion between Different Numeral Systems
B. Convert Denary Numbers into Binary/Hexadecimal Numbers
Example 2.15TConvert the denary number 530(10) into a hexadecimal number.
Solution:16 530
33 … 216
2 … 1
530(10) 212(16)
P. 31
Chapter Summary
2.1 Simplifying Algebraic Expressions Involving Indices
n
nn
b
a
b
a
For positive integers m and n,
1. (am)n amn.
2. (ab)n anbn.
3. , where b 0.
P. 32
Chapter Summary
2.2 Zero and Negative Integral Indices
na
1
For any non-zero number a and positive integer n,
1. a0 1.
2. an .
P. 33
2.3 Simple Exponential Equations
When solving exponential equations, first express all numbers in index notation with the same base, then simplify using the laws of integral indices.
Chapter Summary
P. 34
System Binary Denary Hexadecimal
Digits used
0, 1 0, 1, 2, 3, 4, 5, 6, 7, 8, 9
0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A, B, C, D, E, F
Place values
20, 21, … 100, 101, … 160, 161, …
2.4 Different Numeral Systems
Chapter Summary
P. 35
Inter-conversion of numbers can be done by division or expressing them in the expanded form.
2.5 Inter-conversion between Different Numeral Systems
Chapter Summary
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