2. 3 About this book
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5. 1 Indices and surds
Test this rule for yourself by using
1.1 Integer indices Generally, for any number a, a0 = 1 different values of a.
In the quantity 23, the base is 2 and the index is 3. index
23
23 is evaluated as 23 = 2 ´ 2 ´ 2 = 8 base Consider
3
33 ÷ 35 = 35 = 3×3×3 = 1 = 1
3 3 × 3 × 3 × 3 × 3 3 × 3 32
Generally, for any number a, the quantity an index From the general rule for dividing indices you know that
has base a and index n. an 33 ¸ 35 = 33-5 = 3-2
base
an is evaluated as an = a ´ a ´ . . . ´ a
Hence it follows that 3−2 = 12
3
n times
There are rules to help you perform arithmetic using indices. Indices is the plural of index.
A negative index is equivalent to
Generally, for any number a, a −n = 1n
− a positive reciprocal index.
a
Consider 23 ´ 22 = (2 ´ 2 ´ 2) ´ (2 ´ 2) = 25 5=3+2
3 ´ 33 = 3 ´ (3 ´ 3 ´ 3) = 34 4=1+3
5´ 53 × 52 = 5 ´ (5 ´ 5 ´ 5) ´ (5 ´ 5) = 56 6=1+3+2
Consider (32)3 = (32) ´ (32) ´ (32)
= (3 ´ 3) ´ (3 ´ 3) ´ (3 ´ 3)
This result is true for = 36 6=2´3
Generally, for any number a, am × an = am+n any numbers m and n.
C1
C1
Generally, for any number a, (am)n = am´n Test this rule for yourself by using
2 ÷ 2 = 22 = 2 × 2 × 2 × 2 × 2 = 2 × 2 × 2 = 23
5
5 2
Consider 3=5-2 different values of a, m and n.
2 2×2
54 ÷ 22 = 52 = 5 × 5 × 5 × 5 = 5 × 5 = 52
4
2=4-2
You can use the laws of indices to evaluate numerical quantities.
5 5×5
EXAMPLE 1
Simplify
m n m- n
This result is true for
Generally, for any number a, a ¸ a = a any numbers m and n. a 32 ´ 33
Some important results follow from these two index laws. b 27 ¸ 24
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a Use am ´ an = a m+n: 32 ´ 33 = 32+3 a = 3, m = 2 and n = 3
2 3
2×2×2 = 35
Consider 23 ¸ 23 = 3 = 2×2×2 =1
2
b Use am ¸ an = a m-n: 27 ¸ 24 = 27-4 a = 2, m = 7 and n = 4
From the general rule for dividing indices you know that = 23
23 ¸ 23 = 23-3 = 20 Using am ¸ an = am-n
with a = 2, m = 3 and n = 3.
Hence it follows that 20 = 1
2 3

6. 1 Indices and surds 1 Indices and surds
EXAMPLE 2
EXAMPLE 4
Evaluate Simplify
a (0.2)3 a (3a2)3
(2)
−2
b 3
b (2a3)-2
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a (0.2)3 = (0.2 ´ 0.2) ´ 0.2 Breaking down the calculation a Use the rule (am)n = am´n on both terms:
makes the multiplication easier (3a2)3= (3)3 ´ (a2)3 = 27 ´ a2´3 Splitting the 3 and the a² may
= 0.04 ´ 0.2 help you to simplify the problem.
= 0.008 to do without a calculator. = 27 ´ a6
= 27a6 Remember that both the 3 and
()
−2 the a² need to be cubed.
3
b 2 b Use the rule a-n = 1n :
()
−2 a
3 1
Use a-n = 1n : = a = 3 and n = 2
()
a 2 2 2 (2a3)-2 = 1
3
2 ( 2a3 )2
= 1 Remember that 1 = b Now use the rule (am)n = am´n in the denominator:
(9)
4
()a
b
a
1 = 1 Remember that both the 2 and
(2a ) 3 2 22 × (a3)2 the a3 need to be squared.
= 4 = 2 1 3×2
9
2 ×a
= 1 6
4×a
You can also use the laws of indices to simplify algebraic expressions.
C1
C1
= 16
4a
EXAMPLE 3
Simplify
a p2 ´ p3
EXAMPLE 5
Simplify each expression giving the final answer in the
b t5 ¸ t3 form kxn where k is a constant and n an integer.
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a Use am ´ an = am+n: p2 ´ p3 = p2+3 a 3a3 ´ (4a)2
−3
= p5
b ⎛ 2⎞
3
⎜ ⎟
⎝x ⎠
b Use am ¸ an = am-n: t5 ¸ t3 = t5-3 ••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••
= t2
a 3a3 ´ (4a)2 = 3a3 ´ 16a2 Using the rule (am)n = am´n
= 3 ´ 16 ´ a3 ´ a2 Using the rule am ´ an = am+n
You can apply the index laws to simplify expressions which = 48a5
contain numbers and variables. −3
b ⎛ 32 ⎞
⎜ ⎟ = 1
3
Using the rule a-n = 1n
⎝x ⎠ ⎛ 3 ⎞ a
⎜ 2⎟
⎝x ⎠
= 1 Using the rule (am)n = am´n
27
x6 ( ) to simplify the denominator.
6
=x Since 1 = b
27 ( ba ) a
4 5

7. 1 Indices and surds 1 Indices and surds
Exercise 1.1 5 Simplify these expressions.
1 Evaluate the following quantities. 2
7 5 a ⎛ 5a ⎞
⎜ 2⎟ b (-2a2b)4
a 2 b 3 ⎝b ⎠
c 132 d 23 + 42
c (5t2)3 - (5y2)3 d (2y)3 ´ (3z)3
e 25 + 52 f 92 - 34 3
e (3a)-2 f 2
(4) r2
3
g (22)3 h
3
( 3b )
1 −2
g 2z 2 h (4b)−3 × 1
3 2 2
i (0.4) j (1.01) - (0.99) 3y
k (98)2 - (2)2 l (0.4)2 - (9.6)2 i 4(y2)-3 ¸ (2y)-2 j (3y2t)2 ¸ (3yt2)-2
2 Write each of the following quantities in an index form. 6 Simplify each of the following expressions where possible.
a 32 b 81 c 125 a 4a2 + 9a2 b 9a2 + 4a3
d 0.001 e 25 f 0.04 c 4a2 ´ 9a3 d 4a2 ¸ 9a3
64
e 9a2 - 4a3 f 9a2 ¸ 4a3
g 1 h 441 i 128
10 000 400 243 g 9 ¸ 4a2 h 4a - 9a
C1
C1
3 Evaluate these quantities. i 9a2 ´ 4a - 9a2 j 4a2 ¸ 9a - 4a
a 4-1 b 6-2
INVESTIGATIONS
3
c 70 d (1)
2
7 Which of these numbers can be expressed either as
a power of 3 or as a power of 4?
3 1 1 64 12 1 1 24 81 36
e (2)
3
f (0.5)-2 9 24 16 27
4 8
8 Which of these expressions simplify to give the
g ()
1
3
h ()
1
2 answer, a?
( a25 )
−1
3 3 −1 3
4 Simplify each expression. 5a 4a 2a 3a 4a
a2 (2a)2 2a2 3a −2 2a2 (5a)−2
a p ´ p3 b 3a ´ 4a2 ´ 5a3
c 12b ¸ 3b4 d (3y2)2 9 A saver deposits £80 with a bank giving an interest rate of
5%. Use your knowledge of indices to express as a ´ bn
e 7p2 + 9p2 f (5a)2 ¸ 5a3 the value of the investment
g (-4b3)2 ´ 5b2 h (3a3)2 ¸ (3a2)3 a after five years
i 4p3(3p3 ¸ 2p3) j (-3y)3 ´ (-2y)2 b after n years.
After how many years and months will the investment
reach £100?
6 7

8. 1 Indices and surds
1.2 Fractional indices
Generally, for any number a, (a ) 1 m
n = (am)n = a n
1 m
( n a )m = n
am = a n
m
The index laws also apply to fractional indices.
You can use the index laws to evaluate expressions involving
fractional indices.
Consider 3 × 3 =3
1 1 1+1
You know that 3 2 × 3 2 = 3 2
EXAMPLE 1
2 Using am ´ an = am+n 2
1
Evaluate a (27)3 b (9)−0.5
=3 ••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••
=3
You can deduce that 3 is written as
1
32 in index notation. a Rewrite the expression: (272)3
1
or (27 ) 1 2
3 Choose ( 27 )
1 2
3 as this is simpler
1 1 to work with.
1
Similarly 4 = 42 and 5 = 52 You can follow the steps above You know that 27 3 = 3 27 = 3
to show this for yourself.
= (27 )
2 1 2
Hence (27)3 3
= (3)2 = 9 You should recognise that
Now consider 3
2× 2× 2 =2
3 3 3
2 is the cube root of 2. 1
27 3 = 3
27 = 3
1 1 1 1+1+1 b (9) −0.5
= 11
You know that 23 × 23 × 23 = 23 3 3 You can deduce this for yourself (9)2
1
by considering am ´ an ´ ap and 92 = 9 = 3
= 21
using the index laws. =1
=2 3
C1
C1
1
You can deduce that 3 2 is written as 2 3 in index notation. Sometimes the expression will involve a fraction as the
1 1
Similarly 5 =3
53 and 3
7 = 73 base number.
EXAMPLE 2
Evaluate
( ) ( )
1 2 3
−
Generally, for any number a, n
a = an This is the nth root of a. a 125 3
b 256 4
8 81
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The other index laws also apply to fractional indices.
( 1)
2 m
⎡ 1⎤
( ) ( )
2 m
125 3
⎢ 125 ⎥
3
Using a n = a n
a =
8 ⎣ 8 ⎦
= (5)
3
Consider ( 8 )2 2
Since 3 125 = 5 and 3 8 = 2
This can be rewritten as ( )
1 2
83 Using (am)n = am´n
2
1 ×2 = 25 Remember to square both
4
= 83 the top and bottom numbers.
( ) ( )
−
3 3
2 −n
= 83 b 256 4
= 81 4
Using the rule a = 1n
81 256 a
1 2
( ) =
Similarly 82 3 83 Using (am)n = am´n ⎡
= ⎢ 81 ⎥ ( )
1 ⎤3
4
Using a n =
m
( )
1 m
an = n am
Hence ( )
1 2
83 =
1
( ) =
82 3
2
83
⎣ 256 ⎦
(4)
3
= 3
= 27 Remember to cube both the
64 top and bottom numbers.
8 9