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© Boardworks Ltd 20051 of 38 © Boardworks Ltd 20051 of 38
AS-Level Maths: Core 1for Edexcel
C1.1 Algebra and functions 1
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Co
nte
nts
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Using and manipulating surds
Rationalizing the denominator
The index laws
Zero and negative indices
Fractional indices
Solving equations involving indices
Examination-style questions
The index laws
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Index notation
Simplify:
a × a × a × a × a = a5
a to the power of 5
a5 has been written using index notation.
anThe number a is called the base.
The number n is called the index, power or exponent.
In general:
an = a × a × a × … × a
n of these
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Index notation
Evaluate the following:
0.62 = 0.6 × 0.6 = 0.36
34 = 3 × 3 × 3 × 3 = 81
(–5)3 = –5 × –5 × –5 = –125
27 = 2 × 2 × 2 × 2 × 2 × 2 × 2 = 128
(–1)5 = –1 × –1 × –1 × –1 × –1 = –1
(–4)4 = –4 × –4 × –4 × –4 = 256
When we raise a negative number to an odd power the
answer is negative.
When we raise a negative number to an even power the answer is positive.
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The multiplication rule
For example:
a4 × a2 = (a × a × a × a) × (a × a)
= a × a × a × a × a × a
= a6
When we multiply two terms with the same base the indices are added.
When we multiply two terms with the same base the indices are added.
= a (4 + 2)
In general:
am × an = a(m + n)am × an = a(m + n)
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The division rule
For example:
a5 ÷ a2 =a × a × a × a × a
a × a= a3
4p6 ÷ 2p4 = 2p2
= a (5 – 2)
= 2p(6 – 4)
When we divide two terms with the same base the indices are subtracted.
When we divide two terms with the same base the indices are subtracted.
In general:
am ÷ an = a(m – n)am ÷ an = a(m – n)
24 × p × p × p × p × p × p
2 × p × p × p × p=
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For example:
(y3)2 = (pq2)4 =
The power rule
y3 × y3
= (y × y × y) × (y × y × y)
= y6
pq2 × pq2 × pq2 × pq2
= p4 × q (2 + 2 + 2 + 2)
= p4 × q8
= p4q8
When a term is raised to a power and the result raised to another power, the powers are multiplied.
When a term is raised to a power and the result raised to another power, the powers are multiplied.
In general:
= y3×2
= p1×4q2×4
(am)n = amn(am)n = amn
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Co
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Using and manipulating surds
Rationalizing the denominator
The index laws
Zero and negative indices
Fractional indices
Solving equations involving indices
Examination-style questions
Zero and negative indices
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The zero index
Look at the following division:
y4 ÷ y4 = 1
But using the rule that xm ÷ xn = x(m – n)
y4 ÷ y4 = y(4 – 4) = y0
That means that y0 = 1
In general:
a0 = 1 (for all a ≠ 0)a0 = 1 (for all a ≠ 0)
Any number or term divided by itself is equal to 1.
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Look at the following division:
b2 ÷ b4 =b × b
b × b × b × b=
1b × b
=1b2
But using the rule that am ÷ an = a(m – n)
b2 ÷ b4 = b(2 – 4) = b–2
That means that b–2 = 1b2
In general:
Negative indices
a–n = 1an
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Negative indices
Write the following using fraction notation:
This is the reciprocal of u.
4) 5a(3 – b)–2 =
1) u–1 = 1
u
2) 2n–4 = 4
2
n
3) x2y–3 =
2
3
x
y
2
5
(3 )
a
b
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21) =
a
3
42) =
x
y
2
3) =+ 2
p
q
2 3
34) =
( 5)
m
n
Negative indices
Write the following using negative indices:
2a–1
x3y–4
p2(q + 2)–1
3m(n2 – 5)–3
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Using and manipulating surds
Rationalizing the denominator
The index laws
Zero and negative indices
Fractional indices
Solving equations involving indices
Examination-style questions
Fractional indices
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Indices can also be fractional. For example:
Fractional indices
So
But a a a= ×
a a a1 1 1 1+2 2 2 2× =
= a1
What is the meaning of ?a12
Using the multiplication rule:
= a
a a12 =
is the square root of a.
a12
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Fractional indices
Similarly:a a a a
1 1 1 1 1 1+ +3 3 3 3 3 3× × =
= a1
= a
But a a a a= × ×3 3 3
So a a13 = 3
In general:
nna a1=
is the cube root of a.
a13
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Fractional indices
Using the rule that (am)n = amn, we can write
In general:
What is the meaning of ?a23
or=mn mna a =
mn
mna a
We can write2 13 3
2 as .a a
2 13 32 23( )a a a
We can also write 2 13 3
2 as .a a
2 13 3 2 23( ) ( )a a a
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Fractional indices
Evaluate the following:
541) 16
232) (0.125) 3
23) 36
54 5416 = ( 16)
5= 2
= 32
2 23 3(0.125) = 8
23= ( 8)
2= 2
= 4
32
3
136 =
( 36)
3
1=
6
1=
216
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Here is a summary of the index laws for all rational exponents:
Summary of the index laws
12 =a a
( + )× =m n m na a a
( )÷ =m n m na a a
( ) =m n mna a
1 =a a
0 = 1 (for 0)a a
1
=n na a
=mn n ma a
1=n
na
a
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Co
nte
nts
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Using and manipulating surds
Rationalizing the denominator
The index laws
Zero and negative indices
Fractional indices
Solving equations involving indices
Examination-style questions
Solving equations involving indices
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Solving equations involving indices
We can use the index laws to solve certain types of equation involving indices. For example:
Solve the equation 25x = 1255 – x.
25x = 1255 – x
(52)x = (53)5 – x
52x = 53(5 – x)
2x = 3(5 – x)
2x = 15 – 3x
5x = 15
x = 3
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Co
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Using and manipulating surds
Rationalizing the denominator
The index laws
Zero and negative indices
Fractional indices
Solving equations involving indices
Examination-style questions
Examination-style questions
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Examination-style question 1
6 + 3Show that can be written in the form + 2 where
6 3a b
a and b are integers. Hence find the values of a and b.
Multiplying top and bottom by gives6 + 3
6 + 3 6 + 3=
6 3 6 + 36+2 6 3 +3
6 3
9+2 18=
39+6 2
=3
.18 can be written as 3 2
= 3+2 2
So a = 3 and b = 2
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Examination-style question 2
a) Express 32x in the form 2ax where a is an integer to be determined.
b) Use your answer to part a) to solve the equation2
32 = 2x x
a) 32 = 25
So 32x = (25)x
Using the rule that (am)n = amn32x = 25x
b) Using the answer from part a) this equation can be written as252 = 2x x
5x = x2
5x – x2 = 0x (5 – x) = 0
x = 0 or x = 5