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AS-Level Maths: Core 1for Edexcel

C1.1 Algebra and functions 1

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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


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


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.


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)


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=


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


Using index laws


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The index laws


Fractional indices





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.


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


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


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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The index laws


Fractional indices



Fractional indices


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


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


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


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


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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The index laws


Fractional 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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The index laws


Fractional indices





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


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

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© Boardworks Ltd 2005 1 of 38 © Boardworks Ltd 2005 1 of 38 AS-Level Maths: Core 1 for Edexcel C1.1 Algebra and functions 1 This icon indicates the slide