Cambridge International Mathematics

Exponents and surds (Chapter 6) 129

6 Express the following in simplest form, without brackets:

a (2b^4 )^3 b

μ

3

x^2 y

¶ 2

c (5a^4 b)^2 d

μ

m^3

2 n^2

¶ 4

e

μ

3 a^3

b^5

¶ 3

f (2m^3 n^2 )^5 g

μ

4 a^4

b^2

¶ 2

h (5x^2 y^3 )^3

Consider

23

23

which is obviously 1.

Using the exponent law for division,

23

23

=2^3 ¡^3 =2^0

We therefore conclude that 20 =1.

In general, we can state thezero index law: a^0 =1 for all a 6 =0.

Now consider

24

27

which is

2 £ 2 £ 2 £ 2

2 £ 2 £ 2 £ 2 £ 2 £ 2 £ 2

=

1

23

Using the exponent law of division,

24

27

=2^4 ¡^7 =2¡^3

Consequently, 2 ¡^3 =

1

23

, which means that 2 ¡^3 and 23 arereciprocalsof each other.

C ZERO AND NEGATIVE INDICES [1.9, 2.4]

1

1

In general, we can state thenegative index law:

Ifais any non-zero number andnis an integer, then a¡n=

1

an

.

This means thatananda¡narereciprocalsof one another.

In particular notice that a¡^1 =

1

a

.

Using the negative index law,

¡ 2

3

¢¡ 4

=

1

¡ 2

3

¢ 4

=1¥

24

34

=1£

34

24

=

¡ 3

2

¢ 4

So, in general we can see that:

³a

b

́¡n

=

μ

b

a

¶n

provided a 6 =0, b 6 =0.

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Y:\HAESE\IGCSE01\IG01_06\129IGCSE01_06.CDR Thursday, 18 September 2008 12:18:04 PM PETER