Powers, roots and laws of indices 49

(4) 3^0 =1 and 17^0 = 1

This is the fourth law of indices, which states that

when a number has an index of 0, its value is 1.

(5) 3−^4 =

1

34

and

1

2 −^3

= 23

This isthe fifth law of indices, whichdemonstrates

thata number raised to a negative power is the

reciprocal of that number raised to a positive

power.

(6) 8

32

=

√ 3

82 =( 2 )^2 =4and

25

1

(^2) =^2
√
251 =
√
251 =± 5
(Note that
√
≡^2
√
)
This is the sixth law of indices, which demon-
strates thatwhen a number is raised to a frac-
tional power the denominator of the fraction is
the root of the number and the numerator is the
power.
Here are some worked examples using the laws of
indices.
Problem 6. Evaluate in index form 5^3 × 5 × 52
53 × 5 × 52 = 53 × 51 × 52 (Note that 5 means 5^1 )
= 53 +^1 +^2 from law (1)
= 56
Problem 7. Evaluate
35
34
35
34
= 35 −^4 from law (2)
= 31
= 3
Problem 8. Evaluate
24
24
24
24
= 24 −^4 from law (2)
= 20
But
24
24

2 × 2 × 2 × 2
2 × 2 × 2 × 2

16
16
= 1
Hence, 20 = 1 from law (4)
Any number raised to the power of zero equals 1.For
example, 6^0 = 1 , 1280 = 1 , 137420 =1 and so on.
Problem 9. Evaluate
3 × 32
34
3 × 32
34

31 × 32
34

31 +^2
34

33
34
= 33 −^4 = 3 −^1
from laws (1) and (2)
But
33
34

3 × 3 × 3
3 × 3 × 3 × 3

1 × 1 × 1
1 × 1 × 1 × 3
(by cancelling)

1
3
Hence,
3 × 32
34
= 3 −^1 =
1
3
from law (5)
Similarly, 2−^1 =
1
2
, 2 −^5 =
1
25
,
1
54
= 5 −^4 and so on.
Problem 10. Evaluate
103 × 102
108
103 × 102
108

103 +^2
108

105
108
from law (1)
= 105 −^8 = 10 −^3 from law (2)

1
10 +^3

1
1000
from law (5)
Hence,
103 × 102
108
= 10 −^3 =
1
1000
= 0. 001
Understanding powers of ten is important, especially
when dealing with prefixes in Chapter 8. For example,
102 = 100 , 103 = 1000 , 104 = 10000 ,
105 = 100000 , 106 = 1000000
10 −^1 =
1
10
= 0. 1 , 10 −^2 =
1
102

1
100
= 0. 01
and so on.
Problem 11. Evaluate (a) 5^2 × 53 ÷ 54
(b) (3× 35 )÷( 32 × 33 )
From laws (1) and (2):
(a) 5^2 × 53 ÷ 54 =
52 × 53
54

5 (^2 +^3 )
54

55
54
= 5 (^5 −^4 )= 51 = 5