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
3 4 and
1
2 −3 = 2 3
This is the fifth law of indices, which demonstrates
that a number raised to a negative power is the
reciprocal of that number raised to a positive
power.
(6) 8
2
3 =
3
√
8 2 = (2) 2 = 4 and
25
1
2 =
2
√
25 1 =
√
25 1 = ±5
(Note that
√ ≡
2
√
)
This is the sixth law of indices, which demonstrates that when a number is raised to a fractional 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 × 5 2
5
3
× 5 × 5
2
= 5
3
× 5
1
× 5
2
(Note that 5 means 5
1
)
= 5
3+1+2
from law (1)
= 5
6
Problem 7. Evaluate
3 5
3 4
3 5
3 4 = 3
5−4
from law (2)
= 3
1
= 3
Problem 8. Evaluate
2 4
2 4
2 4
2 4 = 2
4−4
from law (2)
= 2
0
But
2 4
2 4 =
2 × 2 × 2 × 2
2 × 2 × 2 × 2
=
16
16
= 1
Hence,
2
0
= 1
from law (4)
Any number raised to the power of zero equals 1. For
example, 6
0
= 1, 128
0
= 1, 13742
0
= 1 and so on.
Problem 9. Evaluate
3 × 3
2
3 4
3 × 3 2
3 4 =
3 1 × 3 2
3 4 =
3 1+2
3 4 =
3 3
3 4 = 3
3−4
= 3
−1
from laws (1) and (2)
But
3 3
3 4 =
3 × 3 × 3
3 × 3 × 3 × 3
=
1 × 1 × 1
1 × 1 × 1 × 3
(by cancelling)
=
1
3
Hence,
3 × 3 2
3 4 = 3
−1
=
1
3
from law (5)
Similarly, 2
−1
=
1
2
, 2
−5
=
1
2 5 ,
1
5 4 = 5
−4 and so on.
Problem 10. Evaluate
10 3 × 10 2
10 8
10 3 × 10 2
10 8
=
10 3+2
10 8 =
10 5
10 8
from law (1)
= 10
5−8
= 10
−3
from law (2)
=
1
10 +3 =
1
1000
from law (5)
Hence,
10 3 × 10 2
10 8
= 10
−3
=
1
1000
= 0.001
Understanding powers of ten is important, especially
when dealing with prefixes in Chapter 8. For example,
10
2
= 100, 10
3
= 1000, 10
4
= 10 000,
10
5
= 100 000, 10
6
= 1 000 000
10
−1
=
1
10
= 0.1, 10
−2
=
1
10 2 =
1
100
= 0.01
and so on.
Problem 11. Evaluate (a) 5 2 × 5 3 ÷ 5 4
(b) (3 × 3
5
) ÷ (3
2
× 3
3
)
From laws (1) and (2):
(a) 5 2 × 5 3 ÷ 5 4 =
5 2 × 5 3
5 4 =
5 (2+3)
5 4
=
5 5
5 4 = 5 (5−4) = 5 1 = 5
(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
3 4 and
1
2 −3 = 2 3
This is the fifth law of indices, which demonstrates
that a number raised to a negative power is the
reciprocal of that number raised to a positive
power.
(6) 8
2
3 =
3
√
8 2 = (2) 2 = 4 and
25
1
2 =
2
√
25 1 =
√
25 1 = ±5
(Note that
√ ≡
2
√
)
This is the sixth law of indices, which demonstrates that when a number is raised to a fractional 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 × 5 2
5
3
× 5 × 5
2
= 5
3
× 5
1
× 5
2
(Note that 5 means 5
1
)
= 5
3+1+2
from law (1)
= 5
6
Problem 7. Evaluate
3 5
3 4
3 5
3 4 = 3
5−4
from law (2)
= 3
1
= 3
Problem 8. Evaluate
2 4
2 4
2 4
2 4 = 2
4−4
from law (2)
= 2
0
But
2 4
2 4 =
2 × 2 × 2 × 2
2 × 2 × 2 × 2
=
16
16
= 1
Hence,
2
0
= 1
from law (4)
Any number raised to the power of zero equals 1. For
example, 6
0
= 1, 128
0
= 1, 13742
0
= 1 and so on.
Problem 9. Evaluate
3 × 3
2
3 4
3 × 3 2
3 4 =
3 1 × 3 2
3 4 =
3 1+2
3 4 =
3 3
3 4 = 3
3−4
= 3
−1
from laws (1) and (2)
But
3 3
3 4 =
3 × 3 × 3
3 × 3 × 3 × 3
=
1 × 1 × 1
1 × 1 × 1 × 3
(by cancelling)
=
1
3
Hence,
3 × 3 2
3 4 = 3
−1
=
1
3
from law (5)
Similarly, 2
−1
=
1
2
, 2
−5
=
1
2 5 ,
1
5 4 = 5
−4 and so on.
Problem 10. Evaluate
10 3 × 10 2
10 8
10 3 × 10 2
10 8
=
10 3+2
10 8 =
10 5
10 8
from law (1)
= 10
5−8
= 10
−3
from law (2)
=
1
10 +3 =
1
1000
from law (5)
Hence,
10 3 × 10 2
10 8
= 10
−3
=
1
1000
= 0.001
Understanding powers of ten is important, especially
when dealing with prefixes in Chapter 8. For example,
10
2
= 100, 10
3
= 1000, 10
4
= 10 000,
10
5
= 100 000, 10
6
= 1 000 000
10
−1
=
1
10
= 0.1, 10
−2
=
1
10 2 =
1
100
= 0.01
and so on.
Problem 11. Evaluate (a) 5 2 × 5 3 ÷ 5 4
(b) (3 × 3
5
) ÷ (3
2
× 3
3
)
From laws (1) and (2):
(a) 5 2 × 5 3 ÷ 5 4 =
5 2 × 5 3
5 4 =
5 (2+3)
5 4
=
5 5
5 4 = 5 (5−4) = 5 1 = 5