Basic algebra 67
The HCF of each of the three terms is a
1/2 b. Dividing
each term by a 1/2 b gives
a 2 b
ab 2 − a 1/2 b 3 =
a 2 b
a 1/2 b
ab 2
a 1/2 b
−
a 1/2 b 3
a 1/2 b
=
a 3/2
a 1/2 b−b
2
Problem 29. Simplify (a 3 √
b
√
c 5 )(
√
a
3
√
b 2 c 3 )
and evaluate when a =
1
4
, b = 6 and c = 1
Using law (4) of indices, the expression can be written as
(a
3
√
b
c 5 )(
√
a
3
b 2 c
3
) =
a
3 b
1
2 c
5
2
a
1
2 b
2
3 c
3
Using law (1) of indices gives
a
3 b
1
2 c
5
2
a
1
2 b
2
3 c
3
= a
3+
1
2 b
1
2 +
2
3 c
5
2 +3
= a
7
2 b
7
6 c
11
2
It is usual to express the answer in the same form as the
question. Hence,
a
7
2 b
7
6 c
11
2 =
a 7
6
b
7
c 11
When a =
1
4
, b = 64 and c = 1,
a 7
6
b 7
c 11 =
1
4
7
6
√
64 7
√
1 11
=
1
2
7
(2)
7
(1) = 1
Problem 30. Simplify
(x 2 y 1/2 )(
√
x
3
y 2 )
(x 5 y 3 ) 1/2
Using laws (3) and (4) of indices gives
x 2 y 1/2 √
x
3
y 2
(x 5 y 3 ) 1/2
=
x 2 y 1/2
x 1/2 y 2/3
x 5/2 y 3/2
Using laws (1) and (2) of indices gives
x
2+
1
2 −
5
2 y
1
2 +
2
3 −
3
2 = x
0 y
−
1
3 = y
−
1
3 or
1
y 1/3 or
1
3
√ y
from laws (5) and (6) of indices.
Now try the following Practice Exercise
Practice Exercise 38 Laws of indices
(answers on page 343)
1. Simplify
a 3/2 bc −3
a 1/2 b −1/2 c
and evaluate when a = 3, b = 4 and c = 2.
In problems 2 to 5, simplify the given expressions.
2.
a 2 b + a 3 b
a 2 b 2
3. (a 2 ) 1/2 (b 2 ) 3
c 1/2 3
4.
(abc) 2
(a 2 b −1 c −3 ) 3
5.
p 3 q 2
pq 2 − p 2 q
6. (
√
x
y 3 3
√
z 2 )(
√
x
y 3
z 3 )
7. (e
2 f
3
)(e
−3 f
−5
), expressing the answer with
positive indices only.
8.
(a 3 b 1/2 c −1/2 )(ab) 1/3
(
√
a 3
√
b c)