66 Basic Engineering Mathematics
Using law (3) of indices gives
d 2 e 2 f 1/2
(d 3/2 e f 5/2 ) 2 =
d 2 e 2 f 1/2
d 3 e 2 f 5
Using law (2) of indices gives
d
2−3 e
2−2 f
1
2 −5 = d
−1 e
0 f
−
9
2
= d
−1 f
−
9
2
since e
0
= 1 from law
(6) of indices
=
1
df 9/2
from law (5) of indices
Now try the following Practice Exercise
Practice Exercise 37 Laws of indices
(answers on page 343)
In problems 1 to 18, simplify the following, giving
each answer as a power.
1. z 2 × z 6
2. a × a 2 × a 5
3. n 8 × n −5
4. b 4 × b 7
5. b 2 ÷ b 5
6. c 5 × c 3 ÷ c 4
7.
m
5
× m
6
m 4 × m 3
8.
(x
2
)(x)
x 6
9.
x 3 4
10.
y 2 −3
11.
t × t 3 2
12.
c −7 −2
13.
a 2
a 5
3
14.
1
b 3
4
15.
b 2
b 7
−2
16.
1
s 3
3
17. p 3 qr 2 × p 2 q 5 r × pqr 2 18.
x 3 y 2 z
x 5 y z 3
19. Simplify (x 2 y 3 z)(x 3 yz 2 ) and evaluate when
x =
1
2
, y = 2 and z = 3.
20. Simplify
a 5 bc 3
a 2 b 3 c 2 and evaluate when
a =
3
2
, b =
1
2
and c =
2
3
Here are some further worked examples on the laws of
indices
Problem 25. Simplify
p 1/2 q 2 r 2/3
p 1/4 q 1/2 r 1/6 and evaluate
when p = 16, q = 9 and r = 4, taking positive roots
only
Using law (2) of indices gives p
1
2 −
1
4 q
2−
1
2 r
2
3 −
1
6
p
1
2 −
1
4 q
2−
1
2 r
2
3 −
1
6 = p
1
4 q
3
2 r
1
2
When p = 16, q = 9 and r = 4,
p
1
4 q
3
2 r
1
2 = 16
1
4 9
3
2 4
1
2
= (
4
√
16)(
√
9 3 )(
√
4) from law (4) of indices
= (2)(3
3
)(2) = 108
Problem 26. Simplify
x 2 y 3 + x y 2
x y
Algebraic expressions of the form
a + b
c
can be split
into
a
c
+
b
c
. Thus,
x 2 y 3 + x y 2
x y
=
x 2 y 3
x y
+
x y 2
x y
= x
2−1 y
3−1
+ x
1−1 y
2−1
= xy
2
+y
(since x 0 = 1, from law (6) of indices).
Problem 27. Simplify
x 2 y
x y 2 − x y
The highest common factor (HCF) of each of the three
terms comprising the numerator and denominator is x y.
Dividing each term by x y gives
x 2 y
x y 2 − x y
=
x 2 y
x y
x y 2
x y
−
x y
x y
=
x
y − 1
Problem 28. Simplify
a 2 b
ab 2 − a 1/2 b 3