66 Basic Engineering Mathematics

Using law (3) of indices gives

d^2 e^2 f^1 /^2

(d^3 /^2 ef^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) sincee^0 =1 from law
(6) of indices

1
df^9 /^2
from law (5) of indices
Now try the following Practice Exercise
PracticeExercise 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


2. n^8 ×n−^5 4. b^4 ×b^7


3. 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 yz^3

19. Simplify(x^2 y^3 z)(x^3 yz^2 )and evaluate when
    x=

1

2

,y=2andz=3.

20. Simplify

a^5 bc^3

a^2 b^3 c^2

and evaluate when

a=

3

2

,b=

1

2

andc=

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

whenp= 16 ,q=9andr=4, taking positive roots

only

Using law (2) of indices givesp

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
Whenp= 16 ,q=9andr=4,
p
1
(^4) q
3
(^2) r
1
(^2) = 16
1
(^49)
3
(^24)
1
2
=(^4
√
16 )(
√
93 )(
√
4 )from law (4) of indices
=( 2 )( 33 )( 2 )= 108
Problem 26. Simplify
x^2 y^3 +xy^2
xy
Algebraic expressions of the form
a+b
c
can be split
into
a
c

• b
  c

. Thus,

x^2 y^3 +xy^2

xy

=

x^2 y^3

xy

+

xy^2

xy

=x^2 −^1 y^3 −^1 +x^1 −^1 y^2 −^1

=xy^2 +y

(sincex^0 =1,fromlaw(6)ofindices).

Problem 27. Simplify

x^2 y

xy^2 −xy

The highest common factor (HCF) of each of the three

terms comprising the numerator and denominator isxy.

Dividing each term byxygives

x^2 y

xy^2 −xy

=

x^2 y

xy

xy^2

xy

−

xy

xy

=

x

y− 1

Problem 28. Simplify

a^2 b

ab^2 −a^1 /^2 b^3