Basic Engineering Mathematics, Fifth Edition

(Amelia) #1

50 Basic Engineering Mathematics


(b) ( 3 × 35 )÷( 32 × 33 )=

3 × 35
32 × 33

=

3 (^1 +^5 )
3 (^2 +^3 )
=

36
35

= 36 −^5 = 31 = 3

Problem 12. Simplify (a)( 23 )^4 (b)( 32 )^5 ,
expressing the answers in index form

From law (3):
(a) ( 23 )^4 = 23 ×^4 = 212
(b) ( 32 )^5 = 32 ×^5 = 310

Problem 13. Evaluate:

( 102 )^3
104 × 102

From laws (1) to (4):

( 102 )^3
104 × 102

=

10 (^2 ×^3 )
10 (^4 +^2 )

=

106
106

= 106 −^6 = 100 = 1

Problem 14. Find the value of (a)

23 × 24
27 × 25
(b)

( 32 )^3
3 × 39

From the laws of indices:

(a)

23 × 24
27 × 25

=

2 (^3 +^4 )
2 (^7 +^5 )

=

27
212

= 27 −^12

= 2 −^5 =

1
25

=

1
32

(b)

( 32 )^3
3 × 39

=

32 ×^3
31 +^9

=

36
310

= 36 −^10

= 3 −^4 =

1
34

=

1
81

Problem 15. Evaluate (a) 4^1 /^2 (b) 16^3 /^4 (c) 27^2 /^3
(d) 9−^1 /^2

(a) 4^1 /^2 =


4 =± 2
(b) 16^3 /^4 =

√ 4
163 =( 2 )^3 = 8
(Note that it does not matter whether the 4th root
of 16 is found first or whether 16 cubed is found
first – the same answer will result.)
(c) 27^2 /^3 =

√ 3
272 =( 3 )^2 = 9

(d) 9−^1 /^2 =

1
91 /^2

=

1

9

=

1
± 3


1
3

Now try the following Practice Exercise

PracticeExercise 30 Lawsof indices
(answers on page 342)
Evaluate the following without the aid of a
calculator.


  1. 2^2 × 2 × 24 2. 3^5 × 33 × 3
    in index form


3.

27
23

4.

33
35


  1. 7^0 6.


23 × 2 × 26
27

7.

10 × 106
105


  1. 10^4 ÷ 10


9.

103 × 104
109


  1. 5^6 × 52 ÷ 57

  2. (7^2 )^3 in index form 12. (3^3 )^2


13.

37 × 34
35

in 14.

( 9 × 32 )^3
( 3 × 27 )^2

in
index form index form

15.

( 16 × 4 )^2
( 2 × 8 )^3

16.

5 −^2
5 −^4

17.
32 × 3 −^4
33

18.
72 × 7 −^3
7 × 7 −^4

19.

23 × 2 −^4 × 25
2 × 2 −^2 × 26

20.

5 −^7 × 52
5 −^8 × 53

Here are some further worked examples using the laws
of indices.

Problem 16. Evaluate

33 × 57
53 × 34

The laws of indices only apply to termshaving the
same base. Grouping terms having the same base and
then applying the laws of indices to each of the groups
independently gives
33 × 57
53 × 34

=

33
34

×

57
53

= 3 (^3 −^4 )× 5 (^7 −^3 )

= 3 −^1 × 54 =

54
31

=

625
3

= 208

1
3
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