52 Basic Engineering Mathematics
Dividing each term by the HCF (i.e. 2
2 ) gives
2 8 × 3 −4
2 2 × 3 3 − 2 3 =
2 6 × 3 −4
3 3 − 2
=
2 6
3 4 (3 3 − 2)
Problem 23. Simplify
4
3
3
×
3
5
−2
2
5
−3
giving
the answer with positive indices
Raising a fraction to a power means that both the numerator and the denominator of the fraction are raised to that
power, i.e.
4
3
3
=
4 3
3 3
A fraction raised to a negative power has the same value
as the inverse of the fraction raised to a positive power.
Thus,
3
5
−2
=
1
3
5
2 =
1
3 2
5 2
= 1 ×
5 2
3 2 =
5 2
3 2
Similarly,
2
5
−3
=
5
2
3
=
5 3
2 3
Thus,
4
3
3
×
3
5
−2
2
5
−3
=
4 3
3 3 ×
5 2
3 2
5 3
2 3
=
4 3
3 3 ×
5 2
3 2 ×
2 3
5 3 =
(2 2 ) 3 × 2 3
3 (3+2) × 5 (3−2)
=
2 9
3 5 × 5
Now try the following Practice Exercise
Practice Exercise 31 Further problems on
indices (answers on page 342)
In problems 1 to 4, simplify the expressions given,
expressing the answers in index form and with
positive indices.
1.
3 3 × 5 2
5 4 × 3 4
2.
7 −2 × 3 −2
3 5 × 7 4 × 7 −3
3.
4 2 × 9 3
8 3 × 3 4
4.
8 −2 × 5 2 × 3 −4
25 2 × 2 4 × 9 −2
In Problems 5 to 15, evaluate the expressions given.
5.
1
3 2
−1
6. 81 0.25
7. 16
−
1
4
8.
4
9
1/2
9.
9 2 × 7 4
3 4 × 7 4 + 3 3 × 7 2
10.
3 3 × 5 2
2 3 × 3 2 − 8 2 × 9
11.
3 3 × 7 2 − 5 2 × 7 3
3 2 × 5 × 7 2
12.
(2 4 ) 2 − 3 −2 × 4 4
2 3 × 16 2
13.
1
2
3
−
2
3
−2
3
2
2
14.
4
3
4
2
9
2
15.
(3 2 ) 3/2 × (8 1/3 ) 2
(3) 2 × (4 3 ) 1/2 × (9) −1/2