Page 72 - ICSE Math 7

1 a 0 1
–m
–m
For any non-zero rational number a, we have m = m = a 0 – m = a . Thus a = m
a a a
Try This

Find the value of the following.
0
0
0
0
2 5
–2 3
(a) (6 ) (b) –3 5 9 (c) 2 4 ÷ 3 4 (d) a × b × c (e) (9 ) (f) 2 × 3 3
4 3 2
Example 7: Using the laws of exponents, simplify and write the answer in exponential form.
8
5
15
10
2
4
2 3
7
(a) 2 × 2 × 2 (b) 7 ÷ 7 (c) (5 ) × 5 (d) 3 × 3 y
3
3
15
2
1
20
(e) a × b (f) 7 ÷ 7 (g) (2 ÷ 2 ) × 2 4
n
2
5
m
8
Solution: (a) 2 × 2 × 2 = 2 2 + 5 + 8 = 2 15 (Using a × a = a m + n )
m
10
n
15
(b) 7 ÷ 7 = 7 15 – 10 = 7 5 (Using a ÷ a = a m – n )
2 3
4
6
mn
4
m n
4
(c) (5 ) × 5 = 5 2 × 3 × 5 = 5 × 5 (Using (a ) = a )
m
n
= 5 6 + 4 = 5 10 (Using a × a = a m + n )
7
y
m
n
(d) 3 × 3 = 3 7 + y (Using a × a = a m + n )
m
3
3
3
m
m
(e) a × b = (ab) (Using a × b = (ab) )
1
m
n
2
1
–1
(f) 7 ÷ 7 = 7 1 – 2 = 7 = (Using a ÷ a = a m – n )
7
4
4
m
15
n
20
(g) (2 ÷ 2 ) × 2 = (2 20 – 15 ) × 2 (Using a ÷ a = a m – n )
5
4
n
9
m
= 2 × 2 = 2 (Using a × a = a m + n )
Example 8: Simplify each of the following expressions.
× 6x × (–xy )
2
3
(a) (b)
( )y 3 2 3x y
2
× y 3 + 4 y 7
Solution: (a) = = = y 7 – 6 = y
()y 32 y 3 × 2 y 6
3
3
2
3
3
6x × (–xy ) –2x 2 + 1 × y –2x × y
y
(b) = = = – 2x 3 – 2 3 – 1 = –2xy 2
2
2
2
3x y x y x y
Example 9: Simplify and express each of the following in exponential form.
4
3
2 × 3 × 4
0
4
0
2 3
4
(a) (b) {(5 ) × 5 } ÷ 5 3 (c) 125 ÷ 5 3 (d) 3 + 5 + 7 0
2
3 × 32
8
5
5
8 3
(e) 2 × 3 × 4 0 (f) (3 + 2 ) × 5 0 (g) 2 × a (h) 7 × a b
0
0
0
0
3
4 × a 3 7 × a b
5
5 2
4
3
4
3
2 × 3 × 4 2 × 3 × 2
2
0
Solution: (a) = = 2 3 + 2 – 5 × 3 4 – 2 = 2 × 3 = 1 × 9 = 3 2
2
2
3 × 32 3 × 2 5
3
4
3
3
4
2 3
3
4
6
(b) {(5 ) × 5 } ÷ 5 = {5 2 × 3 × 5 } ÷ 5 = {5 × 5 } ÷ 5 = 5 6 + 4 ÷ 5 = 5 10 – 3 = 5 7
3
4
3
12
3
3 4
3
(c) 125 ÷ 5 = (5 ) ÷ 5 = 5 3 × 4 ÷ 5 = 5 ÷ 5 = 5 12 – 3 = 5 9
0
0
0
(d) 3 + 5 + 7 = 1 + 1 + 1 = 3 = 3 1
0
0
0
(e) 2 × 3 × 4 = 1 × 1 × 1 = 1 = 1 1
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