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# NCERT Solutions (Part - 1) - Exponents and Powers, Mathematics Class 8th Class 8 Notes | EduRev

## Class 8 Mathematics by VP Classes

Created by: Full Circle

## Class 8 : NCERT Solutions (Part - 1) - Exponents and Powers, Mathematics Class 8th Class 8 Notes | EduRev

``` Page 1

TRY THESE (Page 194)
Question 1. Find the multiplicative inverse of the following.
(i) 2
–4
(ii) 10
–5
(iii) 7
–2
(iv) 5
–3
(v) 10
–100
Solution: (i) The multiplicative inverse of 2
–4
is 2
4
.
(ii) The multiplicative inverse of 10
–5
is 10
5
(iii) The multiplicative inverse of 7
–2
is 7
2
.
(iv) The multiplicative inverse of 5
–3
is 5
3
.
(v) The multiplicative inverse of 10
–100
is 10
100
.
TRY THESE (Page 194)
Question 1. Expand the following numbers using exponents.
(i) 1025.63 (ii) 1256.249
Solution:
Number Expanded form
(i) 1025.63 1 ¥ 1000 + 0 ¥ 100 + 2 ¥ 10 + 5 ¥ 1 +
6
10
+
3
100
or
1 ¥ 10
3
+ 2 ¥ 10
1
+ 5 ¥ 10
0
+ 6 ¥ 10
–1
+ 3 ¥ 10
–2
(ii) 1256.249 1 ¥ 1000 + 2 ¥ 100 + 5 ¥ 10 + 6 ¥ 1 +
2
10
+
4
100
+
9
1000
or
1 ¥ 10
3
+ 2 ¥ 10
2
+ 5 ¥ 10
1
+ 6 ¥ 10
0
+ 2 ¥ 10
–1
+ 4 ¥ 10
–2
+ 9 ¥ 10
–3
LAWS OF EXPONENTS
So far we have studied laws of exponents by taking exponents as natural numbers. These laws hold
good for negative exponents also. For any non-zero integers ‘a’ and ‘b’ (as bases) and exponents
as m, n (any integer), we have
Law I: a
m
¥ a
n
= a
m+n
Example: (–3)
–2
¥ (–3)
4
= (–3)
–2 + 4
= (–3)
2
TRY THESE (Page 195)
Question 1. Simplify and write in exponential form.
(i) (–2)
–3
¥ (–2)
–4
(ii) p
3
¥ p
–10
(iii) 3
2
¥ 3
–5
¥ 3
6
Solution: (i)(–2)
–3
¥ (–2)
–4
= (–2)
(–3) + (–4)
[ ? a
m
¥a
n
= a
m+n
]
= (–2)
–7
or
1
(– 2)
7
Page 2

TRY THESE (Page 194)
Question 1. Find the multiplicative inverse of the following.
(i) 2
–4
(ii) 10
–5
(iii) 7
–2
(iv) 5
–3
(v) 10
–100
Solution: (i) The multiplicative inverse of 2
–4
is 2
4
.
(ii) The multiplicative inverse of 10
–5
is 10
5
(iii) The multiplicative inverse of 7
–2
is 7
2
.
(iv) The multiplicative inverse of 5
–3
is 5
3
.
(v) The multiplicative inverse of 10
–100
is 10
100
.
TRY THESE (Page 194)
Question 1. Expand the following numbers using exponents.
(i) 1025.63 (ii) 1256.249
Solution:
Number Expanded form
(i) 1025.63 1 ¥ 1000 + 0 ¥ 100 + 2 ¥ 10 + 5 ¥ 1 +
6
10
+
3
100
or
1 ¥ 10
3
+ 2 ¥ 10
1
+ 5 ¥ 10
0
+ 6 ¥ 10
–1
+ 3 ¥ 10
–2
(ii) 1256.249 1 ¥ 1000 + 2 ¥ 100 + 5 ¥ 10 + 6 ¥ 1 +
2
10
+
4
100
+
9
1000
or
1 ¥ 10
3
+ 2 ¥ 10
2
+ 5 ¥ 10
1
+ 6 ¥ 10
0
+ 2 ¥ 10
–1
+ 4 ¥ 10
–2
+ 9 ¥ 10
–3
LAWS OF EXPONENTS
So far we have studied laws of exponents by taking exponents as natural numbers. These laws hold
good for negative exponents also. For any non-zero integers ‘a’ and ‘b’ (as bases) and exponents
as m, n (any integer), we have
Law I: a
m
¥ a
n
= a
m+n
Example: (–3)
–2
¥ (–3)
4
= (–3)
–2 + 4
= (–3)
2
TRY THESE (Page 195)
Question 1. Simplify and write in exponential form.
(i) (–2)
–3
¥ (–2)
–4
(ii) p
3
¥ p
–10
(iii) 3
2
¥ 3
–5
¥ 3
6
Solution: (i)(–2)
–3
¥ (–2)
–4
= (–2)
(–3) + (–4)
[ ? a
m
¥a
n
= a
m+n
]
= (–2)
–7
or
1
(– 2)
7

(ii)p
3
¥ p
–10
= (p)
3 + (–10)
= (p)
–7
or
1
(10)
7
(iii)3
2
¥ 3
–5
¥ 3
6
= 3
2+(–5)+6
= 3
8–5
= 3
3
Law II:
a
a
m
n
= a
m–n
Example: 5
–1
? 5
–2
= 5
–1–(–2)
= 5
–1+2
= 5
1
or 5
Law III: (a
m
)
n
= a
mn
Example: (9
–1
)
–3
= 9
(–1)¥(–3)
= 9
3
Law IV: a
m
¥ b
m
= (ab)
m
Example: 2
–4
¥ 3
–4
= (2 ¥ 3)
–4
= 6
–4
or
1
6
4
Law V:
m
m
a
b
=
a
b
m
F
H
G
I
K
J
Example:
3
–5
7
5 -
=
3
7
5
F
H
G
I
K
J
-
or
7
3
5
F
H
G
I
K
J
Law VI: a
0
= 1
Example: (i) (–38)
0
= 1
(ii) (32456)
0
= 1
REMEMBER
(i) 1
1
= 1
2
= 1
3
= 1
–1
= 1
–2
= … = 1
In general, (1)
n
= 1 for any integer n.
(ii) (–1)
0
= (–1)
2
= (–1)
–2
= (–1)
–4
= … = 1
In general, (–1)
p
= 1 for any even integer p.
EXERCISE 12.1 (Page 197)
Question 1. Evaluate:
(i) 3
–2
(ii) (–4)
–2
(iii)
1
2
5
F
H
G
I
K
J
-
Solution:(i )3
–2
=
1
3
2
=
1
33 ¥
=
1
9
Page 3

TRY THESE (Page 194)
Question 1. Find the multiplicative inverse of the following.
(i) 2
–4
(ii) 10
–5
(iii) 7
–2
(iv) 5
–3
(v) 10
–100
Solution: (i) The multiplicative inverse of 2
–4
is 2
4
.
(ii) The multiplicative inverse of 10
–5
is 10
5
(iii) The multiplicative inverse of 7
–2
is 7
2
.
(iv) The multiplicative inverse of 5
–3
is 5
3
.
(v) The multiplicative inverse of 10
–100
is 10
100
.
TRY THESE (Page 194)
Question 1. Expand the following numbers using exponents.
(i) 1025.63 (ii) 1256.249
Solution:
Number Expanded form
(i) 1025.63 1 ¥ 1000 + 0 ¥ 100 + 2 ¥ 10 + 5 ¥ 1 +
6
10
+
3
100
or
1 ¥ 10
3
+ 2 ¥ 10
1
+ 5 ¥ 10
0
+ 6 ¥ 10
–1
+ 3 ¥ 10
–2
(ii) 1256.249 1 ¥ 1000 + 2 ¥ 100 + 5 ¥ 10 + 6 ¥ 1 +
2
10
+
4
100
+
9
1000
or
1 ¥ 10
3
+ 2 ¥ 10
2
+ 5 ¥ 10
1
+ 6 ¥ 10
0
+ 2 ¥ 10
–1
+ 4 ¥ 10
–2
+ 9 ¥ 10
–3
LAWS OF EXPONENTS
So far we have studied laws of exponents by taking exponents as natural numbers. These laws hold
good for negative exponents also. For any non-zero integers ‘a’ and ‘b’ (as bases) and exponents
as m, n (any integer), we have
Law I: a
m
¥ a
n
= a
m+n
Example: (–3)
–2
¥ (–3)
4
= (–3)
–2 + 4
= (–3)
2
TRY THESE (Page 195)
Question 1. Simplify and write in exponential form.
(i) (–2)
–3
¥ (–2)
–4
(ii) p
3
¥ p
–10
(iii) 3
2
¥ 3
–5
¥ 3
6
Solution: (i)(–2)
–3
¥ (–2)
–4
= (–2)
(–3) + (–4)
[ ? a
m
¥a
n
= a
m+n
]
= (–2)
–7
or
1
(– 2)
7

(ii)p
3
¥ p
–10
= (p)
3 + (–10)
= (p)
–7
or
1
(10)
7
(iii)3
2
¥ 3
–5
¥ 3
6
= 3
2+(–5)+6
= 3
8–5
= 3
3
Law II:
a
a
m
n
= a
m–n
Example: 5
–1
? 5
–2
= 5
–1–(–2)
= 5
–1+2
= 5
1
or 5
Law III: (a
m
)
n
= a
mn
Example: (9
–1
)
–3
= 9
(–1)¥(–3)
= 9
3
Law IV: a
m
¥ b
m
= (ab)
m
Example: 2
–4
¥ 3
–4
= (2 ¥ 3)
–4
= 6
–4
or
1
6
4
Law V:
m
m
a
b
=
a
b
m
F
H
G
I
K
J
Example:
3
–5
7
5 -
=
3
7
5
F
H
G
I
K
J
-
or
7
3
5
F
H
G
I
K
J
Law VI: a
0
= 1
Example: (i) (–38)
0
= 1
(ii) (32456)
0
= 1
REMEMBER
(i) 1
1
= 1
2
= 1
3
= 1
–1
= 1
–2
= … = 1
In general, (1)
n
= 1 for any integer n.
(ii) (–1)
0
= (–1)
2
= (–1)
–2
= (–1)
–4
= … = 1
In general, (–1)
p
= 1 for any even integer p.
EXERCISE 12.1 (Page 197)
Question 1. Evaluate:
(i) 3
–2
(ii) (–4)
–2
(iii)
1
2
5
F
H
G
I
K
J
-
Solution:(i )3
–2
=
1
3
2
=
1
33 ¥
=
1
9

(ii)(–4)
–2
=
1
4 () -
=
1
44 () () -¥-
=
1
16
(iii)
1
2
5
F
H
G
I
K
J
-
=
1
1
2
5
F
H
G
I
K
J
=
1
1
2
1
2
1
2
1
2
1
2
¥¥¥¥
F
H
G
I
K
J
=
1
1
32
F
H
G
I
K
J
= 32
Question 2. Simplify and express the result in power notation with positive exponent.
(i) (–4)
5
? (–4)8 (ii)
1
2
3
2
F
H
G
I
K
J
(iii) (–3)
4
¥
5
3
4
F
H
G
I
K
J
(iv) (3
–7
? 3
–10
) ¥ 3
–5
(v) 2
–3
¥ (–7)
–3
Solution: (i)(–4)
5
? (–4)
8
? a
m
? a
n
= a
m–n
\ (–4)
5
? (–4)
8
= (–4)
5–8
= (–4)
–3
=
1
4
3
() -
(ii)
1
2
3
2
F
H
G
I
K
J
? 1 = 1
3
\
1
2
3
=
1
2
3
3
=
1
2
3
F
H
G
I
K
J
Now
1
2
3
2
F
H
G
I
K
J
=
1
2
3
2
F
H
G
I
K
J
L
N
M
M
O
Q
P
P
=
1
2
32
F
H
G
I
K
J
¥
[Using (a
m
)
n
= a
mn
]
=
1
2
6
F
H
G
I
K
J
=
1
2
6
(iii)(–3)
4
¥
5
3
4
F
H
G
I
K
J
? a
m
¥ b
m
= (ab)
m
\ (–3)
4
¥
5
3
4
F
H
G
I
K
J
= () -¥
L
N
M
O
Q
P
3
5
3
4
= [(–1) ¥ 5]
4
= [(–1)
4
¥ (+5)
4
]
= 1 ¥ (5)
4
= (5)
4
(iv)(3
–7
? 3
–10
) ¥ 3
–5
? a
m
? a
n
= a
m–n
and a
m
¥ a
n
= a
m+n
\ (3
–7
? 3
–10
) ¥ 3
–5
= [3
–7–(–10)
] ¥ 3
–5
= [3
–7+10
] ¥ 3
–5
Page 4

TRY THESE (Page 194)
Question 1. Find the multiplicative inverse of the following.
(i) 2
–4
(ii) 10
–5
(iii) 7
–2
(iv) 5
–3
(v) 10
–100
Solution: (i) The multiplicative inverse of 2
–4
is 2
4
.
(ii) The multiplicative inverse of 10
–5
is 10
5
(iii) The multiplicative inverse of 7
–2
is 7
2
.
(iv) The multiplicative inverse of 5
–3
is 5
3
.
(v) The multiplicative inverse of 10
–100
is 10
100
.
TRY THESE (Page 194)
Question 1. Expand the following numbers using exponents.
(i) 1025.63 (ii) 1256.249
Solution:
Number Expanded form
(i) 1025.63 1 ¥ 1000 + 0 ¥ 100 + 2 ¥ 10 + 5 ¥ 1 +
6
10
+
3
100
or
1 ¥ 10
3
+ 2 ¥ 10
1
+ 5 ¥ 10
0
+ 6 ¥ 10
–1
+ 3 ¥ 10
–2
(ii) 1256.249 1 ¥ 1000 + 2 ¥ 100 + 5 ¥ 10 + 6 ¥ 1 +
2
10
+
4
100
+
9
1000
or
1 ¥ 10
3
+ 2 ¥ 10
2
+ 5 ¥ 10
1
+ 6 ¥ 10
0
+ 2 ¥ 10
–1
+ 4 ¥ 10
–2
+ 9 ¥ 10
–3
LAWS OF EXPONENTS
So far we have studied laws of exponents by taking exponents as natural numbers. These laws hold
good for negative exponents also. For any non-zero integers ‘a’ and ‘b’ (as bases) and exponents
as m, n (any integer), we have
Law I: a
m
¥ a
n
= a
m+n
Example: (–3)
–2
¥ (–3)
4
= (–3)
–2 + 4
= (–3)
2
TRY THESE (Page 195)
Question 1. Simplify and write in exponential form.
(i) (–2)
–3
¥ (–2)
–4
(ii) p
3
¥ p
–10
(iii) 3
2
¥ 3
–5
¥ 3
6
Solution: (i)(–2)
–3
¥ (–2)
–4
= (–2)
(–3) + (–4)
[ ? a
m
¥a
n
= a
m+n
]
= (–2)
–7
or
1
(– 2)
7

(ii)p
3
¥ p
–10
= (p)
3 + (–10)
= (p)
–7
or
1
(10)
7
(iii)3
2
¥ 3
–5
¥ 3
6
= 3
2+(–5)+6
= 3
8–5
= 3
3
Law II:
a
a
m
n
= a
m–n
Example: 5
–1
? 5
–2
= 5
–1–(–2)
= 5
–1+2
= 5
1
or 5
Law III: (a
m
)
n
= a
mn
Example: (9
–1
)
–3
= 9
(–1)¥(–3)
= 9
3
Law IV: a
m
¥ b
m
= (ab)
m
Example: 2
–4
¥ 3
–4
= (2 ¥ 3)
–4
= 6
–4
or
1
6
4
Law V:
m
m
a
b
=
a
b
m
F
H
G
I
K
J
Example:
3
–5
7
5 -
=
3
7
5
F
H
G
I
K
J
-
or
7
3
5
F
H
G
I
K
J
Law VI: a
0
= 1
Example: (i) (–38)
0
= 1
(ii) (32456)
0
= 1
REMEMBER
(i) 1
1
= 1
2
= 1
3
= 1
–1
= 1
–2
= … = 1
In general, (1)
n
= 1 for any integer n.
(ii) (–1)
0
= (–1)
2
= (–1)
–2
= (–1)
–4
= … = 1
In general, (–1)
p
= 1 for any even integer p.
EXERCISE 12.1 (Page 197)
Question 1. Evaluate:
(i) 3
–2
(ii) (–4)
–2
(iii)
1
2
5
F
H
G
I
K
J
-
Solution:(i )3
–2
=
1
3
2
=
1
33 ¥
=
1
9

(ii)(–4)
–2
=
1
4 () -
=
1
44 () () -¥-
=
1
16
(iii)
1
2
5
F
H
G
I
K
J
-
=
1
1
2
5
F
H
G
I
K
J
=
1
1
2
1
2
1
2
1
2
1
2
¥¥¥¥
F
H
G
I
K
J
=
1
1
32
F
H
G
I
K
J
= 32
Question 2. Simplify and express the result in power notation with positive exponent.
(i) (–4)
5
? (–4)8 (ii)
1
2
3
2
F
H
G
I
K
J
(iii) (–3)
4
¥
5
3
4
F
H
G
I
K
J
(iv) (3
–7
? 3
–10
) ¥ 3
–5
(v) 2
–3
¥ (–7)
–3
Solution: (i)(–4)
5
? (–4)
8
? a
m
? a
n
= a
m–n
\ (–4)
5
? (–4)
8
= (–4)
5–8
= (–4)
–3
=
1
4
3
() -
(ii)
1
2
3
2
F
H
G
I
K
J
? 1 = 1
3
\
1
2
3
=
1
2
3
3
=
1
2
3
F
H
G
I
K
J
Now
1
2
3
2
F
H
G
I
K
J
=
1
2
3
2
F
H
G
I
K
J
L
N
M
M
O
Q
P
P
=
1
2
32
F
H
G
I
K
J
¥
[Using (a
m
)
n
= a
mn
]
=
1
2
6
F
H
G
I
K
J
=
1
2
6
(iii)(–3)
4
¥
5
3
4
F
H
G
I
K
J
? a
m
¥ b
m
= (ab)
m
\ (–3)
4
¥
5
3
4
F
H
G
I
K
J
= () -¥
L
N
M
O
Q
P
3
5
3
4
= [(–1) ¥ 5]
4
= [(–1)
4
¥ (+5)
4
]
= 1 ¥ (5)
4
= (5)
4
(iv)(3
–7
? 3
–10
) ¥ 3
–5
? a
m
? a
n
= a
m–n
and a
m
¥ a
n
= a
m+n
\ (3
–7
? 3
–10
) ¥ 3
–5
= [3
–7–(–10)
] ¥ 3
–5
= [3
–7+10
] ¥ 3
–5

= 3
3
¥ 3
–5
= 3
3+(–5)
= 3
–2
=
1
3
2
()
(v)2
–3
¥ (–7)
–3
? a
m
¥ b
m
= (ab)
m
\ 2
–3
¥ (–7)
–3
= [2 ¥ (–7)]
–3
= [–14]
–3
=
1
14
3
() -
Question 3. Find the value of:
(i) (3
0
+ 4
–1
) ¥ 2
2
(ii) (2
–1
¥ 4
–1
) ? 2
–2
(iii)
1
2
1
3
1
4
22 2
F
H
G
I
K
J
+
F
H
G
I
K
J
+
F
H
G
I
K
J
-- -
(iv) (3
–1
+ 4
–1
+ 5
–1
)
0
(v)
- F
H
G
I
K
J
R
S
|
T
|
U
V
|
W
|
-
2
3
2
2
Solution: (i)(3
0
+ 4
–1
) ¥ 2
2
? a
0
= 1 and a
–1
=
1
a
\ (3
0
+ 4
–1
) ¥ 2
2
= 1+
1
4
F
H
G
I
K
J
¥ 2
2
=
5
4
F
H
G
I
K
J
¥ 4 = 5
(ii)(2
–1
¥ 4
–1
) ? 2
–2
? a
m
¥ b
m
= (ab)
m
\ (2
–1
¥ 4
–1
) ? 2
–2
= (2 ¥ 4)
–1
? 2
–2
= (2
1
¥ 2
2
)
–1
? 2
–2
= (2
1+2
)
–1
? 2
–2
= (2
3
)
–1
? 2
–2
= 2
–3
? 2
–2
= (2)
(–3)–(–2)
= (–2)
–3+2
= 2
–1
=
1
2
(iii)
1
2
1
3
1
4
22 2
F
H
G
I
K
J
+
F
H
G
I
K
J
+
F
H
G
I
K
J
-- -
?
1
2
2
F
H
G
I
K
J
-
=
1
1
2
1
1
4
1
4
1
4
2
F
H
G
I
K
J
=
F
H
G
I
K
J
=¥ =
Page 5

TRY THESE (Page 194)
Question 1. Find the multiplicative inverse of the following.
(i) 2
–4
(ii) 10
–5
(iii) 7
–2
(iv) 5
–3
(v) 10
–100
Solution: (i) The multiplicative inverse of 2
–4
is 2
4
.
(ii) The multiplicative inverse of 10
–5
is 10
5
(iii) The multiplicative inverse of 7
–2
is 7
2
.
(iv) The multiplicative inverse of 5
–3
is 5
3
.
(v) The multiplicative inverse of 10
–100
is 10
100
.
TRY THESE (Page 194)
Question 1. Expand the following numbers using exponents.
(i) 1025.63 (ii) 1256.249
Solution:
Number Expanded form
(i) 1025.63 1 ¥ 1000 + 0 ¥ 100 + 2 ¥ 10 + 5 ¥ 1 +
6
10
+
3
100
or
1 ¥ 10
3
+ 2 ¥ 10
1
+ 5 ¥ 10
0
+ 6 ¥ 10
–1
+ 3 ¥ 10
–2
(ii) 1256.249 1 ¥ 1000 + 2 ¥ 100 + 5 ¥ 10 + 6 ¥ 1 +
2
10
+
4
100
+
9
1000
or
1 ¥ 10
3
+ 2 ¥ 10
2
+ 5 ¥ 10
1
+ 6 ¥ 10
0
+ 2 ¥ 10
–1
+ 4 ¥ 10
–2
+ 9 ¥ 10
–3
LAWS OF EXPONENTS
So far we have studied laws of exponents by taking exponents as natural numbers. These laws hold
good for negative exponents also. For any non-zero integers ‘a’ and ‘b’ (as bases) and exponents
as m, n (any integer), we have
Law I: a
m
¥ a
n
= a
m+n
Example: (–3)
–2
¥ (–3)
4
= (–3)
–2 + 4
= (–3)
2
TRY THESE (Page 195)
Question 1. Simplify and write in exponential form.
(i) (–2)
–3
¥ (–2)
–4
(ii) p
3
¥ p
–10
(iii) 3
2
¥ 3
–5
¥ 3
6
Solution: (i)(–2)
–3
¥ (–2)
–4
= (–2)
(–3) + (–4)
[ ? a
m
¥a
n
= a
m+n
]
= (–2)
–7
or
1
(– 2)
7

(ii)p
3
¥ p
–10
= (p)
3 + (–10)
= (p)
–7
or
1
(10)
7
(iii)3
2
¥ 3
–5
¥ 3
6
= 3
2+(–5)+6
= 3
8–5
= 3
3
Law II:
a
a
m
n
= a
m–n
Example: 5
–1
? 5
–2
= 5
–1–(–2)
= 5
–1+2
= 5
1
or 5
Law III: (a
m
)
n
= a
mn
Example: (9
–1
)
–3
= 9
(–1)¥(–3)
= 9
3
Law IV: a
m
¥ b
m
= (ab)
m
Example: 2
–4
¥ 3
–4
= (2 ¥ 3)
–4
= 6
–4
or
1
6
4
Law V:
m
m
a
b
=
a
b
m
F
H
G
I
K
J
Example:
3
–5
7
5 -
=
3
7
5
F
H
G
I
K
J
-
or
7
3
5
F
H
G
I
K
J
Law VI: a
0
= 1
Example: (i) (–38)
0
= 1
(ii) (32456)
0
= 1
REMEMBER
(i) 1
1
= 1
2
= 1
3
= 1
–1
= 1
–2
= … = 1
In general, (1)
n
= 1 for any integer n.
(ii) (–1)
0
= (–1)
2
= (–1)
–2
= (–1)
–4
= … = 1
In general, (–1)
p
= 1 for any even integer p.
EXERCISE 12.1 (Page 197)
Question 1. Evaluate:
(i) 3
–2
(ii) (–4)
–2
(iii)
1
2
5
F
H
G
I
K
J
-
Solution:(i )3
–2
=
1
3
2
=
1
33 ¥
=
1
9

(ii)(–4)
–2
=
1
4 () -
=
1
44 () () -¥-
=
1
16
(iii)
1
2
5
F
H
G
I
K
J
-
=
1
1
2
5
F
H
G
I
K
J
=
1
1
2
1
2
1
2
1
2
1
2
¥¥¥¥
F
H
G
I
K
J
=
1
1
32
F
H
G
I
K
J
= 32
Question 2. Simplify and express the result in power notation with positive exponent.
(i) (–4)
5
? (–4)8 (ii)
1
2
3
2
F
H
G
I
K
J
(iii) (–3)
4
¥
5
3
4
F
H
G
I
K
J
(iv) (3
–7
? 3
–10
) ¥ 3
–5
(v) 2
–3
¥ (–7)
–3
Solution: (i)(–4)
5
? (–4)
8
? a
m
? a
n
= a
m–n
\ (–4)
5
? (–4)
8
= (–4)
5–8
= (–4)
–3
=
1
4
3
() -
(ii)
1
2
3
2
F
H
G
I
K
J
? 1 = 1
3
\
1
2
3
=
1
2
3
3
=
1
2
3
F
H
G
I
K
J
Now
1
2
3
2
F
H
G
I
K
J
=
1
2
3
2
F
H
G
I
K
J
L
N
M
M
O
Q
P
P
=
1
2
32
F
H
G
I
K
J
¥
[Using (a
m
)
n
= a
mn
]
=
1
2
6
F
H
G
I
K
J
=
1
2
6
(iii)(–3)
4
¥
5
3
4
F
H
G
I
K
J
? a
m
¥ b
m
= (ab)
m
\ (–3)
4
¥
5
3
4
F
H
G
I
K
J
= () -¥
L
N
M
O
Q
P
3
5
3
4
= [(–1) ¥ 5]
4
= [(–1)
4
¥ (+5)
4
]
= 1 ¥ (5)
4
= (5)
4
(iv)(3
–7
? 3
–10
) ¥ 3
–5
? a
m
? a
n
= a
m–n
and a
m
¥ a
n
= a
m+n
\ (3
–7
? 3
–10
) ¥ 3
–5
= [3
–7–(–10)
] ¥ 3
–5
= [3
–7+10
] ¥ 3
–5

= 3
3
¥ 3
–5
= 3
3+(–5)
= 3
–2
=
1
3
2
()
(v)2
–3
¥ (–7)
–3
? a
m
¥ b
m
= (ab)
m
\ 2
–3
¥ (–7)
–3
= [2 ¥ (–7)]
–3
= [–14]
–3
=
1
14
3
() -
Question 3. Find the value of:
(i) (3
0
+ 4
–1
) ¥ 2
2
(ii) (2
–1
¥ 4
–1
) ? 2
–2
(iii)
1
2
1
3
1
4
22 2
F
H
G
I
K
J
+
F
H
G
I
K
J
+
F
H
G
I
K
J
-- -
(iv) (3
–1
+ 4
–1
+ 5
–1
)
0
(v)
- F
H
G
I
K
J
R
S
|
T
|
U
V
|
W
|
-
2
3
2
2
Solution: (i)(3
0
+ 4
–1
) ¥ 2
2
? a
0
= 1 and a
–1
=
1
a
\ (3
0
+ 4
–1
) ¥ 2
2
= 1+
1
4
F
H
G
I
K
J
¥ 2
2
=
5
4
F
H
G
I
K
J
¥ 4 = 5
(ii)(2
–1
¥ 4
–1
) ? 2
–2
? a
m
¥ b
m
= (ab)
m
\ (2
–1
¥ 4
–1
) ? 2
–2
= (2 ¥ 4)
–1
? 2
–2
= (2
1
¥ 2
2
)
–1
? 2
–2
= (2
1+2
)
–1
? 2
–2
= (2
3
)
–1
? 2
–2
= 2
–3
? 2
–2
= (2)
(–3)–(–2)
= (–2)
–3+2
= 2
–1
=
1
2
(iii)
1
2
1
3
1
4
22 2
F
H
G
I
K
J
+
F
H
G
I
K
J
+
F
H
G
I
K
J
-- -
?
1
2
2
F
H
G
I
K
J
-
=
1
1
2
1
1
4
1
4
1
4
2
F
H
G
I
K
J
=
F
H
G
I
K
J
=¥ =

1
3
2
F
H
G
I
K
J
-
=
1
1
3
1
1
9
1
9
1
9
2
F
H
G
I
K
J
=
F
H
G
I
K
J
=¥ =
1
4
2
F
H
G
I
K
J
-
=
1
1
4
1
1
16
1
16
1
16
2
F
H
G
I
K
J
=
F
H
G
I
K
J
=¥ =
\
1
2
2
F
H
G
I
K
J
-
+
1
3
2
F
H
G
I
K
J
-
+
1
4
2
F
H
G
I
K
J
-
= 4 + 9 + 16 = 29
(iv)[3
–1
+ 4
–1
+ 5
–1
]
0
? 3
–1
=
1
3
, 4
–1
=
1
4
and 5
–1
=
1
5
\ [3
–1
+ 4
–1
+ 5
–1
]
0
=
1
3
1
4
1
5
0
++
L
N
M
O
Q
P
=
20 15 12
60
0
++ L
N
M
O
Q
P
=
47
60
0
F
H
G
I
K
J
= 1
(v)
- F
H
G
I
K
J
R
S
|
T
|
U
V
|
W
|
-
2
3
2
2
? (a
m
)
n
= a
mn
\
- F
H
G
I
K
J
R
S
|
T
|
U
V
|
W
|
-
2
3
2
2
=
- F
H
G
I
K
J
=
- F
H
G
I
K
J
-¥ -
2
3
2
3
22 4 ()
=
(2)
(3)
a
b
a
b
4
4
m
m
m
- F
H
G
I
K
J
=
L
N
M
M
O
Q
P
P
-
-
?
=
3
2
3333
22 22
4
4
() (– )( )( )( ) -
=
¥¥ ¥
¥ - ¥- ¥-
=
81
16
Question 4. Evaluate (i)
85
2
13
4
-
-
¥
and (ii) (5
–1
¥ 2
–1
) ¥ 6
–1
.
Solution: (i)
85
2
13
4
-
-
¥
=
8 555
2
1
4
-
-
¥¥ ¥ ()
=
1
8
2 125
4
¥¥
=
1
8
2 222 125 ¥¥¥ ¥¥
= 2 ¥ 125 = 250
```

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