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Class 8 Exam > Class 8 Notes > Class 8 Mathematics by VP Classes > NCERT Solutions (Part - 1) - Exponents and Powers, Mathematics Class 8th

NCERT Solutions for Class 7 Maths Chapter 10 - (Part - 1) - Exponents and Powers Class 8th

Q: 2. How do we read and interpret exponential expressions?

Ans. Exponential expressions can be read as 'the base raised to the power of the exponent.' For example, 2^3 is read as '2 raised to the power of 3' or '2 cubed.' This means we need to multiply 2 by itself three times, resulting in 2 * 2 * 2 = 8.

Q: 3. What are the rules of exponents and powers?

Ans. The rules of exponents and powers include:- Multiplying powers with the same base: When multiplying two powers with the same base, add their exponents. For example, a^m * a^n = a^(m+n).- Dividing powers with the same base: When dividing two powers with the same base, subtract their exponents. For example, a^m / a^n = a^(m-n).- Power of a power: When raising a power to another exponent, multiply the exponents. For example, (a^m)^n = a^(m*n).- Power of a product: When raising a product to an exponent, distribute the exponent to each term. For example, (ab)^n = a^n * b^n.

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 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

	Class 8 Mathematics by VP Classes 90 docs\|16 tests

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FAQs on NCERT Solutions for Class 7 Maths Chapter 10 - (Part - 1) - Exponents and Powers Class 8th

1. What are exponents and powers in mathematics?

Ans. Exponents and powers are mathematical operations used to represent repeated multiplication of a number by itself. An exponent is a small superscript number written above and to the right of a base number, indicating the number of times the base should be multiplied by itself. The result of this operation is called a power.

2. How do we read and interpret exponential expressions?

Ans. Exponential expressions can be read as "the base raised to the power of the exponent." For example, 2^3 is read as "2 raised to the power of 3" or "2 cubed." This means we need to multiply 2 by itself three times, resulting in 2 * 2 * 2 = 8.

3. What are the rules of exponents and powers?

Ans. The rules of exponents and powers include: - Multiplying powers with the same base: When multiplying two powers with the same base, add their exponents. For example, a^m * a^n = a^(m+n). - Dividing powers with the same base: When dividing two powers with the same base, subtract their exponents. For example, a^m / a^n = a^(m-n). - Power of a power: When raising a power to another exponent, multiply the exponents. For example, (a^m)^n = a^(m*n). - Power of a product: When raising a product to an exponent, distribute the exponent to each term. For example, (ab)^n = a^n * b^n.

4. How can exponents and powers be used in real-life situations?

Ans. Exponents and powers are used in various real-life situations, such as calculating compound interest, population growth, and measuring large or small quantities. For example, compound interest can be calculated using the formula A = P(1 + r/n)^(nt), where A is the final amount, P is the principal amount, r is the annual interest rate, n is the number of times interest is compounded per year, and t is the number of years. The exponent in this formula helps in determining the growth of the investment over time.

5. What is the importance of understanding exponents and powers in mathematics?

Ans. Understanding exponents and powers is crucial in mathematics as it provides a concise and efficient way to represent repeated multiplication. It simplifies complex calculations and allows for easier manipulation of numbers. Exponents and powers are used in various mathematical concepts, such as algebra, geometry, and calculus. Additionally, they are used in scientific notation to express extremely large or small numbers. A solid understanding of exponents and powers is essential for building a strong foundation in mathematics.

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NCERT Solutions for Class 7 Maths Chapter 10 - (Part - 1) - Exponents and Powers Class 8th

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Class 8 Mathematics by VP Classes

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NCERT Solutions for Class 7 Maths Chapter 10 - (Part - 1) - Exponents and Powers Class 8th

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NCERT Solutions for Class 7 Maths Chapter 10 - (Part - 1) - Exponents and Powers Class 8th

shortcuts and tricks

Viva Questions

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NCERT Solutions for Class 7 Maths Chapter 10 - (Part - 1) - Exponents and Powers Class 8th

Sample Paper

Exam

Summary

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Importance of NCERT Solutions (Part - 1) - Exponents and Powers, Mathematics Class 8th

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

NCERT Solutions (Part - 1) - Exponents and Powers, Mathematics Class 8th Class 8 Questions

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