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Page 1
? ???????? ? ?????????? ? ???
10.1 Introduction
Do you know?
Mass of earth is 5,970,000,000,000, 000, 000, 000, 000 kg. We have
already learnt in earlier class how to write such large numbers more
conveniently using exponents, as, 5.97 × 10
24
kg.
We read 10
24
as 10 raised to the power 24.
We know 2
5
= 2 × 2 × 2 × 2 × 2
and 2
m
= 2 × 2 × 2 × 2 × ... × 2 × 2 ... (m times)
Let us now find what is 2
– 2
is equal to?
10.2 Powers with Negative Exponents
You know that, 10
2
= 10 × 10 = 100
10
1
= 10 =
100
10
10
0
= 1 =
10
10
10
– 1
= ?
Continuing the above pattern we get,10
– 1
=
1
10
Similarly 10
– 2
=
2
1 11 1 1
10
10 10 10 100 10
÷= × = =
10
– 3
=
3
1 11 1 1
10
100 100 10 1000 10
÷= × = =
What is 10
– 10
equal to?
?????????? ???? ??????
? ? ? ? ? ? ?
??
Exponent is a
negative integer.
As the exponent decreases by1, the
value becomes one-tenth of the
previous value.
Reprint 2024-25
Page 2
? ???????? ? ?????????? ? ???
10.1 Introduction
Do you know?
Mass of earth is 5,970,000,000,000, 000, 000, 000, 000 kg. We have
already learnt in earlier class how to write such large numbers more
conveniently using exponents, as, 5.97 × 10
24
kg.
We read 10
24
as 10 raised to the power 24.
We know 2
5
= 2 × 2 × 2 × 2 × 2
and 2
m
= 2 × 2 × 2 × 2 × ... × 2 × 2 ... (m times)
Let us now find what is 2
– 2
is equal to?
10.2 Powers with Negative Exponents
You know that, 10
2
= 10 × 10 = 100
10
1
= 10 =
100
10
10
0
= 1 =
10
10
10
– 1
= ?
Continuing the above pattern we get,10
– 1
=
1
10
Similarly 10
– 2
=
2
1 11 1 1
10
10 10 10 100 10
÷= × = =
10
– 3
=
3
1 11 1 1
10
100 100 10 1000 10
÷= × = =
What is 10
– 10
equal to?
?????????? ???? ??????
? ? ? ? ? ? ?
??
Exponent is a
negative integer.
As the exponent decreases by1, the
value becomes one-tenth of the
previous value.
Reprint 2024-25
???? ? ? ? ????? ????
TRY THESE
TRY THESE
Now consider the following.
3
3
= 3 × 3 × 3 = 27
3
2
= 3 × 3 = 9 =
27
3
3
1
= 3 =
9
3
3° = 1 =
3
3
So looking at the above pattern, we say
3
– 1
= 1 ÷ 3 =
1
3
3
– 2
=
1
3
3
÷
=
1
33 ×
=
2
1
3
3
– 3
=
2
1
3
3
÷ =
2
1
3
×
1
3
=
3
1
3
You can now find the value of 2
– 2
in a similar manner .
We have, 10
– 2
=
2
1
10
or 10
2
=
2
1
10
-
10
– 3
=
3
1
10
or 10
3
=
3
1
10
-
3
– 2
=
2
1
3
or 3
2
=
2
1
3
-
etc.
In general, we can say that for any non-zero integer a, a
– m
=
1
m
a
, where m is a
positive integer. a
–m
is the multiplicative inverse of a
m
.
Find the multiplicative inverse of the following.
(i) 2
– 4
(ii) 10
– 5
(iii) 7
– 2
(iv) 5
– 3
(v) 10
– 100
We learnt how to write numbers like 1425 in expanded form using exponents as
1 × 10
3
+ 4 × 10
2
+ 2 × 10
1
+ 5 × 10°.
Let us see how to express 1425.36 in expanded form in a similar way .
We have 1425.36 = 1 × 1000 + 4 × 100 + 2 × 10 + 5 × 1 +
36
10 100
+
= 1 × 10
3
+ 4 × 10
2
+ 2 × 10 + 5 × 1 + 3 × 10
– 1
+ 6 × 10
– 2
The previous number is
divided by the base 3.
Expand the following numbers using exponents.
(i) 1025.63 (ii) 1256.249
10
– 1
=
1
10
, 10
– 2
=
2
11
100 10
=
Reprint 2024-25
Page 3
? ???????? ? ?????????? ? ???
10.1 Introduction
Do you know?
Mass of earth is 5,970,000,000,000, 000, 000, 000, 000 kg. We have
already learnt in earlier class how to write such large numbers more
conveniently using exponents, as, 5.97 × 10
24
kg.
We read 10
24
as 10 raised to the power 24.
We know 2
5
= 2 × 2 × 2 × 2 × 2
and 2
m
= 2 × 2 × 2 × 2 × ... × 2 × 2 ... (m times)
Let us now find what is 2
– 2
is equal to?
10.2 Powers with Negative Exponents
You know that, 10
2
= 10 × 10 = 100
10
1
= 10 =
100
10
10
0
= 1 =
10
10
10
– 1
= ?
Continuing the above pattern we get,10
– 1
=
1
10
Similarly 10
– 2
=
2
1 11 1 1
10
10 10 10 100 10
÷= × = =
10
– 3
=
3
1 11 1 1
10
100 100 10 1000 10
÷= × = =
What is 10
– 10
equal to?
?????????? ???? ??????
? ? ? ? ? ? ?
??
Exponent is a
negative integer.
As the exponent decreases by1, the
value becomes one-tenth of the
previous value.
Reprint 2024-25
???? ? ? ? ????? ????
TRY THESE
TRY THESE
Now consider the following.
3
3
= 3 × 3 × 3 = 27
3
2
= 3 × 3 = 9 =
27
3
3
1
= 3 =
9
3
3° = 1 =
3
3
So looking at the above pattern, we say
3
– 1
= 1 ÷ 3 =
1
3
3
– 2
=
1
3
3
÷
=
1
33 ×
=
2
1
3
3
– 3
=
2
1
3
3
÷ =
2
1
3
×
1
3
=
3
1
3
You can now find the value of 2
– 2
in a similar manner .
We have, 10
– 2
=
2
1
10
or 10
2
=
2
1
10
-
10
– 3
=
3
1
10
or 10
3
=
3
1
10
-
3
– 2
=
2
1
3
or 3
2
=
2
1
3
-
etc.
In general, we can say that for any non-zero integer a, a
– m
=
1
m
a
, where m is a
positive integer. a
–m
is the multiplicative inverse of a
m
.
Find the multiplicative inverse of the following.
(i) 2
– 4
(ii) 10
– 5
(iii) 7
– 2
(iv) 5
– 3
(v) 10
– 100
We learnt how to write numbers like 1425 in expanded form using exponents as
1 × 10
3
+ 4 × 10
2
+ 2 × 10
1
+ 5 × 10°.
Let us see how to express 1425.36 in expanded form in a similar way .
We have 1425.36 = 1 × 1000 + 4 × 100 + 2 × 10 + 5 × 1 +
36
10 100
+
= 1 × 10
3
+ 4 × 10
2
+ 2 × 10 + 5 × 1 + 3 × 10
– 1
+ 6 × 10
– 2
The previous number is
divided by the base 3.
Expand the following numbers using exponents.
(i) 1025.63 (ii) 1256.249
10
– 1
=
1
10
, 10
– 2
=
2
11
100 10
=
Reprint 2024-25
? ???????? ? ?????????? ? ???
TRY THESE
10.3 Laws of Exponents
We have learnt that for any non-zero integer a, a
m
× a
n
= a
m
+ n
, where m and n are natural
numbers. Does this law also hold if the exponents are negative? Let us explore.
(i) We know that 2
– 3
=
3
1
2
and 2
– 2
=
2
1
2
Therefore,
32
22
--
× =
3 2 3 2 32
11 1 1
2 22 22
+
×= = =
×
2
– 5
(ii) Take (–3)
– 4
× (–3)
–3
(–3)
– 4
×(–3)
–3
=
43
11
(3) (3)
×
--
=
4 3 43
11
(3) (3) (3)
+
=
- ×- -
= (–3)
–7
(iii) Now consider 5
–2
× 5
4
5
–2
× 5
4
=
4
4 42
2 2
15
55
55
-
×= = = 5
(2)
(iv) Now consider (–5)
– 4
× (–5)
2
(–5)
– 4
× (–5)
2
=
2
2
4 44 2
1 ( 5) 1
( 5)
(5) (5) (5) (5)
-
-
×- = =
- - - ×-
=
42
1
( 5)
-
-
= (–5)
– (2)
In general, we can say that for any non-zero integer a,
a
m
× a
n
= a
m + n
, where m and n are integers.
Simplify and write in exponential form.
(i) (–2)
–3
× (–2)
– 4
(ii) p
3
× p
–10
(iii) 3
2
× 3
–5
× 3
6
On the same lines you can verify the following laws of exponents, where a and b are non
zero integers and m, n are any integers.
(i)
m
mn
n
a
a
a
-
= (ii) (a
m
)
n
= a
mn
(iii) a
m
× b
m
= (ab)
m
(iv)
a
b
a
b
m
m
m
=
?
?
?
?
?
? (v) a
0
= 1
Let us solve some examples using the above Laws of Exponents.
1
m
m
a
a
-
=
for any non-zero integer a.
In Class VII, you have learnt that for any
non-zero integer a,
m
mn
n
a
a
a
-
= , where
m and n are natural numbers and m > n.
These laws you have studied
in Class VII for positive
exponents only.
–5 is the sum of two exponents – 3 and – 2
(– 4) + (–3) = – 7
(–2) + 4 = 2
(– 4) + 2 = –2
Reprint 2024-25
Page 4
? ???????? ? ?????????? ? ???
10.1 Introduction
Do you know?
Mass of earth is 5,970,000,000,000, 000, 000, 000, 000 kg. We have
already learnt in earlier class how to write such large numbers more
conveniently using exponents, as, 5.97 × 10
24
kg.
We read 10
24
as 10 raised to the power 24.
We know 2
5
= 2 × 2 × 2 × 2 × 2
and 2
m
= 2 × 2 × 2 × 2 × ... × 2 × 2 ... (m times)
Let us now find what is 2
– 2
is equal to?
10.2 Powers with Negative Exponents
You know that, 10
2
= 10 × 10 = 100
10
1
= 10 =
100
10
10
0
= 1 =
10
10
10
– 1
= ?
Continuing the above pattern we get,10
– 1
=
1
10
Similarly 10
– 2
=
2
1 11 1 1
10
10 10 10 100 10
÷= × = =
10
– 3
=
3
1 11 1 1
10
100 100 10 1000 10
÷= × = =
What is 10
– 10
equal to?
?????????? ???? ??????
? ? ? ? ? ? ?
??
Exponent is a
negative integer.
As the exponent decreases by1, the
value becomes one-tenth of the
previous value.
Reprint 2024-25
???? ? ? ? ????? ????
TRY THESE
TRY THESE
Now consider the following.
3
3
= 3 × 3 × 3 = 27
3
2
= 3 × 3 = 9 =
27
3
3
1
= 3 =
9
3
3° = 1 =
3
3
So looking at the above pattern, we say
3
– 1
= 1 ÷ 3 =
1
3
3
– 2
=
1
3
3
÷
=
1
33 ×
=
2
1
3
3
– 3
=
2
1
3
3
÷ =
2
1
3
×
1
3
=
3
1
3
You can now find the value of 2
– 2
in a similar manner .
We have, 10
– 2
=
2
1
10
or 10
2
=
2
1
10
-
10
– 3
=
3
1
10
or 10
3
=
3
1
10
-
3
– 2
=
2
1
3
or 3
2
=
2
1
3
-
etc.
In general, we can say that for any non-zero integer a, a
– m
=
1
m
a
, where m is a
positive integer. a
–m
is the multiplicative inverse of a
m
.
Find the multiplicative inverse of the following.
(i) 2
– 4
(ii) 10
– 5
(iii) 7
– 2
(iv) 5
– 3
(v) 10
– 100
We learnt how to write numbers like 1425 in expanded form using exponents as
1 × 10
3
+ 4 × 10
2
+ 2 × 10
1
+ 5 × 10°.
Let us see how to express 1425.36 in expanded form in a similar way .
We have 1425.36 = 1 × 1000 + 4 × 100 + 2 × 10 + 5 × 1 +
36
10 100
+
= 1 × 10
3
+ 4 × 10
2
+ 2 × 10 + 5 × 1 + 3 × 10
– 1
+ 6 × 10
– 2
The previous number is
divided by the base 3.
Expand the following numbers using exponents.
(i) 1025.63 (ii) 1256.249
10
– 1
=
1
10
, 10
– 2
=
2
11
100 10
=
Reprint 2024-25
? ???????? ? ?????????? ? ???
TRY THESE
10.3 Laws of Exponents
We have learnt that for any non-zero integer a, a
m
× a
n
= a
m
+ n
, where m and n are natural
numbers. Does this law also hold if the exponents are negative? Let us explore.
(i) We know that 2
– 3
=
3
1
2
and 2
– 2
=
2
1
2
Therefore,
32
22
--
× =
3 2 3 2 32
11 1 1
2 22 22
+
×= = =
×
2
– 5
(ii) Take (–3)
– 4
× (–3)
–3
(–3)
– 4
×(–3)
–3
=
43
11
(3) (3)
×
--
=
4 3 43
11
(3) (3) (3)
+
=
- ×- -
= (–3)
–7
(iii) Now consider 5
–2
× 5
4
5
–2
× 5
4
=
4
4 42
2 2
15
55
55
-
×= = = 5
(2)
(iv) Now consider (–5)
– 4
× (–5)
2
(–5)
– 4
× (–5)
2
=
2
2
4 44 2
1 ( 5) 1
( 5)
(5) (5) (5) (5)
-
-
×- = =
- - - ×-
=
42
1
( 5)
-
-
= (–5)
– (2)
In general, we can say that for any non-zero integer a,
a
m
× a
n
= a
m + n
, where m and n are integers.
Simplify and write in exponential form.
(i) (–2)
–3
× (–2)
– 4
(ii) p
3
× p
–10
(iii) 3
2
× 3
–5
× 3
6
On the same lines you can verify the following laws of exponents, where a and b are non
zero integers and m, n are any integers.
(i)
m
mn
n
a
a
a
-
= (ii) (a
m
)
n
= a
mn
(iii) a
m
× b
m
= (ab)
m
(iv)
a
b
a
b
m
m
m
=
?
?
?
?
?
? (v) a
0
= 1
Let us solve some examples using the above Laws of Exponents.
1
m
m
a
a
-
=
for any non-zero integer a.
In Class VII, you have learnt that for any
non-zero integer a,
m
mn
n
a
a
a
-
= , where
m and n are natural numbers and m > n.
These laws you have studied
in Class VII for positive
exponents only.
–5 is the sum of two exponents – 3 and – 2
(– 4) + (–3) = – 7
(–2) + 4 = 2
(– 4) + 2 = –2
Reprint 2024-25
???? ? ? ? ????? ????
Example 1: Find the value of
(i) 2
–3
(ii)
2
1
3
-
Solution:
(i)
3
3
11
2
8 2
-
==
(ii)
2
2
1
3 33 9
3
-
= = ×=
Example 2: Simplify
(i) (– 4)
5
× (– 4)
–10
(ii) 2
5
÷ 2
– 6
Solution:
(i) (– 4)
5
× (– 4)
–10
= (– 4)
(5 – 10)
= (– 4)
–5
=
5
1
( 4) -
( a
m
× a
n
= a
m + n
,
1
m
m
a
a
-
=
)
(ii) 2
5
÷ 2
– 6
= 2
5 – (– 6)
= 2
11
(a
m
÷ a
n
= a
m – n
)
Example 3: Express 4
– 3
as a power with the base 2.
Solution: We have, 4 = 2 × 2 = 2
2
Therefore, (4)
– 3
= (2 × 2)
– 3
= (2
2
)
– 3
= 2
2 × (– 3)
= 2
– 6
[(a
m
)
n
= a
mn
]
Example 4: Simplify and write the answer in the exponential form.
(i) (2
5
÷ 2
8
)
5
× 2
– 5
(ii) (– 4)
– 3
× (5)
– 3
× (–5)
– 3
(iii)
3
1
(3)
8
-
×
(iv) () -×
?
?
?
?
?
?
3
5
3
4
4
Solution:
(i) (2
5
÷ 2
8
)
5
× 2
– 5
= (2
5
– 8
)
5
× 2
– 5
= (2
– 3
)
5
× 2
– 5
= 2
– 15 – 5
= 2
–20
=
20
1
2
(ii) (– 4)
– 3
× (5)
– 3
× (–5)
–3
= [(– 4) × 5 × (–5)]
– 3
= [100]
– 3
=
3
1
100
[using the law a
m
× b
m
= ( ab)
m
,
a
–m
=
1
m
a
]
(iii)
3 3 33 33
3 3
11 1
(3) (3) 2 3 (2 3) 6
8 2 6
- - -- --
× = × =× = × ==
(iv) () -×
?
?
?
?
?
?
3
5
3
4
4
=
4
4
4
5
( 1 3)
3
-× × = (–1)
4
× 3
4
×
4
4
5
3
= (–1)
4
× 5
4
= 5
4
[(–1)
4
= 1]
Example 5: Find m so that (–3)
m + 1
× (–3)
5
= (–3)
7
Solution: (–3)
m + 1
× (–3)
5
= (–3)
7
(–3)
m + 1+ 5
= (–3)
7
(–3)
m + 6
= (–3)
7
On both the sides powers have the same base different from 1 and – 1, so their exponents
must be equal.
Reprint 2024-25
Page 5
? ???????? ? ?????????? ? ???
10.1 Introduction
Do you know?
Mass of earth is 5,970,000,000,000, 000, 000, 000, 000 kg. We have
already learnt in earlier class how to write such large numbers more
conveniently using exponents, as, 5.97 × 10
24
kg.
We read 10
24
as 10 raised to the power 24.
We know 2
5
= 2 × 2 × 2 × 2 × 2
and 2
m
= 2 × 2 × 2 × 2 × ... × 2 × 2 ... (m times)
Let us now find what is 2
– 2
is equal to?
10.2 Powers with Negative Exponents
You know that, 10
2
= 10 × 10 = 100
10
1
= 10 =
100
10
10
0
= 1 =
10
10
10
– 1
= ?
Continuing the above pattern we get,10
– 1
=
1
10
Similarly 10
– 2
=
2
1 11 1 1
10
10 10 10 100 10
÷= × = =
10
– 3
=
3
1 11 1 1
10
100 100 10 1000 10
÷= × = =
What is 10
– 10
equal to?
?????????? ???? ??????
? ? ? ? ? ? ?
??
Exponent is a
negative integer.
As the exponent decreases by1, the
value becomes one-tenth of the
previous value.
Reprint 2024-25
???? ? ? ? ????? ????
TRY THESE
TRY THESE
Now consider the following.
3
3
= 3 × 3 × 3 = 27
3
2
= 3 × 3 = 9 =
27
3
3
1
= 3 =
9
3
3° = 1 =
3
3
So looking at the above pattern, we say
3
– 1
= 1 ÷ 3 =
1
3
3
– 2
=
1
3
3
÷
=
1
33 ×
=
2
1
3
3
– 3
=
2
1
3
3
÷ =
2
1
3
×
1
3
=
3
1
3
You can now find the value of 2
– 2
in a similar manner .
We have, 10
– 2
=
2
1
10
or 10
2
=
2
1
10
-
10
– 3
=
3
1
10
or 10
3
=
3
1
10
-
3
– 2
=
2
1
3
or 3
2
=
2
1
3
-
etc.
In general, we can say that for any non-zero integer a, a
– m
=
1
m
a
, where m is a
positive integer. a
–m
is the multiplicative inverse of a
m
.
Find the multiplicative inverse of the following.
(i) 2
– 4
(ii) 10
– 5
(iii) 7
– 2
(iv) 5
– 3
(v) 10
– 100
We learnt how to write numbers like 1425 in expanded form using exponents as
1 × 10
3
+ 4 × 10
2
+ 2 × 10
1
+ 5 × 10°.
Let us see how to express 1425.36 in expanded form in a similar way .
We have 1425.36 = 1 × 1000 + 4 × 100 + 2 × 10 + 5 × 1 +
36
10 100
+
= 1 × 10
3
+ 4 × 10
2
+ 2 × 10 + 5 × 1 + 3 × 10
– 1
+ 6 × 10
– 2
The previous number is
divided by the base 3.
Expand the following numbers using exponents.
(i) 1025.63 (ii) 1256.249
10
– 1
=
1
10
, 10
– 2
=
2
11
100 10
=
Reprint 2024-25
? ???????? ? ?????????? ? ???
TRY THESE
10.3 Laws of Exponents
We have learnt that for any non-zero integer a, a
m
× a
n
= a
m
+ n
, where m and n are natural
numbers. Does this law also hold if the exponents are negative? Let us explore.
(i) We know that 2
– 3
=
3
1
2
and 2
– 2
=
2
1
2
Therefore,
32
22
--
× =
3 2 3 2 32
11 1 1
2 22 22
+
×= = =
×
2
– 5
(ii) Take (–3)
– 4
× (–3)
–3
(–3)
– 4
×(–3)
–3
=
43
11
(3) (3)
×
--
=
4 3 43
11
(3) (3) (3)
+
=
- ×- -
= (–3)
–7
(iii) Now consider 5
–2
× 5
4
5
–2
× 5
4
=
4
4 42
2 2
15
55
55
-
×= = = 5
(2)
(iv) Now consider (–5)
– 4
× (–5)
2
(–5)
– 4
× (–5)
2
=
2
2
4 44 2
1 ( 5) 1
( 5)
(5) (5) (5) (5)
-
-
×- = =
- - - ×-
=
42
1
( 5)
-
-
= (–5)
– (2)
In general, we can say that for any non-zero integer a,
a
m
× a
n
= a
m + n
, where m and n are integers.
Simplify and write in exponential form.
(i) (–2)
–3
× (–2)
– 4
(ii) p
3
× p
–10
(iii) 3
2
× 3
–5
× 3
6
On the same lines you can verify the following laws of exponents, where a and b are non
zero integers and m, n are any integers.
(i)
m
mn
n
a
a
a
-
= (ii) (a
m
)
n
= a
mn
(iii) a
m
× b
m
= (ab)
m
(iv)
a
b
a
b
m
m
m
=
?
?
?
?
?
? (v) a
0
= 1
Let us solve some examples using the above Laws of Exponents.
1
m
m
a
a
-
=
for any non-zero integer a.
In Class VII, you have learnt that for any
non-zero integer a,
m
mn
n
a
a
a
-
= , where
m and n are natural numbers and m > n.
These laws you have studied
in Class VII for positive
exponents only.
–5 is the sum of two exponents – 3 and – 2
(– 4) + (–3) = – 7
(–2) + 4 = 2
(– 4) + 2 = –2
Reprint 2024-25
???? ? ? ? ????? ????
Example 1: Find the value of
(i) 2
–3
(ii)
2
1
3
-
Solution:
(i)
3
3
11
2
8 2
-
==
(ii)
2
2
1
3 33 9
3
-
= = ×=
Example 2: Simplify
(i) (– 4)
5
× (– 4)
–10
(ii) 2
5
÷ 2
– 6
Solution:
(i) (– 4)
5
× (– 4)
–10
= (– 4)
(5 – 10)
= (– 4)
–5
=
5
1
( 4) -
( a
m
× a
n
= a
m + n
,
1
m
m
a
a
-
=
)
(ii) 2
5
÷ 2
– 6
= 2
5 – (– 6)
= 2
11
(a
m
÷ a
n
= a
m – n
)
Example 3: Express 4
– 3
as a power with the base 2.
Solution: We have, 4 = 2 × 2 = 2
2
Therefore, (4)
– 3
= (2 × 2)
– 3
= (2
2
)
– 3
= 2
2 × (– 3)
= 2
– 6
[(a
m
)
n
= a
mn
]
Example 4: Simplify and write the answer in the exponential form.
(i) (2
5
÷ 2
8
)
5
× 2
– 5
(ii) (– 4)
– 3
× (5)
– 3
× (–5)
– 3
(iii)
3
1
(3)
8
-
×
(iv) () -×
?
?
?
?
?
?
3
5
3
4
4
Solution:
(i) (2
5
÷ 2
8
)
5
× 2
– 5
= (2
5
– 8
)
5
× 2
– 5
= (2
– 3
)
5
× 2
– 5
= 2
– 15 – 5
= 2
–20
=
20
1
2
(ii) (– 4)
– 3
× (5)
– 3
× (–5)
–3
= [(– 4) × 5 × (–5)]
– 3
= [100]
– 3
=
3
1
100
[using the law a
m
× b
m
= ( ab)
m
,
a
–m
=
1
m
a
]
(iii)
3 3 33 33
3 3
11 1
(3) (3) 2 3 (2 3) 6
8 2 6
- - -- --
× = × =× = × ==
(iv) () -×
?
?
?
?
?
?
3
5
3
4
4
=
4
4
4
5
( 1 3)
3
-× × = (–1)
4
× 3
4
×
4
4
5
3
= (–1)
4
× 5
4
= 5
4
[(–1)
4
= 1]
Example 5: Find m so that (–3)
m + 1
× (–3)
5
= (–3)
7
Solution: (–3)
m + 1
× (–3)
5
= (–3)
7
(–3)
m + 1+ 5
= (–3)
7
(–3)
m + 6
= (–3)
7
On both the sides powers have the same base different from 1 and – 1, so their exponents
must be equal.
Reprint 2024-25
? ???????? ? ?????????? ? ???
Therefore, m + 6 = 7
or m = 7 – 6 = 1
Example 6: Find the value of
2
3
2
?
?
?
?
?
?
-
.
Solution:
2
3
2
3
3
2
9
4
2
2
2
2
2
?
?
?
?
?
?
= ==
-
-
-
Example 7: Simplify (i)
1
3
1
2
1
4
23 2
?
?
?
?
?
?
-
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
÷
?
?
?
?
?
?
-- -
(ii)
–7 –5
58
85
?? ??
×
?? ??
?? ??
Solution:
(i)
1
3
1
2
1
4
23 2
?
?
?
?
?
?
-
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
÷
?
?
?
?
?
?
-- -
=
1
3
1
2
1
4
2
2
3
3
2
2
-
-
-
-
-
-
-
?
?
?
?
?
?
÷
=
3
1
2
1
4
1
9 8 16
1
16
2
2
3
3
2
2
-
?
?
?
?
?
?
÷ = -÷ = { }
(ii)
5
8
8
5
75
?
?
?
?
?
?
×
?
?
?
?
?
?
--
=
757 5
(7) – (5) (5) ( 7)
755 7
5 858
58
855 8
-- --
- - - --
--- -
×= ×= ×
=
2
22
2
8 64
58
25 5
-
×= =
EXERCISE 10.1
1. Evaluate.
(i) 3
–2
(ii) (– 4)
– 2
(iii)
1
2
5
?
?
?
?
?
?
-
2. Simplify and express the result in power notation with positive exponent.
(i) (– 4)
5
÷ (– 4)
8
(ii)
1
2
3
2
?
?
?
?
?
?
(iii) () -×
?
?
?
?
?
?
3
5
3
4
4
(iv) (3
– 7
÷ 3
– 10
) × 3
– 5
(v) 2
– 3
× (–7)
– 3
3. Find the value of.
(i) (3° + 4
– 1
) × 2
2
(ii) (2
– 1
× 4
– 1
) ÷ 2
– 2
(iii)
1
2
1
3
1
4
2 2 2
?
?
?
?
?
?
+
?
?
?
?
?
?
+
?
?
?
?
?
?
-- -
2
3
2
3
3
2
3
2
2
2
2
2
2
2
?
?
?
?
?
?
= ==
?
?
?
?
?
?
-
-
-
In general,
a
b
b
a
mm
?
?
?
?
?
?
=
?
?
?
?
?
?
-
a
n
= 1 only if n = 0. This will work for any a.
For a = 1, 1
1
= 1
2
= 1
3
= 1
– 2
= ... = 1 or (1)
n
=
1 for infinitely many n.
For a = –1,
(–1)
0
= (–1)
2
= (–1)
4
= (–1)
–2
= ... = 1 or
(–1)
p
= 1 for any even integer p.
Reprint 2024-25
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FAQs on NCERT Textbook: Exponents & Powers - NCERT Textbooks (Class 6 to Class 12) - CTET & State TET
| 1. What is an exponent? |  |
Ans. An exponent is a mathematical operation that represents the number of times a number is multiplied by itself. It is written as a small number or a superscript to the right of a larger number, like 2³, where 3 is the exponent and 2 is the base.
| 2. What are the rules of exponents? | |
Ans. The rules of exponents are as follows:
- When multiplying two numbers with the same base, add the exponents.
- When dividing two numbers with the same base, subtract the exponents.
- When raising a power to another power, multiply the exponents.
- When raising a product to an exponent, raise each factor to that exponent.
- When raising a quotient to an exponent, raise each factor to that exponent.
| 3. How do you simplify expressions with exponents? | |
Ans. To simplify expressions with exponents, use the rules of exponents to combine like terms and reduce the expression to its simplest form. For example, if you have the expression 3² x 3³, you can add the exponents to get 3^(2+3) = 3⁵. If you have the expression (2x²)³, you can raise each factor to the exponent to get (2³ x (x²)³) = 8x⁶.
| 4. What is scientific notation? | |
Ans. Scientific notation is a way of expressing very large or very small numbers using powers of 10. It is written as a number between 1 and 10 multiplied by 10 raised to a power, such as 6.02 x 10²³. The power of 10 indicates the number of times the decimal point is moved to the left or right to convert the number to standard form.
| 5. How do you convert a number to scientific notation? | |
Ans. To convert a number to scientific notation, move the decimal point to the left or right until there is only one non-zero digit to the left of the decimal point. Count the number of places you moved the decimal point, and write this as the exponent of 10. For example, the number 0.0000345 can be written in scientific notation as 3.45 x 10⁻⁵.
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