NCERT Textbook- Exponents and Powers Class 8 Notes | EduRev

Mathematics (Maths) Class 8

Created by: Indu Gupta

Class 8 : NCERT Textbook- Exponents and Powers Class 8 Notes | EduRev

 Page 1


EXPONENTS AND POWERS  193
12.1  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?
12.2  Powers with Negative Exponents
Y ou 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
1111 1
10
10 10 10 100 10
÷= × = =
10
– 3
 =
3
1111 1
10
100 100 10 1000 10
÷= × = =
What is 10
– 10
 equal to?
Exponents and Powers
CHAPTER
12
Exponent is a
negative integer.
As the exponent decreases by1, the
value becomes one-tenth of the
previous value.
Page 2


EXPONENTS AND POWERS  193
12.1  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?
12.2  Powers with Negative Exponents
Y ou 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
1111 1
10
10 10 10 100 10
÷= × = =
10
– 3
 =
3
1111 1
10
100 100 10 1000 10
÷= × = =
What is 10
– 10
 equal to?
Exponents and Powers
CHAPTER
12
Exponent is a
negative integer.
As the exponent decreases by1, the
value becomes one-tenth of the
previous value.
194  MATHEMATICS
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
Y ou can now find the value of 2
– 2
 in a similar manner.
W e 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
W e 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
=
Page 3


EXPONENTS AND POWERS  193
12.1  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?
12.2  Powers with Negative Exponents
Y ou 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
1111 1
10
10 10 10 100 10
÷= × = =
10
– 3
 =
3
1111 1
10
100 100 10 1000 10
÷= × = =
What is 10
– 10
 equal to?
Exponents and Powers
CHAPTER
12
Exponent is a
negative integer.
As the exponent decreases by1, the
value becomes one-tenth of the
previous value.
194  MATHEMATICS
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
Y ou can now find the value of 2
– 2
 in a similar manner.
W e 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
W e 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
=
EXPONENTS AND POWERS  195
TRY THESE
12.3  Laws of Exponents
W e 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
--
× = 
32 3 2 32
11 1 1
22 2 2 2
+
×= = =
×
2 
– 5
(ii) Take (–3)
– 4
 × (–3)
–3
(–3)
– 4
 ×(–3)
–3
 =
43
11
(3) ( 3)
×
--
=
43 43
11
(3) ( 3) ( 3)
+
=
-×- -
 = (–3)
–7
(iii) Now consider 5
–2
 × 5
4
5
–2 
× 5
4 
= 
4
442
22
15
55
55
-
×= = = 5
(2)
(iv) Now consider (–5)
– 4
 × (–5)
2
(–5)
– 4
 × (–5)
2
 =
2
2
444 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)
m
m
m
aa
b b
??
=
??
??
(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
Page 4


EXPONENTS AND POWERS  193
12.1  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?
12.2  Powers with Negative Exponents
Y ou 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
1111 1
10
10 10 10 100 10
÷= × = =
10
– 3
 =
3
1111 1
10
100 100 10 1000 10
÷= × = =
What is 10
– 10
 equal to?
Exponents and Powers
CHAPTER
12
Exponent is a
negative integer.
As the exponent decreases by1, the
value becomes one-tenth of the
previous value.
194  MATHEMATICS
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
Y ou can now find the value of 2
– 2
 in a similar manner.
W e 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
W e 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
=
EXPONENTS AND POWERS  195
TRY THESE
12.3  Laws of Exponents
W e 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
--
× = 
32 3 2 32
11 1 1
22 2 2 2
+
×= = =
×
2 
– 5
(ii) Take (–3)
– 4
 × (–3)
–3
(–3)
– 4
 ×(–3)
–3
 =
43
11
(3) ( 3)
×
--
=
43 43
11
(3) ( 3) ( 3)
+
=
-×- -
 = (–3)
–7
(iii) Now consider 5
–2
 × 5
4
5
–2 
× 5
4 
= 
4
442
22
15
55
55
-
×= = = 5
(2)
(iv) Now consider (–5)
– 4
 × (–5)
2
(–5)
– 4
 × (–5)
2
 =
2
2
444 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)
m
m
m
aa
b b
??
=
??
??
(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
196  MATHEMATICS
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
33 3 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)
4
4
5
(3)
3
??
-×
??
??
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)
3333 33
33
11 1
(3) (3) 2 3 (2 3) 6
826
---- --
×= ×= × = × = =
(iv)
4
4
5
(3)
3
??
-×
??
??
=
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.
Page 5


EXPONENTS AND POWERS  193
12.1  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?
12.2  Powers with Negative Exponents
Y ou 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
1111 1
10
10 10 10 100 10
÷= × = =
10
– 3
 =
3
1111 1
10
100 100 10 1000 10
÷= × = =
What is 10
– 10
 equal to?
Exponents and Powers
CHAPTER
12
Exponent is a
negative integer.
As the exponent decreases by1, the
value becomes one-tenth of the
previous value.
194  MATHEMATICS
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
Y ou can now find the value of 2
– 2
 in a similar manner.
W e 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
W e 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
=
EXPONENTS AND POWERS  195
TRY THESE
12.3  Laws of Exponents
W e 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
--
× = 
32 3 2 32
11 1 1
22 2 2 2
+
×= = =
×
2 
– 5
(ii) Take (–3)
– 4
 × (–3)
–3
(–3)
– 4
 ×(–3)
–3
 =
43
11
(3) ( 3)
×
--
=
43 43
11
(3) ( 3) ( 3)
+
=
-×- -
 = (–3)
–7
(iii) Now consider 5
–2
 × 5
4
5
–2 
× 5
4 
= 
4
442
22
15
55
55
-
×= = = 5
(2)
(iv) Now consider (–5)
– 4
 × (–5)
2
(–5)
– 4
 × (–5)
2
 =
2
2
444 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)
m
m
m
aa
b b
??
=
??
??
(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
196  MATHEMATICS
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
33 3 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)
4
4
5
(3)
3
??
-×
??
??
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)
3333 33
33
11 1
(3) (3) 2 3 (2 3) 6
826
---- --
×= ×= × = × = =
(iv)
4
4
5
(3)
3
??
-×
??
??
=
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.
EXPONENTS AND POWERS  197
Therefore, m + 6 = 7
or m = 7 – 6  = 1
Example 6: Find the value of 
2
2
3
-
??
??
??
.
Solution:  
2
22
22
22 3 9
34 32
-
-
-
??
== =
??
??
Example 7: Simplify (i)  
23 2
11 1
32 4
-- -
??
?? ?? ? ? ? ?
-÷
?? ? ? ????
?? ? ? ??
??
??
(ii)  
–7–5
58
85
?? ? ?
×
?? ? ?
?? ? ?
Solution:
 (i)
23 2
11 1
32 4
-- -
??
?? ? ? ?? ??
-÷
?? ? ? ????
?? ? ? ??
??
??
 = 
23 2
23 2
11 1
32 4
-- -
-- -
??
-÷
??
??
=
23 2
23 2
32 4 1
{9 8} 16
16 11 1
??
-÷ = - ÷ =
??
??
(ii)
75
58
85
--
?? ??
×
?? ??
?? ??
= 
75 7 5
( 7)–(5) ( 5) (7)
755 7
58 5 8
58
855 8
-- - -
-- ---
--- -
×= × = ×
=
2
22
2
864
58
25 5
-
×= =
EXERCISE 12.1
1. Evaluate.
(i) 3
–2
(ii) (– 4)
– 2
(iii)
5
1
2
-
??
??
??
2. Simplify and express the result in power notation with positive exponent.
(i) (– 4)
5
 ÷ (– 4)
8
(ii)
2
3
1
2
??
??
??
(iii)
4
4
5
(3)
3
??
-×
??
??
(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)
22 2
111
23 4
-- -
?? ? ? ??
++
?? ?? ??
?? ?? ??
22
22
22
22 3 3
32 32
-
-
-
?? ? ?
== =
?? ??
?? ??
In general, 
mm
ab
ba
-
?? ??
=
?? ??
?? ??
a
n
 = 1 only if n = 0. This will work for any a
except a = 1 or a = –1. 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.
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