Selina Solutions Concise Maths Class 7 Chapter 5 Exponents (Including Laws of Exponents) Exercise 5A has problems on index or exponent and exponential form of a number. Numerous solved examples are present before each exercise to help students analyse the type of problems that would appear in the annual exam. Scoring good marks in a subject like Mathematics requires regular practise. Students gain a hold on the concepts, which are covered in the chapter, based on the latest syllabus. Selina Solutions Concise Maths Class 7 Chapter 5 Exponents (Including Laws of Exponents) Exercise 5A PDF links are given here with a free download option.
Selina Solutions Concise Maths Class 7 Chapter 5: Exponents (Including Laws of Exponents) Exercise 5A Download PDF
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1. Find the value of:
(i) 62
(ii) 73
(iii) 44
(iv) 55
(v) 83
(vi) 75
Solution:
(i) 62
It can be written as
= 6 × 6
= 36
(ii) 73
It can be written as
= 7 × 7 × 7
= 343
(iii) 44
It can be written as
= 4 × 4 × 4 × 4
= 256
(iv) 55
It can be written as
= 5 × 5 × 5 × 5 × 5
= 3125
(v) 83
It can be written as
= 8 × 8 × 8
= 512
(vi) 75
It can be written as
= 7 × 7 × 7 × 7 × 7
= 16807
2. Evaluate:
(i) 23 × 42
(ii) 23 × 52
(iii) 33 × 52
(iv) 22 × 33
(v) 32 × 53
(vi) 53 × 24
(vii) 32 × 42
(viii) (4 × 3)3
(ix) (5 × 4)2
Solution:
(i) 23 × 42
It can be written as
= 2 × 2 × 2 × 4 × 4
On further calculation
= 8 × 16
= 128
(ii) 23 × 52
It can be written as
= 2 × 2 × 2 × 5 × 5
On further calculation
= 8 × 25
= 200
(iii) 33 × 52
It can be written as
= 3 × 3 × 3 × 5 × 5
On further calculation
= 27 × 25
= 675
(iv) 22 × 33
It can be written as
= 2 × 2 × 3 × 3 × 3
On further calculation
= 4 × 27
= 108
(v) 32 × 53
It can be written as
= 3 × 3 × 5 × 5 × 5
On further calculation
= 9 × 125
= 1125
(vi) 53 × 24
It can be written as
= 5 × 5 × 5 × 2 × 2 × 2 × 2
On further calculation
= 125 × 16
= 2000
(vii) 32 × 42
It can be written as
= 3 × 3 × 4 × 4
On further calculation
= 9 × 16
= 144
(viii) (4 × 3)3
It can be written as
= 4 × 4 × 4 × 3 × 3 × 3
On further calculation
= 64 × 27
= 1728
(ix) (5 × 4)2
It can be written as
= 5 × 5 × 4 × 4
On further calculation
= 25 × 16
= 400
3. Evaluate:
(i) (3/4)4
(ii) (-5/6)5
(iii) (-3/-5)3
Solution:
(i) (3/4)4
It can be written as
= (3/4) × (3/4) × (3/4) × (3/4)
On further calculation
= (3 × 3 × 3 × 3)/ (4 × 4 × 4 × 4)
= 81/256
(ii) (-5/6)5
It can be written as
= (-5/6) × (-5/6) × (-5/6) × (-5/6) × (-5/6)
On further calculation
= [(-5) × (-5) × (-5) × (-5) × (-5)]/ (6 × 6 × 6 × 6 × 6)
= -3125/776
(iii) (-3/-5)3
It can be written as
= (-3/-5) × (-3/-5) × (-3/-5)
On further calculation
= [(-3) × (-3) × (-3)]/ [(-5) × (-5) × (-5)]
= 27/125
4. Evaluate:
(i) (2/3)3 × (3/4)2
(ii) (-3/4)3 × (2/3)4
(iii) (3/5)2 × (-2/3)3
Solution:
(i) (2/3)3 × (3/4)2
It can be written as
= (2/3) × (2/3) × (2/3) × (3/4) × (3/4)
On further calculation
= 8/27 × 9/16
= 1/6
(ii) (-3/4)3 × (2/3)4
It can be written as
= (-3/4) × (-3/4) × (-3/4) × (2/3) × (2/3) × (2/3) × (2/3)
On further calculation
= -27/64 × 16/81
= -1/12
(iii) (3/5)2 × (-2/3)3
It can be written as
= (3/5) × (3/5) × (-2/3) × (-2/3) × (-2/3)
On further calculation
= 9/25 × (-8/27)
= -8/75
5. Which is greater:
(i) 23 or 32
(ii) 25 or 52
(iii) 43 or 34
(iv) 54 or 45
Solution:
(i) 23 or 32
It can be written as
23 = 2 × 2 × 2 = 8
32 = 3 × 3 = 9
Hence, 9 is greater than 8 i.e. 32 > 23.
(ii) 25 or 52
It can be written as
25 = 2 × 2 × 2 × 2 × 2 = 32
52 = 5 × 5 = 25
Hence, 32 is greater than 25 i.e. 25 > 52.
(iii) 43 or 34
It can be written as
43 = 4 × 4 × 4 = 64
34 = 3 × 3 × 3 × 3 = 81
Hence, 81 is greater than 64 i.e. 34 > 43.
(iv) 54 or 45
It can be written as
54 = 5 × 5 × 5 × 5 = 625
45 = 4 × 4 × 4 × 4 × 4 = 1024
Hence, 1024 is greater than 625 i.e. 45 > 54.
6. Express each of the following in exponential form:
(i) 512
(ii) 1250
(iii) 1458
(iv) 3600
(v) 1350
(vi) 1176
Solution:
(i) 512
It can be written as
So we get
512 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 29
(ii) 1250
It can be written as
So we get
1250 = 2 × 5 × 5 × 5 × 5 = 2 × 54
(iii) 1458
It can be written as
So we get
1458 = 2 × 3 × 3 × 3 × 3 × 3 × 3 = 2 × 36
(iv) 3600
It can be written as
So we get
3600 = 2 × 2 × 2 × 2 × 3 × 3 × 5 × 5 = 24 × 32 × 52
(v) 1350
It can be written as
So we get
1350 = 2 × 3 × 3 × 3 × 5 × 5 = 2 × 33 × 52
(vi) 1176
It can be written as
So we get
1176 = 2 × 2 × 2 × 3 × 7 × 7 = 23 × 3 × 72
7. If a = 2 and b = 3, find the value of:
(i) (a + b)2
(ii) (b – a)3
(iii) (a × b)a
(iv) (a × b)b
Solution:
(i) (a + b)2
By substituting the values of a and b
= (2 + 3)2
On further calculation
= 52
= 5 × 5
= 25
(ii) (b – a)3
By substituting the values of a and b
= (3 – 2)3
On further calculation
= 13
= 1 × 1 × 1
= 1
(iii) (a × b)a
By substituting the values of a and b
= (2 × 3)2
On further calculation
= 62
= 6 × 6
= 36
(iv) (a × b)b
By substituting the values of a and b
= (2 × 3)3
On further calculation
= 63
= 6 × 6 × 6
= 216
8. Express:
(i) 1024 as a power of 2.
(ii) 343 as a power of 7.
(iii) 729 as a power of 3.
Solution:
(i) 1024 as a power of 2.
It can be written as
So we get
1024 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 210
(ii) 343 as a power of 7.
It can be written as
So we get
343 = 7 × 7 × 7 = 73
(iii) 729 as a power of 3.
It can be written as
So we get
729 = 3 × 3 × 3 × 3 × 3 × 3 = 36
9. If 27 × 32 = 3x × 2y; find the values of x and y.
Solution:
It is given that
27 × 32 = 3x × 2y
So we get
27 = 3x
Here
27 = 3 × 3 × 3 = 33 = 3x
We get
x = 3x
Similarly
32 = 2y
Here
32 = 2 × 2 × 2 × 2 × 2 = 25 = 2y
We get
y = 5
10. If 64 × 625 = 2a × 5b; find: (i) the values of a and b. (ii) 2b × 5a.
Solution:
(i) the values of a and b
It is given that
64 × 625 = 2a × 5b
We know that
64 = 2a
We can write it as
64 = 2 × 2 × 2 × 2 × 2 × 2
So we get
64 = 26
a = 6
Similarly
625 = 5b
We can write it as
625 = 5 × 5 × 5 × 5
So we get
625 = 54
b = 4
(ii) 2b × 5a
Substituting the values of a and b
= 24 × 56
It can be written as
= 2 × 2 × 2 × 2 × 5 × 5 × 5 × 5 × 5 × 5
So we get
= 16 × 15625
= 250000
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