# Solutions of Ncert (Cbse) Mathematics Textbook Exercise 13.1 | Class 7, Chapter-13, Exponents and Powers

Class VII, NCERT (CBSE) Mathematics
EXPONENTS AND POWERS
Solutions of NCERT Math Exercise 13.1
Q1: Find the value of:
(i) 26 (ii) 93 (iii) 112 (iv)54
Solution:
(i) 26 = 2 × 2 × 2 × 2 × 2 × 2 = 64
(ii) 93 = 9 × 9 × 9 = 729
(iii) 112 = 11 × 11 = 121
(iv) 54 = 5 × 5 × 5 × 5 = 625

Q2: Express the following in exponential form:
(i) 6 × 6 × 6 × 6 (ii) t × t (iii) b × b × b × b (iv) 5 × 5 × 7 ×7 × 7
(v) 2 × 2 × a × a (vi) a × a × a × c × c × c × c × d
Solution:
(i) 6 × 6 × 6 × 6 = 64
(ii) t × t= t2
(iii) b × b × b × b = b4
(iv) 5 × 5 × 7 × 7 × 7 = 52 × 73
(v) 2 × 2 × a × a = 22 × a2
(vi) a × a × a × c × c × c × c × d = a3 c4 d

Q3: Express the following numbers using exponential notation:
(i) 512 (ii) 343 (iii) 729 (iv) 3125
Solution:
(i) 512 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 29
(ii) 343 = 7 × 7 × 7 = 73
(iii) 729 = 3 × 3 × 3 × 3 × 3 × 3 = 36
(iv) 3125 = 5 × 5 × 5 × 5 × 5 = 55

Q4: Identify the greater number, wherever possible, in each of the following?
(i) 43 or 34 (ii) 53 or 35 (iii) 28 or 82 (iv) 1002 or 2100 (v) 210 or 102
Solution:
(i) 43 = 4 × 4 × 4 = 64
34 = 3 × 3 × 3 × 3 = 81
So, 34 > 43
(ii) 53 = 5 × 5 × 5 =125
35 = 3 × 3 × 3 × 3 × 3 = 243
So, 35 > 53
(iii) 28 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 256
82 = 8 × 8 = 64
Therefore, 28 > 82
(iv)1002 or 2100
210 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 1024
2100 = 1024 × 1024 × 1024 × 1024 × 1024 × 1024 × 1024 × 1024 ×1024 × 1024
1002 = 100 × 100 = 10000
Therefore, 2100 > 1002
(v) 210 and 102
210 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 1024
102 = 10 × 10 = 100
Therefore, 210 > 102

Q5: Express each of the following as product of powers of their prime factors:
(i) 648 (ii) 405 (iii) 540 (iv) 3,600
Solution:
(i) 648 = 2 × 2 × 2 × 3 × 3 × 3 × 3 = 23. 34
(ii) 405 = 3 × 3 × 3 × 3 × 5 = 34 . 5
(iii) 540 = 2 × 2 × 3 × 3 × 3 × 5 = 22. 33. 5
(iv) 3600 = 2 × 2 × 2 × 2 × 3 × 3 × 5 × 5 = 24. 32. 52

Q6: Simplify:
(i) 2 × 103 (ii) 72 × 22 (iii) 23 × 5 (iv) 3 × 44
(v) 0 × 102 ­­­­­ (vi) 52 × 33 (vii) 24 × 32 (viii) 32 × 104
Solution:
(i) 2 × 103 = 2 × 10 × 10 × 10 = 2 × 1000 = 2000
(ii) 72 × 22 = 7 × 7 × 2 × 2 = 49 × 4 = 196
(iii) 23 × 5 = 2 × 2 × 2 × 5 = 8 × 5 = 40
(iv) 3 × 44 = 3 × 4 × 4 × 4 × 4 = 3 × 256 = 768
(v) 0 × 102 = 0 × 10 × 10 = 0
(vi) 52 × 33 = 5 × 5 × 3 × 3 × 3 = 25 × 27 = 675
(vii) 24 × 32 = 2 × 2 × 2 × 2 × 3 × 3 = 16 × 9 = 144
(viii) 32 × 104 = 3 × 3 × 10 × 10 × 10 × 10 = 9 × 10000 = 90000

Q7: Simplify:
(i) (− 4)3 (ii) (− 3) × (− 2)3 (iii) (− 3)2 × (− 5)2 (iv)(− 2)3 × (−10)3
Solution:
(i) (−4)3 = (−4) × (−4) × (−4) = −64
(ii) (−3) × (−2)3
= (−3) × (−2) × (−2) × (−2) = 24
(iii) (−3)2 × (−5)2 = (−3) × (−3) × (−5) × (−5)
= 9 × 25 = 225
(iv) (−2)3 × (−10)3
= (−2) × (−2) × (−2) × (−10) × (−10) × (−10)
= (−8) × (−1000)
= 8000

Q8: Compare the following numbers:
(i) 2.7 × 1012; 1.5 × 108
(ii) 4 × 1014; 3 × 1017
Solution:
(i) 2.7 × 1012 > 1.5 × 108
(ii) 3 × 1017 > 4 × 1014