**Class VII, NCERT (CBSE) Mathematics**

**EXPONENTS AND POWERS**

__Solutions of NCERT Math Exercise 13.1__

Q1: Find the value of:

(i) 2

^{6}(ii) 9^{3}(iii) 11^{2}(iv)5^{4}
Solution:

(i)
2

^{6}= 2 × 2 × 2 × 2 × 2 × 2 = 64
(ii)
9

^{3}= 9 × 9 × 9 = 729
(iii)
11

^{2}= 11 × 11 = 121
(iv)
5

^{4}= 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 = 6

^{4}
(ii)

*t*×*t*=*t*^{2}
(iii)

*b*×*b*×*b*×*b*=*b*^{4}
(iv)
5 × 5 × 7 × 7 × 7 = 5

^{2}× 7^{3 }
(v)
2 × 2 ×

*a*×*a*= 2^{2}×*a*^{2}
(vi)

*a*×*a*×*a*×*c*×*c*×*c*×*c*×*d*=*a*^{3}*c*^{4}*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 = 2

^{9}
(ii)
343 = 7 × 7 × 7 = 7

^{3}
(iii)
729 = 3 × 3 × 3 × 3 × 3 × 3 = 3

^{6}
(iv)
3125 = 5 × 5 × 5 × 5 × 5 = 5

^{5 }
Q4: Identify the greater number, wherever possible, in each of
the following?

(i) 4

^{3}or 3^{4}(ii) 5^{3}or 3^{5}(iii) 2^{8}or 8^{2}(iv) 100^{2 }or 2^{100}(v) 2^{10}or 10^{2}
Solution:

(i) 4

^{3}= 4 × 4 × 4 = 64
3

^{4}= 3 × 3 × 3 × 3 = 81
So,
3

^{4}> 4^{3}
(ii)
5

^{3}= 5 × 5 × 5 =125
3

^{5}= 3 × 3 × 3 × 3 × 3 = 243
So,
3

^{5}> 5^{3}
(iii)
2

^{8}= 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 256
8

^{2}= 8 × 8 = 64
Therefore,
2

^{8}> 8^{2}
(iv)100

^{2}or 2^{100}
2

^{10}= 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 1024
2

^{100}= 1024 × 1024 × 1024 × 1024 × 1024 × 1024 × 1024 × 1024 ×1024 × 1024
100

^{2}= 100 × 100 = 10000
Therefore,
2

^{100}> 100^{2}
(v)
2

^{10}and 10^{2}
2

^{10}= 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 1024
10

^{2}= 10 × 10 = 100
Therefore,
2

^{10}> 10^{2}
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 = 2

^{3}. 3^{4}
(ii)
405 = 3 × 3 × 3 × 3 × 5 = 3

^{4}. 5
(iii)
540 = 2 × 2 × 3 × 3 × 3 × 5 = 2

^{2}. 3^{3}. 5
(iv)
3600 = 2 × 2 × 2 × 2 × 3 × 3 × 5 × 5 = 2

^{4}. 3^{2}. 5^{2}
Q6: Simplify:

(i) 2 × 10

^{3}(ii) 7^{2}× 2^{2}(iii) 2^{3}× 5 (iv) 3^{ }× 4^{4}
(v) 0 × 10

^{2 }_{}(vi) 5^{2}× 3^{3}(vii) 2^{4 }× 3^{2}(viii) 3^{2}× 10^{4}
Solution:

(i)
2 × 10

^{3}= 2 × 10 × 10 × 10 = 2 × 1000 = 2000
(ii)
7

^{2}× 2^{2}= 7 × 7 × 2 × 2 = 49 × 4 = 196
(iii)
2

^{3}× 5 = 2 × 2 × 2 × 5 = 8 × 5 = 40
(iv)
3 × 4

^{4}= 3 × 4 × 4 × 4 × 4 = 3 × 256 = 768
(v)
0 × 10

^{2}= 0 × 10 × 10 = 0
(vi)
5

^{2}× 3^{3}= 5 × 5 × 3 × 3 × 3 = 25 × 27 = 675
(vii)
2

^{4}× 3^{2}= 2 × 2 × 2 × 2 × 3 × 3 = 16 × 9 = 144
(viii)
3

^{2}× 10^{4}= 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 × 10

^{12}; 1.5 × 10^{8}
(ii) 4 × 10

^{14}; 3 × 10^{17}
Solution:

(i)
2.7 × 10

^{12}> 1.5 × 10^{8}
(ii)
3 × 10

^{17}> 4 × 10^{14}
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