NCERT Solutions for Class 7 Math Chapter 13 Exponents And Powers are provided here with simple step-by-step explanations. These solutions for Exponents And Powers are extremely popular among Class 7 students for Math Exponents And Powers Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the NCERT Book of Class 7 Math Chapter 13 are provided here for you for free. You will also love the ad-free experience on Meritnationâ€™s NCERT Solutions. All NCERT Solutions for class Class 7 Math are prepared by experts and are 100% accurate.

#### Page No 252:

#### Question 1:

Find the value of:

(i) 2^{6} (ii) 9^{3}

(iii) 11^{2} (iv)5^{4}

#### Answer:

(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

#### Page No 252:

#### Question 2:

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*

#### Answer:

(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*

#### Page No 253:

#### Question 3:

Express the following numbers using exponential notation:

(i) 512 (ii) 343

(iii) 729 (iv) 3125

#### Answer:

(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}

#### Page No 253:

#### Question 4:

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}

#### Answer:

(i) 4^{3}
= 4 × 4 × 4 = 64

3^{4} = 3 × 3 × 3 × 3 = 81

Therefore, 3^{4} > 4^{3}

(ii) 5^{3}
= 5 × 5 × 5 =125

3^{5} = 3 × 3 × 3 × 3 × 3 = 243

Therefore, 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}

#### Page No 253:

#### Question 5:

Express each of the following as product of powers of their prime factors:

(i) 648 (ii) 405

(iii) 540 (iv) 3,600

#### Answer:

(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}

â€‹â€‹â€‹â€‹â€‹â€‹â€‹â€‹â€‹â€‹â€‹â€‹â€‹â€‹â€‹â€‹

##### Video Solution for Exponents and Powers (Page: 253 , Q.No.: 5)

NCERT Solution for Class 7 math - Exponents and Powers 253 , Question 5

#### Page No 253:

#### Question 6:

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}

#### Answer:

(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

#### Page No 253:

#### Question 7:

Simplify:

(i) (−
4)^{3} (ii) (− 3) ×
(− 2)^{3}

(iii) (−
3)^{2} × (−
5)^{2} (iv)(− 2)^{3} ×
(−10)^{3}

#### Answer:

(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

#### Page No 253:

#### Question 8:

Compare the following numbers:

(i) 2.7 × 10^{12}; 1.5 × 10^{8}

(ii) 4 × 10^{14}; 3 × 10^{17}

#### Answer:

(i) 2.7 × 10^{12}; 1.5 × 10^{8}

2.7 × 10^{12} > 1.5 × 10^{8}

(ii) 4 × 10^{14}; 3 × 10^{17}

3 × 10^{17} > 4 × 10^{14}

â€‹â€‹â€‹â€‹â€‹â€‹â€‹â€‹â€‹â€‹â€‹â€‹â€‹â€‹â€‹â€‹

##### Video Solution for Exponents and Powers (Page: 253 , Q.No.: 8)

NCERT Solution for Class 7 math - Exponents and Powers 253 , Question 8

#### Page No 260:

#### Question 1:

Using laws of exponents, simplify and write the answer in exponential form:

(i) 3^{2}
× 3^{4} ×
3^{8} (ii) 6^{15} ÷
6^{10} (iii) *a*^{3} ×
*a*^{2}

(iv) 7^{x}× 7^{2} (v) (vi) 2^{5}
× 5^{5}

(vii) *a*^{4}
× *b*^{4} (viii) (3^{4})^{3}

(ix) (x) 8^{t}
÷ 8^{2}

#### Answer:

(i) 3^{2}
× 3^{4} × 3^{8} = (3)^{2 + 4 + 8}
(*a*^{m} × *a*^{n}
= *a*^{m}^{+}^{n})

= 3^{14}

(ii) 6^{15}
÷ 6^{10} = (6)^{15 − 10} (*a*^{m}
÷ *a*^{n}
= *a*^{m}^{−}^{n})

= 6^{5}

(iii) *a*^{3}
× *a*^{2 } = *a*^{(3 + 2) } (*a*^{m}
× *a*^{n} = *a*^{m}^{+}^{n})

= *a*^{5}

(iv) 7^{x}
+ 7^{2} = 7^{x}^{ + 2} (*a*^{m}
× *a*^{n} = *a*^{m}^{+}^{n})

(v) (5^{2})^{3}
÷ 5^{3}

= 5^{2 × 3} ÷ 5^{3 } (*a*^{m})^{n}
= *a*^{mn}

= 5^{6} ÷ 5^{3}

= 5^{(6 − 3) }(*a*^{m} ÷
*a*^{n} = *a*^{m}^{−}^{n})

= 5^{3}

(vi) 2^{5}
× 5^{5}

= (2 × 5)^{5} [*a*^{m} ×
*b*^{m} = (a × *b*)^{m}]

= 10^{5}

(vii) *a*^{4}
× *b*^{4}

= (*ab*)^{4} [*a*^{m} × *b*^{m}
= (a × *b*)^{m}]

(viii) (3^{4})^{3}
= 3^{4 × 3} = 3^{12} (*a*^{m})^{n}
= *a*^{mn}

(ix) (2^{20}
÷ 2^{15}) × 2^{3}

= (2^{20 − 15})^{}× 2^{3} (*a*^{m}
÷ *a*^{n}
= *a*^{m}^{−}^{n})

= 2^{5} × 2^{3}

= (2^{5 + 3}) (*a*^{m} × *a*^{n}
= *a*^{m}^{+}^{n})

= 2^{8}

(x) 8^{t}
÷ 8^{2} = 8^{(}^{t}^{ −
2)} (*a*^{m} ÷
*a*^{n} = *a*^{m}^{−}^{n})

#### Page No 260:

#### Question 2:

Simplify and express each of the following in exponential form:

(i) (ii) (iii)

(iv) (v) (vi) 2^{0}
+ 3^{0} + 4^{0}

(vii) 2^{0}
× 3^{0} ×
4^{0} (viii) (3^{0} + 2^{0}) ×
5^{0} (ix)

(x) (xi) (xii)

#### Answer:

(i)

(ii) [(5^{2})^{3}
× 5^{4}] ÷ 5^{7}

= [5^{2 × 3} × 5^{4}] ÷ 5^{7} (*a*^{m})^{n}
= *a*^{mn}

= [5^{6} × 5^{4}] ÷ 5^{7}

= [5^{6 + 4}] ÷ 5^{7} (*a*^{m}
× *a*^{n} = *a*^{m}^{+}^{n})

= 5^{10} ÷ 5^{7}

= 5^{10 − 7} (*a*^{m} ÷
*a*^{n} = *a*^{m}^{−}^{n})

= 5^{3}

(iii) 25^{4}
÷ 5^{3} = (5 ×5)^{4} ÷ 5^{3}

= (5^{2})^{4} ÷ 5^{3}

= 5^{2 × 4} ÷ 5^{3} (*a*^{m})^{n}
= *a*^{mn}

= 5^{8} ÷ 5^{3}

= 5^{8 − 3} (*a*^{m} ÷
*a*^{n} = *a*^{m}^{−}^{n})

= 5^{5}

(iv)

= 1 × 7 × 11^{5} = 7 × 11^{5}

(v)

(vi) 2^{0}
+ 3^{0} + 4^{0} = 1 + 1 + 1 = 3

(vii) 2^{0}
× 3^{0} × 4^{0} = 1 × 1 × 1 =
1

(viii) (3^{0}
+ 2^{0}) × 5^{0} = (1 + 1) × 1 = 2

(ix)

(x)

(xi)

(xii) (2^{3}
× 2)^{2} =
^{} (*a*^{m}
× *a*^{n} = *a*^{m}^{+}^{n})

= (2^{4})^{2} = 2^{4 × 2} (*a*^{m})^{n}
= *a*^{mn}

= 2^{8}

#### Page No 260:

#### Question 3:

Say true or false and justify your answer:

(i) 10 ×
10^{11} = 100^{11} (ii) 2^{3} > 5^{2}

(iii) 2^{3}
× 3^{2} =
6^{5 }(iv) 3^{0} = (1000)^{0}

#### Answer:

(i) 10 ×
10^{11} = 100^{11}

L.H.S. = 10 × 10^{11} = 10^{11 + 1 } (*a*^{m}
× *a*^{n} = *a*^{m}^{+}^{n})

= 10^{12}

R.H.S. = 100^{11} = (10 ×10)^{11}= (10^{2})^{11}

= 10^{2 × 11} = 10^{22} (*a*^{m})^{n}
= *a*^{mn}

As L.H.S. ≠ R.H.S.,

Therefore, the given statement is false.

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

L.H.S. = 2^{3} = 2 × 2 × 2 = 8

R.H.S. = 5^{2} = 5 × 5 = 25

As 25 > 8,

Therefore, the given statement is false.

(iii) 2^{3}
× 3^{2} = 6^{5}

L.H.S. = 2^{3} × 3^{2} = 2 × 2 × 2
× 3 × 3 = 72

R.H.S. = 6^{5} = 7776

As L.H.S. ≠ R.H.S.,

Therefore, the given statement is false.

(iv) 3^{0}
= (1000)^{0}

L.H.S. = 3^{0} = 1

R.H.S. = (1000)^{0} = 1 = L.H.S.

Therefore, the given statement is true.

#### Page No 261:

#### Question 4:

Express each of the following as a product of prime factors only in exponential form:

(i) 108 × 192 (ii) 270

(iii) 729 × 64^{}(iv) 768

#### Answer:

(i) 108 × 192

= (2 × 2 × 3 × 3 × 3) × (2 × 2 × 2 × 2 × 2 × 2 × 3)

= (2^{2} × 3^{3}) × (2^{6} × 3)^{}

= 2^{6 + 2} × 3^{3 + 1} (*a*^{m} × *a*^{n} = *a*^{m}^{+}^{n})

= 2^{8} × 3^{4}

â€‹â€‹â€‹â€‹â€‹â€‹â€‹â€‹â€‹â€‹â€‹â€‹â€‹â€‹â€‹â€‹

(ii) 270 = 2 × 3 × 3 × 3 × 5 = 2 × 3^{3 }× 5

(iii) 729 × 64 = (3 × 3 × 3 × 3 × 3 × 3) × (2 × 2 × 2 × 2 × 2 × 2)

= 3^{6} × 2^{6}

(iv) 768 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3 = 2^{8} × 3

##### Video Solution for Exponents and Powers (Page: 261 , Q.No.: 4)

NCERT Solution for Class 7 math - Exponents and Powers 261 , Question 4

#### Page No 261:

#### Question 5:

Simplify:

(i) (ii) (iii)

#### Answer:

(i)

(ii)

(iii)

#### Page No 263:

#### Question 1:

Write the following numbers in the expanded forms:

279404, 3006194, 2806196, 120719, 20068

#### Answer:

279404 = 2
× 10^{5} + 7 × 10^{4} + 9 × 10^{3}
+ 4 × 10^{2} + 0 × 10^{1} + 4 × 10^{0}

3006194 =
3 × 10^{6} + 0 × 10^{5} + 0 × 10^{4}
+ 6 × 10^{3} + 1 × 10^{2} + 9 × 10^{1}
+ 4 × 10^{0}

2806196 =
2 × 10^{6} + 8 × 10^{5} + 0 × 10^{4}
+ 6 × 10^{3} + 1 × 10^{2} + 9 × 10^{1}
+ 6 × 10^{0}

120719 = 1
× 10^{5} + 2 × 10^{4} + 0 × 10^{3}
+ 7 × 10^{2} + 1 × 10^{1} + 9 × 10^{0}

20068 = 2
× 10^{4} + 0 × 10^{3} + 0 × 10^{2}
+ 6 × 10^{1} + 8 × 10^{0}

#### Page No 263:

#### Question 2:

Find the number from each of the following expanded forms:

(a) 8 ×
10^{4} + 6 × 10^{3}
+ 0 × 10^{2} + 4 ×
10^{1} + 5 × 10^{0}

(b) 4 ×
10^{5 }+ 5 × 10^{3
}+ 3 × 10^{2 }+
2 × 10^{0}

(c) 3 ×
10^{4} + 7 × 10^{2}
+ 5 × 10^{0}

(d) 9 ×
10^{5} + 2 × 10^{2}
+ 3 × 10^{1}

#### Answer:

(a) 8 ×
10^{4} + 6 × 10^{3} + 0 × 10^{2}
+ 4 × 10^{1} + 5 × 10^{0}

= 86045

(b) 4 ×
10^{5} + 5 × 10^{3} + 3 × 10^{2}
+ 2 × 10^{0}

= 405302

(c) 3 ×
10^{4} + 7 × 10^{2} + 5 × 10^{0}

= 30705

(d) 9 ×
10^{5} + 2 × 10^{2} + 3 × 10^{1}

= 900230

#### Page No 263:

#### Question 3:

Express the following numbers in standard form:

(i) 5, 00, 00, 000 (ii) 70, 00, 000

(iii) 3, 18, 65, 00, 000 (iv) 3, 90, 878

(v) 39087.8 (vi) 3908.78

#### Answer:

(i) 50000000
= 5 × 10^{7}

(ii) 7000000
= 7 × 10^{6}

(iii) 3186500000
= 3.1865 × 10^{9}

(iv) 390878
= 3.90878 × 10^{5}

(v) 39087.8
= 3.90878 × 10^{4}

(vi) 3908.78
= 3.90878 × 10^{3}

#### Page No 263:

#### Question 4:

Express the number appearing in the following statements in standard form.

(a) The distance between Earth and Moon is 384, 000, 000 m.

(b) Speed of light in vacuum is 300, 000, 000 m/s.

(c) Diameter of the Earth is 1, 27, 56, 000 m.

(d) Diameter of the Sun is 1, 400, 000, 000 m.

(e) In a galaxy there are on an average 100, 000, 000, 000 stars.

(f) The universe is estimated to be about 12, 000, 000, 000 years old.

(g) The distance of the Sun from the centre of the Milky Way Galaxy is estimated to be 300, 000, 000, 000, 000, 000, 000 m.

(h) 60, 230, 000, 000, 000, 000, 000, 000 molecules are contained in a drop of water weighing 1.8 gm.

(i) The earth has 1, 353, 000, 000 cubic km of sea water.

(j) The population of India was about 1, 027, 000, 000 in March, 2001.

#### Answer:

(a) 3.84 ×
10^{8} m

(b) 3 ×
10^{8} m/s

(c) 1.2756
× 10^{7} m

(d) 1.4 ×
10^{9} m

(e) 1 ×
10^{11} stars

(f) 1.2 ×
10^{10} years

(g) 3 ×
10^{20} m

(h) 6.023
× 10^{22}

(i) 1.353
× 10^{9} cubic km

(j) 1.027
× 10^{9}

View NCERT Solutions for all chapters of Class 7