NCERT Solutions for Class 7 Math Chapter 13 - Exponents And Powers

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Mathematics NCERT Grade 8, Chapter 13: Exponents and Powers- This chapter highlights all the concepts related to exponents and powers. Initially, the concept of exponents will be discussed where a student will study how to read, write and represent exponents and powers.
Students will know about the importance and use of exponents.

The short notation 10⁴ stands for the product 10×10×10×10. Here ‘10’ is called the base and ‘4’ the exponent. The number 10⁴ is read as 10 raised to the power of 4 or simply as the fourth power of 10. 10⁴is called the exponential form of 10,000.

The other portion of the chapter gives insight about Laws of exponents. This particular section is divided into several sub-sections:

Multiplying Powers with the Same Base
Dividing Powers with the Same Base
Taking Power of a Power
Multiplying Powers with the Same Exponents
Dividing Powers with the Same Exponents
Numbers with exponent zero

Once the student has come across various laws of exponents miscellaneous examples using the laws of exponents can be studied.
This is followed by Decimal Number System and expressing large numbers in the standard form.

Any number can be expressed as a decimal number between 1.0 and 10.0 including 1.0 multiplied by a power of 10. Such a form of a number is called its standard form.

The chapter includes 3 unsolved exercises and various solved examples.
The chapter comprises a summary in which all the important concepts of the chapter- Exponents and Powers are mentioned.

Page No 252:

Question 1:

Find the value of:

(i) 2⁶ (ii) 9³

(iii) 11² (iv)5⁴

Answer:

(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

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 = 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

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 = 2⁹

(ii) 343 = 7 × 7 × 7 = 7³

(iii) 729 = 3 × 3 × 3 × 3 × 3 × 3 = 3⁶

(iv) 3125 = 5 × 5 × 5 × 5 × 5 = 5⁵

Page No 253:

Question 4:

Identify the greater number, wherever possible, in each of the following?

(i) 4³ or 3⁴ (ii) 5³ or 3⁵

(iii) 2⁸ or 8² (iv) 100²or 2¹⁰⁰

(v) 2¹⁰ or 10²

Answer:

(i) 4³ = 4 × 4 × 4 = 64

3⁴ = 3 × 3 × 3 × 3 = 81

Therefore, 3⁴ > 4³

(ii) 5³ = 5 × 5 × 5 =125

3⁵ = 3 × 3 × 3 × 3 × 3 = 243

Therefore, 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

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 = 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

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 × 10³ (ii) 7² × 2²

(iii) 2³ × 5 (iv) 3× 4⁴

(v) 0 × 10² (vi) 5² × 3³

(vii) 2⁴× 3² (viii) 3² × 10⁴

Answer:

(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

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 × 10¹²; 1.5 × 10⁸

(ii) 4 × 10¹⁴; 3 × 10¹⁷

Answer:

(i) 2.7 × 10¹²; 1.5 × 10⁸

2.7 × 10¹² > 1.5 × 10⁸

(ii) 4 × 10¹⁴; 3 × 10¹⁷

3 × 10¹⁷ > 4 × 10¹⁴

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) 3² × 3⁴ × 3⁸ (ii) 6¹⁵ ÷ 6¹⁰ (iii) a³ × a²

(iv) 7^x× 7² (v) (vi) 2⁵ × 5⁵

(vii) a⁴ × b⁴ (viii) (3⁴)³

(ix) (x) 8^t ÷ 8²

Answer:

(i) 3² × 3⁴ × 3⁸ = (3)^{2 + 4 + 8} (a^m × aⁿ = a^m⁺ⁿ)

= 3¹⁴

(ii) 6¹⁵ ÷ 6¹⁰ = (6)^{15 − 10} (a^m ÷ aⁿ = a^m⁻ⁿ)

= 6⁵

(iii) a³ × a² = a^{(3 + 2)} (a^m × aⁿ = a^m⁺ⁿ)

= a⁵

(iv) 7^x + 7² = 7^x^{+ 2} (a^m × aⁿ = a^m⁺ⁿ)

(v) (5²)³ ÷ 5³

= 5^{2 × 3} ÷ 5³ (a^m)ⁿ = a^mn

= 5⁶ ÷ 5³

= 5^{(6 − 3)}(a^m ÷ aⁿ = a^m⁻ⁿ)

= 5³

(vi) 2⁵ × 5⁵

= (2 × 5)⁵ [a^m × b^m = (a × b)^m]

= 10⁵

(vii) a⁴ × b⁴

= (ab)⁴ [a^m × b^m = (a × b)^m]

(viii) (3⁴)³ = 3^{4 × 3} = 3¹² (a^m)ⁿ = a^mn

(ix) (2²⁰ ÷ 2¹⁵) × 2³

= (2^{20 − 15})× 2³ (a^m ÷ aⁿ = a^m⁻ⁿ)

= 2⁵ × 2³

= (2^{5 + 3}) (a^m × aⁿ = a^m⁺ⁿ)

= 2⁸

(x) 8^t ÷ 8² = 8⁽^t^−
2) (a^m ÷ aⁿ = a^m⁻ⁿ)

Page No 260:

Question 2:

Simplify and express each of the following in exponential form:

(i) (ii) (iii)

(iv) (v) (vi) 2⁰ + 3⁰ + 4⁰

(vii) 2⁰ × 3⁰ × 4⁰ (viii) (3⁰ + 2⁰) × 5⁰ (ix)

(x) (xi) (xii)

Answer:

(i)

(ii) [(5²)³ × 5⁴] ÷ 5⁷

= [5^{2 × 3} × 5⁴] ÷ 5⁷ (a^m)ⁿ = a^mn

= [5⁶ × 5⁴] ÷ 5⁷

= [5^{6 + 4}] ÷ 5⁷ (a^m × aⁿ = a^m⁺ⁿ)

= 5¹⁰ ÷ 5⁷

= 5^{10 − 7} (a^m ÷ aⁿ = a^m⁻ⁿ)

= 5³

(iii) 25⁴ ÷ 5³ = (5 ×5)⁴ ÷ 5³

= (5²)⁴ ÷ 5³

= 5^{2 × 4} ÷ 5³ (a^m)ⁿ = a^mn

= 5⁸ ÷ 5³

= 5^{8 − 3} (a^m ÷ aⁿ = a^m⁻ⁿ)

= 5⁵

(iv)

= 1 × 7 × 11⁵ = 7 × 11⁵

(v)

(vi) 2⁰ + 3⁰ + 4⁰ = 1 + 1 + 1 = 3

(vii) 2⁰ × 3⁰ × 4⁰ = 1 × 1 × 1 = 1

(viii) (3⁰ + 2⁰) × 5⁰ = (1 + 1) × 1 = 2

(ix)

(x)

(xi)

(xii) (2³ × 2)² = (a^m × aⁿ = a^m⁺ⁿ)

= (2⁴)² = 2^{4 × 2} (a^m)ⁿ = a^mn

= 2⁸

Page No 260:

Question 3:

Say true or false and justify your answer:

(i) 10 × 10¹¹ = 100¹¹ (ii) 2³ > 5²

(iii) 2³ × 3² = 6⁵(iv) 3⁰ = (1000)⁰

Answer:

(i) 10 × 10¹¹ = 100¹¹

L.H.S. = 10 × 10¹¹ = 10^{11 + 1} (a^m × aⁿ = a^m⁺ⁿ)

= 10¹²

R.H.S. = 100¹¹ = (10 ×10)¹¹= (10²)¹¹

= 10^{2 × 11} = 10²² (a^m)ⁿ = a^mn

As L.H.S. ≠ R.H.S.,

Therefore, the given statement is false.

(ii) 2³ > 5²

L.H.S. = 2³ = 2 × 2 × 2 = 8

R.H.S. = 5² = 5 × 5 = 25

As 25 > 8,

Therefore, the given statement is false.

(iii) 2³ × 3² = 6⁵

L.H.S. = 2³ × 3² = 2 × 2 × 2 × 3 × 3 = 72

R.H.S. = 6⁵ = 7776

As L.H.S. ≠ R.H.S.,

Therefore, the given statement is false.

(iv) 3⁰ = (1000)⁰

L.H.S. = 3⁰ = 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)

= (2² × 3³) × (2⁶ × 3)

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

= 2⁸ × 3⁴

(ii) 270 = 2 × 3 × 3 × 3 × 5 = 2 × 3³× 5

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

= 3⁶ × 2⁶

(iv) 768 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3 = 2⁸ × 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 × 10⁵ + 7 × 10⁴ + 9 × 10³ + 4 × 10² + 0 × 10¹ + 4 × 10⁰

3006194 = 3 × 10⁶ + 0 × 10⁵ + 0 × 10⁴ + 6 × 10³ + 1 × 10² + 9 × 10¹ + 4 × 10⁰

2806196 = 2 × 10⁶ + 8 × 10⁵ + 0 × 10⁴ + 6 × 10³ + 1 × 10² + 9 × 10¹ + 6 × 10⁰

120719 = 1 × 10⁵ + 2 × 10⁴ + 0 × 10³ + 7 × 10² + 1 × 10¹ + 9 × 10⁰

20068 = 2 × 10⁴ + 0 × 10³ + 0 × 10² + 6 × 10¹ + 8 × 10⁰

Page No 263:

Question 2:

Find the number from each of the following expanded forms:

(a) 8 × 10⁴ + 6 × 10³ + 0 × 10² + 4 × 10¹ + 5 × 10⁰

(b) 4 × 10⁵+ 5 × 10³+ 3 × 10²+ 2 × 10⁰

(d) 9 × 10⁵ + 2 × 10² + 3 × 10¹

Answer:

(a) 8 × 10⁴ + 6 × 10³ + 0 × 10² + 4 × 10¹ + 5 × 10⁰

= 86045

(b) 4 × 10⁵ + 5 × 10³ + 3 × 10² + 2 × 10⁰

= 405302

= 30705

(d) 9 × 10⁵ + 2 × 10² + 3 × 10¹

= 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 × 10⁷

(ii) 7000000 = 7 × 10⁶

(iii) 3186500000 = 3.1865 × 10⁹

(iv) 390878 = 3.90878 × 10⁵

(v) 39087.8 = 3.90878 × 10⁴

(vi) 3908.78 = 3.90878 × 10³

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.

(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 × 10⁸ m

(b) 3 × 10⁸ m/s

(d) 1.4 × 10⁹ m

(e) 1 × 10¹¹ stars

(f) 1.2 × 10¹⁰ years

(g) 3 × 10²⁰ m

(h) 6.023 × 10²²

(i) 1.353 × 10⁹ cubic km

(j) 1.027 × 10⁹

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