Class 7 Maths - Exponents and Powers NCERT Solutions

Chapter 13: Exponents and Powers (NCERT Solutions)

Exercise 13.1

Q1. Find the value of:
(i) 2⁶
(ii) 9³
(iii) 11²
(iv) 5⁴

(i) 2⁶ = 2 × 2 × 2 × 2 × 2 × 2 = 64

(ii) 9³ = 9 × 9 × 9 = 81 × 9 = 729

(iii) 11² = 11 × 11 = 121

(iv) 5⁴ = 5 × 5 × 5 × 5 = 25 × 25 = 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

(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 each of the following numbers using exponential notation:
(i) 512
(ii) 343
(iii) 729
(iv) 3125

(i) 512
By prime factorization:
512 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 2⁹

(ii) 343
By prime factorization:
343 = 7 × 7 × 7 = 7³

(iii) 729
By prime factorization:
729 = 3 × 3 × 3 × 3 × 3 × 3 = 3⁶
(Alternatively, 9³)

(iv) 3125
By prime factorization:
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

(i) 4³ = 4 × 4 × 4 = 64
3⁴ = 3 × 3 × 3 × 3 = 81
Since 81 > 64, 3⁴ is greater.

(ii) 5³ = 5 × 5 × 5 = 125
3⁵ = 3 × 3 × 3 × 3 × 3 = 243
Since 243 > 125, 3⁵ is greater.

(iii) 2⁸ = 256
8² = 8 × 8 = 64
Since 256 > 64, 2⁸ is greater.

(iv) 100² = 10000
2¹⁰ = 1024, so 2¹⁰⁰ will be much larger than 10000.
Therefore, 2¹⁰⁰ is greater.

(v) 2¹⁰ = 1024
10² = 100
Since 1024 > 100, 2¹⁰ is greater.

Q5. Express each of the following as product of powers of their prime factors:
(i) 648
(ii) 405
(iii) 540
(iv) 3,600

(i) 648
648 = 2 × 324 = 2 × 2 × 162 = 2 × 2 × 2 × 81 = 2³ × 3⁴
648 = 2³ × 3⁴

(ii) 405
405 = 3 × 135 = 3 × 3 × 45 = 3 × 3 × 3 × 15 = 3 × 3 × 3 × 3 × 5
405 = 3⁴ × 5

(iii) 540
540 = 2 × 270 = 2 × 2 × 135 = 2 × 2 × 3 × 45 = 2² × 3³ × 5
540 = 2² × 3³ × 5

(iv) 3,600
3600 = 36 × 100 = (2² × 3²) × (2² × 5²)
= 2²⁺² × 3² × 5²
3600 = 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

(i) 2 × 10³ = 2 × 1000 = 2000

(ii) 7² × 2² = 49 × 4 = 196

(iii) 2³ × 5 = 8 × 5 = 40

(iv) 3 × 4⁴ = 3 × 256 = 768

(v) 0 × 10² = 0 × 100 = 0

(vi) 5² × 3³ = 25 × 27 = 675

(vii) 2⁴ × 3² = 16 × 9 = 144

(viii) 3² × 10⁴ = 9 × 10000 = 90000

Q7. Simplify:
(i) (-4)³
(ii) (-3) × (-2)³
(iii) (-3)² × (-5)²
(iv) (-2)³ × (-10)³

(i) (-4)³ = (-4) × (-4) × (-4) = -64

(ii) (-3) × (-2)³ = (-3) × (-8) = 24

(iii) (-3)² × (-5)² = (9) × (25) = 225

(iv) (-2)³ × (-10)³ = (-8) × (-1000) = 8000

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

(i) 2.7 × 10¹² and 1.5 × 10⁸
Comparing the powers of 10, 10¹² > 10⁸.
Therefore, 2.7 × 10¹² > 1.5 × 10⁸

(ii) 4 × 10¹⁴ and 3 × 10¹⁷
Comparing the powers of 10, 10¹⁷ > 10¹⁴.
Therefore, 3 × 10¹⁷ > 4 × 10¹⁴

Exercise 13.2

Q1. 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) (5²)³ ÷ 5³
(vi) 2⁵ × 5⁵
(vii) a⁴ × b⁴
(viii) (3⁴)³
(ix) (2²⁰ ÷ 2¹⁵) × 2³
(x) 8^t ÷ 8²

(i) 3² × 3⁴ × 3⁸ = 3^(2+4+8) = 3¹⁴

(ii) 6¹⁵ ÷ 6¹⁰ = 6^(15-10) = 6⁵

(iii) a³ × a² = a⁽³⁺²⁾ = a⁵

(iv) 7^x × 7² = 7^(x+2)

(v) (5²)³ ÷ 5³ = 5^(2×3) ÷ 5³ = 5⁶ ÷ 5³ = 5^(6-3) = 5³

(vi) 2⁵ × 5⁵ = (2 × 5)⁵ = 10⁵

(vii) a⁴ × b⁴ = (ab)⁴

(viii) (3⁴)³ = 3^(4×3) = 3¹²

(ix) (2²⁰ ÷ 2¹⁵) × 2³ = (2^20-15) × 2³ = 2⁵ × 2³ = 2⁵⁺³ = 2⁸

(x) 8^t ÷ 8² = 8^(t-2)

Q2. Simplify and express each of the following in exponential form:
(i) (2³ × 3⁴ × 4) / (3 × 32)
(ii) ((5²)³ × 5⁴) ÷ 5⁷
(iii) 25⁴ ÷ 5³
(iv) (3 × 7² × 11⁸) / (21 × 11³)
(v) 3⁷ / (3⁴ × 3³)
(vi) 2⁰ + 3⁰ + 4⁰
(vii) 2⁰ × 3⁰ × 4⁰
(viii) (3⁰ + 2⁰) × 5⁰
(ix) (2⁸ × a⁵) / (4³ × a³)
(x) (a⁵ / a³) × a⁸
(xi) (4⁵ × a⁸ b³) / (4⁵ × a⁵ b²)
(xii) (2³ × 2)²

(i) (2³ × 3⁴ × 2²) / (3 × 2⁵)
= (2³⁺² × 3⁴) / (2⁵ × 3¹)
= (2⁵ × 3⁴) / (2⁵ × 3¹)
= 2^5-5 × 3^4-1 = 2⁰ × 3³ = 1 × 3³ = 3³

(ii) (5⁶ × 5⁴) ÷ 5⁷
= 5⁽⁶⁺⁴⁾ ÷ 5⁷ = 5¹⁰ ÷ 5⁷ = 5^(10-7) = 5³

(iii) (5²)⁴ ÷ 5³
= 5⁸ ÷ 5³ = 5^(8-3) = 5⁵

(iv) (3 × 7² × 11⁸) / (3 × 7 × 11³)
= 3^(1-1) × 7^(2-1) × 11^(8-3)
= 3⁰ × 7¹ × 11⁵ = 1 × 7 × 11⁵ = 7 × 11⁵

(v) 3⁷ / 3⁽⁴⁺³⁾ = 3⁷ / 3⁷ = 3^(7-7) = 3⁰ = 1

(vi) 1 + 1 + 1 = 3

(vii) 1 × 1 × 1 = 1

(viii) (1 + 1) × 1 = 2 × 1 = 2

(ix) (2⁸ × a⁵) / ((2²)³ × a³)
= (2⁸ × a⁵) / (2⁶ × a³)
= 2^(8-6) × a^(5-3) = 2² × a² = (2a)²

(x) a^(5-3) × a⁸ = a² × a⁸ = a⁽²⁺⁸⁾ = a¹⁰

(xi) 4^(5-5) × a^(8-5) × b^(3-2)
= 4⁰ × a³ × b¹ = 1 × a³ × b = a³b

(xii) (2⁽³⁺¹⁾)² = (2⁴)² = 2^(4×2) = 2⁸

Q3. Say true or false and justify your answer:
(i) 10 × 10¹¹ = 100¹¹
(ii) 2³ > 5²
(iii) 2³ × 3² = 6⁵
(iv) 3⁰ = (1000)⁰

(i) False. LHS = 10¹ × 10¹¹ = 10¹². RHS = (10²)¹¹ = 10²². Thus, 10¹² ≠ 10²².

(ii) False. LHS = 8. RHS = 25. Thus, 8 is not greater than 25.

(iii) False. LHS = 8 × 9 = 72. RHS = 6⁵ = 7776. Thus, 72 ≠ 7776.

(iv) True. 3⁰ = 1. Also, (1000)⁰ = 1. Thus, 1 = 1.

Q4. 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

(i) 108 = 2² × 3³ and 192 = 2⁶ × 3
So, 108 × 192 = (2² × 3³) × (2⁶ × 3) = 2⁽²⁺⁶⁾ × 3⁽³⁺¹⁾ = 2⁸ × 3⁴

(ii) 270 = 2 × 135 = 2 × 3 × 45 = 2 × 3 × 3 × 15 = 2 × 3 × 3 × 3 × 5 = 2 × 3³ × 5

(iii) 729 = 3⁶ and 64 = 2⁶
So, 729 × 64 = 3⁶ × 2⁶ (or (2 × 3)⁶ = 6⁶)

(iv) 768 = 2 × 384 = 2 × 2 × 192 = 2² × 2⁶ × 3 = 2⁸ × 3

Q5. Simplify:
(i) (2⁵)² × 7³ / (8³ × 7)
(ii) (25 × 5² × t⁸) / (10³ × t⁴)
(iii) (3⁵ × 10⁵ × 25) / (5⁷ × 6⁵)

(i) (2¹⁰ × 7³) / ((2³)³ × 7¹)
= (2¹⁰ × 7³) / (2⁹ × 7¹)
= 2^(10-9) × 7^(3-1) = 2¹ × 7² = 2 × 49 = 98

(ii) (5² × 5² × t⁸) / ((2 × 5)³ × t⁴)
= (5⁴ × t⁸) / (2³ × 5³ × t⁴)
= (5^(4-3) × t^(8-4)) / 2³ = (5¹ × t⁴) / 8 = (5t⁴) / 8

(iii) (3⁵ × (2 × 5)⁵ × 5²) / (5⁷ × (2 × 3)⁵)
= (3⁵ × 2⁵ × 5⁵ × 5²) / (5⁷ × 2⁵ × 3⁵)
= (3⁵ × 2⁵ × 5⁷) / (3⁵ × 2⁵ × 5⁷) = 1

Exercise 13.3

Q1. Write the following numbers in the expanded forms:
279404, 3006194, 2806196, 120719, 20068

279404 = 2 × 100000 + 7 × 10000 + 9 × 1000 + 4 × 100 + 0 × 10 + 4 × 1
= 2 × 10⁵ + 7 × 10⁴ + 9 × 10³ + 4 × 10² + 0 × 10¹ + 4 × 10⁰

3006194 = 3 × 1000000 + 0 × 100000 + 0 × 10000 + 6 × 1000 + 1 × 100 + 9 × 10 + 4 × 1
= 3 × 10⁶ + 0 × 10⁵ + 0 × 10⁴ + 6 × 10³ + 1 × 10² + 9 × 10¹ + 4 × 10⁰

2806196 = 2 × 1000000 + 8 × 100000 + 0 × 10000 + 6 × 1000 + 1 × 100 + 9 × 10 + 6 × 1
= 2 × 10⁶ + 8 × 10⁵ + 0 × 10⁴ + 6 × 10³ + 1 × 10² + 9 × 10¹ + 6 × 10⁰

120719 = 1 × 100000 + 2 × 10000 + 0 × 1000 + 7 × 100 + 1 × 10 + 9 × 1
= 1 × 10⁵ + 2 × 10⁴ + 0 × 10³ + 7 × 10² + 1 × 10¹ + 9 × 10⁰

20068 = 2 × 10000 + 0 × 1000 + 0 × 100 + 6 × 10 + 8 × 1
= 2 × 10⁴ + 0 × 10³ + 0 × 10² + 6 × 10¹ + 8 × 10⁰

Q2. 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
(c) 3 × 10⁴ + 7 × 10² + 5 × 10⁰
(d) 9 × 10⁵ + 2 × 10² + 3 × 10¹

(a) = 8 × 10000 + 6 × 1000 + 0 × 100 + 4 × 10 + 5 × 1 = 86045

(b) = 4 × 100000 + 0 × 10000 + 5 × 1000 + 3 × 100 + 0 × 10 + 2 × 1 = 405302

(c) = 3 × 10000 + 0 × 1000 + 7 × 100 + 0 × 10 + 5 × 1 = 30705

(d) = 9 × 100000 + 0 × 10000 + 0 × 1000 + 2 × 100 + 3 × 10 + 0 × 1 = 900230

Q3. 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

(i) 5,00,00,000 = 5.0 × 10,000,000 = 5 × 10⁷

(ii) 70,00,000 = 7.0 × 1,000,000 = 7 × 10⁶

(iii) 3,18,65,00,000 = 3.1865 × 1,000,000,000 = 3.1865 × 10⁹

(iv) 3,90,878 = 3.90878 × 100,000 = 3.90878 × 10⁵

(v) 39087.8 = 3.90878 × 10,000 = 3.90878 × 10⁴

(vi) 3908.78 = 3.90878 × 1,000 = 3.90878 × 10³

Q4. 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.

(a) 384,000,000 m = 3.84 × 10⁸ m

(b) 300,000,000 m/s = 3 × 10⁸ m/s

(c) 12,756,000 m = 1.2756 × 10⁷ m

(d) 1,400,000,000 m = 1.4 × 10⁹ m

(e) 100,000,000,000 stars = 1 × 10¹¹ stars

(f) 12,000,000,000 years = 1.2 × 10¹⁰ years

(g) 300,000,000,000,000,000,000 m = 3 × 10²⁰ m

(h) 60,230,000,000,000,000,000,000 = 6.023 × 10²²

(i) 1,353,000,000 cubic km = 1.353 × 10⁹ cubic km

(j) 1,027,000,000 = 1.027 × 10⁹

Class 7 Maths - Exponents and Powers Practice

Chapter 13: Exponents and Powers (Practice Questions)

RD Sharma / Extra Practice

Q1. Evaluate: (-2)⁵ × (-3)²

(-2)⁵ = -32
(-3)² = 9
(-32) × 9 = -288

Q2. Simplify: (4² × 3³) / (2³ × 9)

= ((2²)² × 3³) / (2³ × 3²)
= (2⁴ × 3³) / (2³ × 3²)
= 2^(4-3) × 3^(3-2) = 2¹ × 3¹ = 2 × 3 = 6

Q3. Write the standard form of 0.00000000405

Move the decimal point 9 places to the right.
4.05 × 10^-9

Q4. Express 4.5 × 10⁶ in usual form.

Move the decimal point 6 places to the right.
4,500,000

Q5. If 2ⁿ = 64, find the value of (2^n-2 + 2^n-1)

2ⁿ = 64 ⇒ 2ⁿ = 2⁶ ⇒ n = 6
Now substitute n = 6:
2^6-2 + 2^6-1 = 2⁴ + 2⁵
= 16 + 32 = 48

Q6. Find the value of x if 3^x × 3⁴ = 3¹⁰

Using base multiplication rule: 3^(x+4) = 3¹⁰
Equating powers: x + 4 = 10
x = 10 - 4 = 6

Q7. Simplify: ( (1/2)² × (2/3)³ × (3/4)² )

= (1/4) × (8/27) × (9/16)
= (1 × 8 × 9) / (4 × 27 × 16)
= 72 / 1728 = 1/24

Q8. Find the reciprocal of (-3/8)³

The reciprocal of (a/b)ⁿ is (b/a)ⁿ.
So, the reciprocal is (-8/3)³.

Q9. Simplify: (6² + 8²)^1/2

6² = 36
8² = 64
(36 + 64)^1/2 = (100)^(1/2)
This implies the square root of 100, which is 10.

Q10. What is the value of 5⁰ × 4⁰ × 8⁰ ?

Any non-zero number to the power of 0 is 1.
1 × 1 × 1 = 1

Q11. Which is greater: 10³ or 3¹⁰?

10³ = 1000
3¹⁰ = (3⁵)² = (243)² = 59049
3¹⁰ is greater.

Q12. Express 343 × 64 as a product of prime factors in exponential form.

343 = 7³
64 = 2⁶
Result: 7³ × 2⁶

Q13. Simplify: ((1/3)^-2 - (1/2)^-3) ÷ (1/4)^-2

= (3² - 2³) ÷ 4²
= (9 - 8) ÷ 16
= 1 ÷ 16 = 1/16

Q14. Find x if (5/3)^-5 × (5/3)¹¹ = (5/3)^8x

(5/3)^{(-5 + 11)} = (5/3)^8x
(5/3)⁶ = (5/3)^8x
6 = 8x ⇒ x = 6/8 = 3/4

Q15. Write 5,340,000,000 in standard form.

Move the decimal 9 places to the left.
5.34 × 10⁹

Q16. Simplify: 2³ + (-2)³ + (-2)⁴

= 8 + (-8) + 16
= 0 + 16 = 16

Q17. Express 1.2 × 10^-5 in decimal form.

Move the decimal 5 places to the left.
0.000012

Q18. Find m if 7³ × 7^m = 7⁹

7^{(3 + m)} = 7⁹
3 + m = 9
m = 9 - 3 = 6

Q19. Evaluate: (2² × 3³ × 5¹) / (4 × 27)

= (4 × 27 × 5) / (4 × 27)
The 4 and 27 cancel out.
Remaining = 5

Q20. If a = 2, b = 3, compile the value of a^b + b^a

= 2³ + 3²
= 8 + 9 = 17

Class 7 Maths - Exponents and Powers Summary

Chapter 13: Exponents and Powers (Concepts & Summary)

1. Introduction to Exponents

Very large numbers are difficult to read, understand, compare, and operate upon. So, we use exponents to write them compactly.
For example, 10,000 = 10 × 10 × 10 × 10 = 10⁴.
In 10⁴, 10 is the base and 4 is the exponent or power. It is read as "10 raised to the power 4".
Numbers in this form are said to be in exponential form.

2. Laws of Exponents

For any non-zero integers a and b and whole numbers m and n:

Multiplying powers with same base: a^m × aⁿ = a^{(m + n)}
Dividing powers with same base: a^m ÷ aⁿ = a^{(m
- n)}, where m > n
Taking power of a power: (a^m)ⁿ = a^{(m ×
n)}
Multiplying powers with same exponents: a^m × b^m = (ab)^m
Dividing powers with same exponents: a^m ÷ b^m = (a/b)^m
Zero Exponent Rule: Any number (except 0) raised to the power 0 is 1. i.e., a⁰ = 1

3. Scientific Notation (Standard Form)

Numbers can be expressed in standard form to make them easier to comprehend.
A number expressed as a decimal between 1.0 and 10.0 multiplied by a power of 10 is said to be in standard form.
Example: 5,985,000,000,000,000,000,000,000 kg (mass of Earth) is written as 5.985 × 10²⁴ kg.
Shift the decimal point to the left to determine the positive power of 10.

4. Negative Bases

(-1)^{even number} = 1
(-1)^{odd number} = -1
Example: (-2)⁴ = 16, whereas (-2)³ = -8.

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