NCERT Solutions for Class 7 Maths Chapter 13 Exponents and Powers Exercise 13.1
Ex 13.1 Class 7 Maths Question 1.
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

Ex 13.1 Class 7 Maths Question 2.
Exress 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 = 6³
(ii) t × t = t²
(iii) b × b × b × b = b⁴
(iv) 5 × 5× 7 × 7 × 7 = 5² × 7³ = 5² · 7³
(v) 2 × 2 × a × a = 2² × a² = 2² · a²
(vi) a × a ×a × c × c × c × c × d = a³ × c⁴ × d = a³ · c⁴ · d

Ex 13.1 Class 7 Maths Question 3.
Express each of the following numbers using exponential notation:
(i) 512
(ii) 343
(iii) 729
(iv) 3125
Solution:

Ex 13.1 Class 7 Maths 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) 100² or 2¹⁰⁰
(v) 2¹⁰ or 10²
Solution:
(i) 4³ or 3⁴
4³ = 4 × 4 × 4 = 64,
3⁴ = 3 × 3 × 3 × 3 = 81
Since 81 > 64
∴ 34 is greater than 43.

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

(iii) 2⁸ or 8²
2⁸ =2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 256
8² = 8 × 8 = 64
Since 256 > 64
∴ 28 is greater than 28.

(iv) 100² or 2¹⁰⁰
100² = 100 × 100 = 10000
2¹⁰⁰ = 2 × 2 × 2 × … 100 times
Here 2 × 2 × 2 ×2 × 2 × 2 × 2 ×2 × 2 × 2 × 2 × 2 × 2 × 2 = 214 = 16384
Since 16384 > 10,000
∴ 2100 is greater than 1002.

(v) 2¹⁰ or 10²
2¹⁰ = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 1024
10² = 10 × 10 = 100
Since 1024 > 100
∴ 210 is greater than 102.

Ex 13.1 Class 7 Maths Question 5.
Express each of the following as the product of powers of their prime
(i) 648
(ii) 405
(iii) 540
(iv) 3600
Solution:

Ex 13.1 Class 7 Maths 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
Solution:
(i) 2 × 10³ = 2 × 10 × 10 × 10 = = 2000
(ii) 7² × 2² = = 7 × 7 × 2 × 2 = 196
(iii) 2³ × 5 = 2 × 2 × 2 × 5 = 40
(iv) 3 × 4⁴ = 3 × 4 × 4 × 4 × 4 = 768
(v) 0 × 10² = 0 × 10 × 10 = = 0
(vi) 5² × 3³ = 5 × 5 × 3 × 3 × 3 = 675
(vii) 2⁴ × 3² = 2 × 2 × 2 × 2 × 3 × 3 = 144
(viii) 3² × 10⁴ = 3 × 3 × 10 × 10 × 10 × 10 = 90000

Ex 13.1 Class 7 Maths Question 7.
Simplify:
(i) (-4)³
(ii) (-3) × (-2)³
(iii) (-3)² × (-5)²
(iv) (-2)³ × (-10)³
Solution:
(i) (-4)² = (-4) × (-4) × (-4) = -64 [∵ (-a)^{odd number} = -a^{odd number}]
(ii) (-3) × (-2)³ = (-3) × (-2) × (-2) × (-2)
= (-3) × (-8) = 24
(iii) (-3)² × (-5)² = [(-3) × (-5)]²
= 15² = 225 [∵ a^m × b^m = (ab)^m)
(iv) (-2)³ × (-10)³ = [(-2) × (-10)]³
= 20² = 8000 [∵ a^m × b^m = (ab)^m]

Ex 13.1 Class 7 Maths Question 8.
Compare the following:
(i) 2.7 × 10¹²; 1.5 × 10⁸
(ii) 4 × 10¹⁴; 3 × 10¹⁴
Solution:
(i) 2.7 × 10¹²; 1.5 × 10⁸
Here, 10¹² > 10⁸
∴ 2.7 × 10¹²> 1.5 × 10⁸
(ii) 4 × 10¹⁴; 3 × 10¹⁷
Here, 10¹⁷ > 10¹⁴
∴ 4 × 10¹⁴ < 3 × 10¹⁷

NCERT Solutions for Class 7 Maths Chapter 13 Exponents and Powers Exercise 13.2
Ex 13.2 Class 7 Maths Question 1.
Using laws of e×ponents, simplify and write the answer in e×ponential 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
Solution:
(i) 3² × 3⁴ × 3⁸ = 3²⁺⁴⁺⁸ = 3¹⁴ [a^m ÷ aⁿ = a^m+n]
(ii) 6¹⁵ ÷ 6¹⁰ = 6^15-10 = 6⁵ [a^m ÷ aⁿ = a^m-n]
(iii) a³ × a² = a³⁺² = a⁵ [a^m × aⁿ = a^m+n]
(iv) 7^x × 7² = 7^x+2 [a^m × aⁿ = a^m+n]
(v) (5²)³ ÷ 5³ = 5^2×3 ÷ 5³ = 5⁶ ÷ 5³ = 5^6-3 = 5³ [(a³)ⁿ = a^mn, a^m ÷ aⁿ = a^m-n]
(vi) 2⁵ × 5⁵ = (2 × 5)⁵ = 10⁵ [a^m × b^m = (ab)^m]
(vii) a⁴ × b⁴ = (ab)⁴ [a^m × b^m = (ab)⁴]
(ix) (2²⁰ ÷ 2¹⁵) × 2³ = 2^20-15 × 2³
=2⁵ × 2³ = 2⁵⁺³ = 2⁸
(x) 8^t ÷ 8² = 8^t-2 [a^m ÷ aⁿ = a^m-n]

Ex 13.2 Class 7 Maths Question 2.
Simplify and express each of the following in exponential form:

Solution:

Ex 13.2 Class 7 Maths Question 3.
Say true or false and justify your answer:
(i) 10 × 10¹¹ = 100¹¹
(ii) 2³ > 5²
(iii) 2³ × 3² = 6⁵
(iv) 3²⁰ = (1000)⁰
Solution:
(i) 10 × 10¹¹ = 10¹⁺¹¹ = 10¹²
RHS = 100¹¹ = (10²)¹¹ = 10²²
10¹² ≠ 10²²
∴ Statement is false.

(ii) 2³ > 5²
LHS = 2³ = 8
RHS = 5²2 = 25
8 < 25
∴ 2³ < 5²
Thus, the statement is false.

(iii) 2³ × 3² = 6⁵
LHS = 2³3 × 3² = 8 × 9 = 72
RHS = 6⁵ = 6 × 6 × 6 × 6 × 6 = 7776
∴ 72 ≠ 7776
∴ The statement is false.

(iv) 3⁰ = (1000)⁰
⇒ 1 = 1 True [∵ a⁰ = 1]

Ex 13.2 Class 7 Maths 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
Solution:
(i) 108 × 192 = 2 × 2 × 3 × 3 × 3 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3
=2⁸ × 3⁴

(iii) 729 × 64 = 3 × 3 × 3 × 3 × 3 × 3 × 2 × 2 × 2 × 2 × 2 × 2
=3⁶ × 2⁶

Ex 13.2 Class 7 Maths Question 5.
Simplify:

Solution:

NCERT Solutions for Class 7 Maths Chapter 13 Exponents and Powers Exercise 13.3
Ex 13.3 Class 7 Maths Question 1.
Write the following numbers in the e×panded forms:
279404, 3006194, 2806196, 120719, 20068
Solution:
(i) 279404 = 2 × 100000 + 7 × 10000 + 9 × 1000 + 4 × 100 + 0 × 10 + 4
= 2 × 10⁵ + 7 × 10⁴ + 9 × 10³² + 4 × 10² + 0 × 10¹ + 4 × 10⁰
(ii) 3006194 = 3 × 1000000 + 0 × 100000 + 0 × 10000 + 6 × 1000 + 1 × 100 + 9 × 10 + 4
= 3 × 10⁶ + 0 × 10⁵ + 0 × 10⁴ + 6 × 10³ + 1 × 10² + 9 × 10¹ + 4 × 10⁰
(iii) 2806196 = 2 × 1000000 + 8 × 100000 + 0 × 10000 + 6 × 1000 + 1 × 100 + 9 × 10 + 6
= 2 × 10⁶ + 8 × 10⁵ + 0 × 10⁴ + 6 × 10³ + 1 × 10² + 9 × 10¹ + 6 × 10⁰
(iv) 120719 = 1 × 100000 + 2 × 10000 + 0 × 1000 + 7 × 100 + 1 × 10 + 9
= 1 × 10⁵ + 2 × 10⁴ + 0 × 10³ + 7 × 10² + 1 × 10¹ + 9 × 10⁰
(v) 20068 = 2 × 10000 + 0 × 1000 + 0 × 100 + 6 × 10 + 8
= 2 × 10⁴ + 0 × 10³ + 0 × 10² + 6 × 10¹ + 8 × 10⁰

Ex 13.3 Class 7 Maths 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
(c) 3 × 10⁴ + 7 × 10² + 5 × 10⁰
(d) 9 × 10⁵ + 2 × 10² + 3 × 10¹
Solution:
(a) 8 × 10⁴ + 6 × 10³ + 0 × 10² + 4 × 101 + 5 × 10⁰
= 8 × 10000 + 6 × 1000 + 0 × 100 + 4 × 10 + 5 × 1
= 80000 + 6000 + 0 + 40 + 5 = 86045

(b) 4 × 10⁵ + 5 × 10³ + 3 × 10² + 2 × 10⁰
= 4 × 100000 + 5 × 1000 + 3 × 100 + 2 × 1
= 400000 + 5000 + 300 + 2 = 405302

(c) 3 × 10⁴ + 7 × 10² + 5 × 10⁰
= 3 × 10000 + 7 × 100 + 5 × 1
= 30000 + 700 + 5 = 30705

(d) 9 × 10⁵ + 2 × 10² + 3 × 10¹
= 9 × 100000 + 2 × 100 + 3 × 10
= 900000 + 200 + 30 = 900230

Ex 13.3 Class 7 Maths 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
Solution:
(i) 5,00,00,000 = 5 × 107⁷
(ii) 70,00,000 = 7 × 10⁶
(iii) 3,18,65,00,000 = 3.1865 × 10⁹
(iv) 3,90,878 = 3.90878 × 10⁵
(v) 39087.8 = 3.90878 × 10⁴
(vi) 3908.7 8 = 3.90878 × 10³

Ex 13.3 Class 7 Maths Question 4.
Express the number appearing in the following statements in standard form:
(a) The distance between Earth and Moon is 384.0. 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 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.
Solution:
(a) 384,000,000 m = 3.84 × 10⁸ m
(b) 300,000,000 m/s = 3 × 10⁸ m/s
(c) 1,27,56,000 m = 1.2756 × 10⁷² 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 old = 1.2 × 10¹⁰ years old
(g) 300,000,000,000,000,000,000 m = 3 × 10²⁰ m
(h) 60, 230, 000, 000, 000, 000, 000, 000 molecules = 6.023 × 10²² molecules
(i) 1,353,000,000 cubic km = 1.353 × 10⁹ cubic km
(j) 1,0,27,000,000 = 1.027 × 10⁹