Chapter – Chapter 13 Exponents and Powers | Class 7th |NCERT Maths Solutions | Edugrown

Question 1.
Find the value of:

(i) 2⁶
Solution:
2⁶ = 2 × 2 × 2 × 2 × 2 × 2 = 64

(ii) 9³
Solution:
9² = 9 × 9 × 9 = 729

(iii) 11²
Solution:
11² = 11 × 11 = 121

(iv) 5⁴
Solution:
5⁴ = 5 × 5 × 5 × 5 = 625

Question 2.
Express the following in exponential form :

(i) 6 × 6 × 6 × 6
Solution:
6 × 6 × 6 × 6 = 6⁴

(ii) t x t
Solution:
t × t = t²

(iii) b × b × b × b
Solution:
b × b × b × b = b⁴

(iv) 5 × 5 × 7 × 7 × 7
Solution:
5 × 5 × 7 × 7 × 7 = 5² × 7³

(v) 2 × 2 × a × a
Solution:
2 × 2 × a × a = 2² × a²

(vi) a × a × a × c × c × c × c × d
Solution:
a × a × a × c × c × c × c  d = a³ × c⁴ × d

Question 3.
Express each of the following numbers using the exponential notation:

(i) 512
Solution:
We have

(ii) 343
Solution:
We have

(iii) 729
Solution:
We have

(vi) 3125
Solution:
We have

Question 4.
Identify the greater number, wherever possible, in each of the following?

(i) 4³ or 3⁴
Solution:

(ii) 5³ or 3⁵
Solution:

(iii) 2⁸ or 8²
Solution:

(iv) 100² or 2¹⁰⁰
Solution:

(v) 2¹⁰ or 10²
Solution:

Question 5.
Express each of the following as a product of powers of their prime factors:

(i) 648
Solution:
To determine the prime factorization of 648, we use the division method, as shown below:

(ii) 405
Solution:
We use the division method as shown under:

(iii) 540
Solution:
We use the division method as shown under:

(iv) 3600
Solution:
We use the division method as shown under:

Question 6.
Simplify:

(i) 2 × 10³
Solution:
2 × 10³ = 2 × 1000 = 2000

(ii) 7² x 2²
Solution:
7² × 2² = (7 × 2)² = (14)³ = 14 × 14 = 196

(iii) 2³ × 5
Solution:
2³ × 5 = 2 × 2 × 2 × 5
= 8 × 5 = 40

(iv) 3 × 4⁴
Solution:
3 × 4⁴ = 3 × 4 × 4 × 4 × 4
= 3 × 256 = 768

(v) 0 × 10²
Solution:
0 × 10² = 0

(vi) 5² × 3³
Solution:
5² × 3³ = 5 × 5 × 3 × 3 × 3 = 25 × 27 = 675

(vii) 24 × 32
Solution:
2⁴ × 3² = 2 × 2 × 2 × 2 × 3 × 3 = 16 × 9 = 144

(viii) 3² × 10⁴
Solution:
3² × 10⁴ = 3 × 3 × 10 × 10 × 10 × 10 = 9 × 10000 = 90000

Question 7.
Simplify:

(i) (-4)³
Solution:
(- 4)³ = (- 4) × (- 4) × (- 4) = – 64

(ii) (-3) × (-2)³
Solution:
(-3) × (-2)³ = (-3) × (- 2) × (- 2) × (- 2) = (- 3) × (- 8) = 24

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

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

Question 8.
Compare the following numbers:

(i) 2.7 × 10¹² ; 1.5 × 10⁸
Solution:
We have,
2.7 × 10¹² = 2.7 × 10 × 10¹¹
= 27 × 10¹¹, contains 13 digits and 1.5 × 10⁸ = 1.5 × 10 × 10⁷
= 15 × 10⁷ contains 9 digits
Clearly, 2.7 × 10¹² > 1.5 × 10⁸

(ii) 4 × 10¹⁴; 3 × 10¹⁷
Solution:
We have, 4 × 10¹⁴, contains 15 digits and, 3 × 10¹⁷, contains 18 digits 4 × 10¹⁴ < 3 × 10¹⁷

Question 1.
Using the law of exponents, simplify and write the answer in exponential form:

(i) 3² × 3⁴ × 3⁸
Solution:
3² × 3⁴ × 3⁸ = 3² + 4 + 8

(ii) 6¹⁵ ÷ 6¹⁰
Solution:
6¹⁵ ÷ 6¹⁰ = 6^15 – 10 = 6⁵

(iii) a³ × a²
Solution:
a³ × a² = a^3 + 2 = a⁵

(iv) 7^x x 7²
Solution:
7^x × 7² = 7^x + 2

(v) (5²)³ ÷ 5³
Solution:
(5²)³ ÷ 5³ = 5^2 × 3 ÷ 5³
= 5⁶ ÷ 5³ = 5^6 – 3 = 5³

(vi) 2⁵ × 5⁵
Solution:
2⁵ × 5⁵ = (2 × 5)⁵ = (10)⁵

(vii) a⁴ × b⁴
Solution:
a⁴ × b⁴ = (ab)⁴

(viii) (3⁴)³
Solution:
(3⁴)³ = 3^4 × 3 = 3¹²

(ix) (2²⁰ ÷ 2¹⁵) × 2³
Solution:
(2²⁰ ÷ 2¹⁵) × 2³ = (2^10 – 15) × 2³
= 2⁵ × 2³ = 2^5 + 3 = 2⁸

(x) 8^t ÷ 8²
Solution:
8^t ÷ 8² = 8^t – 2

Question 2.
Simplify and express each of the following in exponential form:
(i) 23×34×43×32
Solution:
23×34×43×32 = 23×34×223×25
[∵ 4 = 2 × 2 = 2² , 32 = 2 × 2 × 2 × 2 × 2 = 2⁵]
= 23+2×3431×25 = 25×3431×25
= 2^5 – 5 x 3^4 – 1 = 2⁰ × 3³ = 1 × 3³ = 3³

(ii) [(5²)³ × 5⁴] ÷ 5⁷
Solution:
[(5²)³ × 5⁴] ÷ 5⁷ = (5 ^{2 × 3} × 5⁴) ÷ 5⁷
=(5⁶ × 5⁴) ÷ 5⁷ = 5^6 + 4 ÷ 5⁷
= 5¹⁰ ÷ 5⁷ = 5^10 – 7 = 5³

(iii) 25⁴ ÷ 5³
Solution:
25⁴ ÷ 5³ = (5²)⁴ ÷ 5³
= 5^2 × 4 ÷ 5³
= 5⁸ ÷ 5³ = 5^8 – 3
= 5⁵

(iv) 3×72×11821×113
Solution:
3×72×11821×113 = 3×72×1183×7×113 = 3^1 – 1 × 7^2 – 1 × 11^8 – 3
= 3⁰ × 7¹ × 11⁵
= 1 × 7 × 11⁵
= 7 × 11⁵
= 7 × 11⁵

(v) 3734×33
Solution:
3734×33 = 3734+3 = 3737
3^7 – 7 = 3⁰ = 1

(vi) 2⁰ + 3⁰ + 4⁰
Solution:
2⁰ + 3⁰ + 4⁰ = 1 + 1 + 1 = 3

(vii) 2⁰ x 3⁰ x 4⁰
Solution:
2⁰ × 3⁰ × 4⁰ = 1 × 1 × 1 = 3

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

(ix) 28×a543×a3
Solution:
28×a543×a3 = 28×a5(22)3×a3 = 28×a526×a3
= 2^8 – 6 × a^5 – 3 = 2² × a² = (2a)²

(x) (a5a3) x a⁸
Solution:
(a5a3) × a⁸ = (a^5 – 3) × a⁸
= a² × a⁸
= a^2 + 8
= a¹⁰

(xi) 45×a8b345×a5b2
Solution:
45×a8b345×a5b2 = 4^5 – 5 × a^8 – 5 × b^3 – 2
= 4⁰ × a³ × b¹
= 1 × a × b
= a³b

(xii) (2³ × 2)²
Solution:
(2³ × 2)² = (2³ + 1)² = (2⁴)² = 2⁸

Question 3.
Say true or false and justify your answer:

(i) 10 × 10¹¹ = 100¹¹
Solution:

(ii) 2³ > 5²
Solution:

(iii) 2³ × 3² = 6⁵
Solution:

(iv) 3⁰ = (1000)⁰
Solution:

Question 4.
Express each of the following as a product of prime factors only in exponential form:

(i) 108 × 192
Solution:
We have

∴ 108 × 192 = (2 × 2 × 3 × 3 × 3) x (2 × 2 × 2 × 2 × 2 × 2 × 3)
= 2² × 3³ × 2⁶ × 3¹
= 2^2 + 6 × 3^3 + 1
= 2⁸ × 3⁴

(ii) 270
Solution:
We have

∴ 270 = 2 × 3 × 3 ×  3 × 5 = 2 × 3³ × 5

(iii) 729 x 64
Solution:
We have

∴ 729 = 3 × 3 × 3 × 3 × 3 × 3 = 3²
64 = 2 × 2 × 2 × 2 × 2 × 2 = 2⁶
∴ 729 × 64 = 3⁶ × 2⁶

(iv) 768
Solution:
We have

∴ 768 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3 = 2⁸ × 3

Question 5.
Simplify:

(i) (25)2×7383×7
Solution:
(25)2×7383×7 = 210×73(23)3×7 = 210×7329×7
= 2^10 – 9  × 7^3 – 1
= 2¹ × 7²
= 2 × 49
= 98

(ii) 25×52×t8103×t4
Solution:
25×52×t8103×t4 = 52×52×t8(2×5)3×t4 = 54×t823×53×t4 = 54−3×t8−423 = 5×t423 = 5t48

(iii) 35×105×2557×65
Solution:
35×105×2557×65 = 35×(2×5)5×5257×(2×3)5 = 35×25×55×5257×25×35
= 3^5 – 5 × 2^5 – 5 × 5^{5 + 2 – 7}
= 30 × 20 × 50
= 1 × 1 × 1
= 1

Question 1.
Write the following numbers in the expanded forms :
Solution:

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

(ii) 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 × 10°

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

(iv) 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

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

Question 2.
Find the number from each of the following expanded forms:

(a) 8 × 10⁴ + 6 × 10³ + 0 × 10² + 4 × 10¹ + 5 × 10⁰
Solution:
8 × 10⁴ + 6 × 10³ + 0 × 10² + 4 × 10¹ + 5 × 10⁰
= 8 × 10000 + 6 × 1000 + 0 × 100 + 4 × 10 + 5 × 1
= 86045

(b) 4 × 10⁵ + 5 × 10³ + 3 × 10² + 2 × 10⁰
Solution:
4 × 10⁵ + 5 × 10³ + 3 × 10² + 2 × 10⁰
= 4 × 100000 + 5 × 1000 + 3 × 100 + 2
= 405302

(c) 3 × 10⁴ + 7 × 10² + 5 × 10⁰
Solution:
3 × 10⁴ + 7 × 10² + 5 × 1
= 3 × 10000 + 7 × 100 + 5 × 1
= 30705

(d) 9 × 10⁵ + 2 × 10² + 3 × 10¹
Solution:
9 × 10⁵ + 2 × 10² + 3 × 10¹
= 9 × 100000 + 2 × 100 + 30
= 900230

Question 3.
Express the following numbers in standard form:

(i) 5,00,00,000
Solution:
5,00,00,000 = 5 × 10000000 = 5 × 10⁷

(ii) 70,00,000
Solution:
70,00,000 = 7 × 1000000 = 7 × 10⁶

(iii) 3,18,65,00,000
Solution:
3,18,65,00,000 = 3.1865 × 1000000000 = 3.1865 × 10⁹

(iv) 3,90,878
Solution:
3,90,878 = 3.90878 × 100000 = 3.90878 × 10⁵

(v) 39087.8
Solution:
39087.8 = 3.90878 × 10000 = 3.90878 × 10⁴

(vi) 3908.78
Solution:
3908.78 = 3.90878 × 1000 = 3.90878 × 10³

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.
Solution:
The mean distance between Earth and Moon = 384,000,000 m = 3.84 × 100000000 m = 3.84 × 10⁸ m

(b) Speed of light in vaccum is 300,000,000 m/s.
Solution:
Speed of light in vaccum
= 300,000,000 m/s
= 3.0 × 100000000 m/s
= 3.0 × 10⁸ m/s

(c) Diameter of the Earth is 1,27,56,000 m.
Solution:
Diameter of the Earth 1,27,56,000 m
= 1.2756 × 10000000 m = 1.2756 × 10⁷

(d) Diameter of the Sun is 1,400,000,000 m.
Solution:
Diameter of the Sun
= 1,400,000,000 m = 1.4 × 1000000000 m = 1.4 × 10⁹ m

(e) In a galaxy there are on4m average 100,000,000,000 stars.
Solution:
In a galaxy there are on an average
= 100,000,000,000 stars = 1 × 100000000000 stars = 1 × 10¹¹ stars

(f) The universe is estimated to be 12,000,000,000 years old.
Solution:
The universe is estimated to be 12,000,000,000 years old
= 12 × 1,000,000,000 = 1.2 × 10¹⁰ year

(g) The distance of the Sun from the center of the Milky Way Galaxy is estimated to be 300,000,000,000,000,000,000 m.
Solution:
The distance of Sun from the centre of the Milky Way Galaxy is estimated to be
= 300,000,000,000,000,000,000
= 3 × 100000000000000000000
= 3 × 10²⁰ m

(h) 60,230,000,000,000,000,000,000 molecules are contained in a drop of water weighing 1.8 gm.
Solution:
Number of molecules contained in a drop of water weighing 1.8 gm = 60,230,000,000,000,000,000,000
= 6.023 × 10000000000000000000000 = 6.023 × 10²⁰

(i) The Earth has 1,353,000,000 cubic km of seawater.
Solution:
The Earth has 1,353,000,000 cubic km of seawater i.e.,
1.353 × 1,000,000,000 = 1.353 × 10⁹ cubic km.

(j) The population of India was about 1,027,000,000 in March, 2001.
Solution:
The population of India was about in March 2001 = 1,027,000,000 = 1.027 × 1000000000 = 1.027 × 10⁹