Gujarat Board GSEB Textbook Solutions Class 7 Maths Chapter 13 Exponents and Powers Ex 13.1 Textbook Questions and Answers.

Gujarat Board Textbook Solutions Class 7 Maths Chapter 13 Exponents and Powers Ex 13.1

Question 1.
Find the value of:
(i) 2⁶
(ii) 9³
(iii) 11²
(iv) 5⁴
Solution:
(i) 2⁶
We have: 2⁶ = 2 x 2 x 2 x 2 x 2 x 2 = 64
Thus, the value of 2⁶ is 64.

(ii) 9³
We have: 9³ = 9 x 9 x 9 = 729
Thus, the value of 9³ is 729.

(iii) 11²
We have: 11² = 11 x 11 = 121
Thus, the value of 11² is 121.

(iv) 5⁴
We have: 5⁴ = 5 x 5 x 5 x 5 = 625
Thus, the value of 5⁴ is 625.

Question 2.
Express the following in exponential form:
(i) 6 x 6 x 6 x 6
(ii) t x t
(iii) b x b x h x b
(iv) 5 x 5 x 7 x 7 x 7
(v) 2 x 2 x a x a
(vi) a x a x a x c x c x c x c x d
Solution:
(i) We have: 6 x 6 x 6 x 6 = 64
(ii) We have: t x t = t²
(iii) We have: b x b x b x b = b⁴
(iv) We have: 5 x 5 x 7 x 7 x 7 = 5² x 7³
(v) We have: 2 x 2 x a x a = 2² x a²
(vi) We have: a x a x a x c x c x c x c x d
= (a x a x a) x (c x c x c x c) x d
= (a³) x (c⁴) x d
= a³ x c⁴ x d

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

We have: 512 = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 = 2⁹
Thus, exponential form of 512 is 2⁹.

(ii) 343

We have: 343 = 7 x 7 x 7
= 7³
∴ The exponential notation of 343 is 7³

(iii) 729

We have: 729 = 3 x 3 x 3 x 3 = 3⁶
∴ The exponential form of 729 is 3⁶.

(iv) 3125

We have: 3125 = 5 x 5 x 5 x 5 x 5 = 5⁵
Thus, the exponential form of 3125 is 5⁵.

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
Since, 4³ = 4 x 4 x 4 = 64
and 3⁴ = 3 x 3 x 3 x 3 = 81
∵ 81 > 64
∴ 3⁵ > 4³

(ii) 5³ or 3⁵
We have: 5³ = 5 x 5 x 5 = 125
3⁵ = 3 x 3 x 3 x 3 x 3 = 243
Since, 243 > 125
i.e. 3⁵ > 5³

(iii) 2⁸ or 8²
We have: 2⁸ = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 = 256
and 8² = 8 x 8 = 64
Since, 256 > 64
i.e. 28 > 82

(iv) 100² or 2¹⁰⁰
We have: 100² = 100 x 100 = 10000
2¹⁰⁰ = (2¹⁰)¹⁰
= (2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2)¹⁰
= (1024)¹⁰ = [1024²]5
= [1024 x 1024]⁵ = (1048576)⁵
Since 1048576 > 10000
∴ (1048576)⁵ > 100²
or (2¹⁰)¹⁰ > 100²
or 2100 > 100²

(v) 2¹⁰ or 10²
We have: 210 = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 = 1024
and 10² = 10 x 10 = 100
Since, 10² > 100
∴ 210 > 10²

Question 5.
Express each of the following as product of powers of their prime factors:
(i) 648
(ii) 405
(iii) 540
(iv) 3,600
Solution:
(i) 648.
We have: 648
= 2 x 324
= 2 x 2 x 162
= 2 x 2 x 2 x 81
= 2 x 2 x 2 x 3 x 27
= 2 x 2 x 2 x 3 x 3 x 9
= 2 x 2 x 2 x 3 x 3 x 3 x 3
= 2³ x 3⁴
Thus, 648 = 2³ x 3⁴

(ii) 405
We have: 405 = 3 x 135
= 3 x 3 x 45
= 3 x 3 x 3 x 15
= 3 x 3 x 3 x 3 x 5
= 3⁴ x 5¹
∴ 405 = 3⁴ x 5

(iii) 540
We have: 540 = 2 x 270
= 2 x 2 x 135
= 2 x 2 x 3 x 45
= 2 x 2 x 3 x 3 x 15
= 2 x 2 x 3 x 3 x 3 x 5
= 22 x 33 x 5
∴ 540 = 2² x 3³ x 5

(iv) 3600
We have: 3600
= 2 x 1800
= 2 x 2 x 900
= 2 x 2 x 2 x 450
= 2 x 2 x 2 x 2 x 225
= 2 x 2 x 2 x 2 x 3 x 75
= 2 x 2 x 2 x 2 x 3 x 3 x 25
= 2 x 2 x 2 x 2 x 3 x 3 x 5 x 5
= 2⁴ x 3² x 5²
Thus, 3600 = 2⁴ x 3² x 5²

Question 6.
Simplify:
(i) 2 x 10³
(ii) 7² x 2²
(iii) 2³ x 5
(iv) 3 x 4⁴
(v) 0 x 10²
(vi) 5² x 3³
(vii) 2⁴ x 3²
(viii) 3² x 10⁴
Solution:
(i) 2 x 10³
∵ 10³ = 10 x 10 x 10 = 1000
∴ 2 x 10³ = 2 x 1000 = 2000

(ii) 7² x 2²:
∵ 7² = 7 x 7 = 49 and 2² = 2 x 2 = 4
∴ 7² x 2² = 49 x 4 = 196

(iii) 2³ x 5:
∵ 2³ = 2 x 2 x 2 = 8
∴ 2³ x 5 = 8 x 5 = 40

(iv) 3 x 4⁴:
∵ 4⁴ = 4 x 4 x 4 x 4 = 256
∴ 3 x 4⁴ = 3 x 256 = 768

(v) 0 x 10²:
∵ 10² = 10 x 10 = 100
∴ 0 x 100 = 0
Thus, 0 x 10² = 0

(vi) 5² x 3³:
∵ 5² = 5 x 5 = 25 and 3³ = 3 x 3 x 3 = 27
∴ 5² x 3³ = 25 x 27 = 675

(vii) 2⁴ x 3²:
∵ 2⁴ = 2 x 2 x 2 x 2 = 16 and 3² = 3 x 3 = 9
∴ 2⁴ x 3² = 16 x 9 = 144

(viii) 3² x 10⁴:
∵ 3² = 3 x 3 = 9
And 10⁴ = 10 x 10 x 10 x 10 = 10000
∴ 3² x 10⁴ = 9 x 10000 = 90000.

Question 7.
Simplify:
(i) (4)³
(ii) (-3) x (-2)³
(iii) (-3)² x (-5)²
(iv) (-2)³ x (-10)³
Solution:
(i) (-4)³
We have: (-4)³ = (-4) x (-4) x (-4)
= (-1)³ x 4 x 4 x 4
= -1 x 64 = -64

(ii) (-3) x (-2)³:
We have: (-3) x (-2)³
= (-3) x [(-2) x (-2) x (-2)] = (-3)[-8]
= (-3) x (-8) = 24

(iii) (-3)² x (-5)²
We have: (-3)² = (-3) x (-3) = 9
and (-5)² = (-5) x (-5) = 25
(-3)² x (-5)² = 9 x 25 = 225

(iv) (-2)³ x (-10)³
We have:
(-2)³ x (-10)³ = [(-2) x (-2) x (-2)] x [(-10) x (-10) x (-10)]
= [-8] X (-1000) = 8000

Question 8.
Compare the following numbers:
(i) 2.7 x 10¹²; 1.5 x 10⁸
(ii) 4 x 10¹⁴; 3 x 10¹⁷
Solution:
(i) 2.7 x 10¹²; 1.5 x 10⁸
Since 2.7 x 10¹² = (frac { 27 }{ 10 }) x 10¹² = 27 x 10¹¹
Also, 1.5 x 10⁸ = (frac { 15 }{ 10 }) x 10⁸ = 15 x 10⁷
Now, 27 x 10¹¹ = 27 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 = 27,00,00,00,00,000
and 15 x 10⁷ = 15 x 10 x 10 x 10 x 10 x 10 x 10 x 10 = 150000000
Since, 2700000000000 > 150000000
∴ 2.7 x 1012 > 1.5 x 10⁸

(ii) 4 x 10¹⁴; 3 x 10¹⁷:
Since 4 x 10¹⁴ contains 15 digits, and 3 x 10¹⁷ contains 18 digits.
Obviously, a number containing more digits is greater.
∴ 3 x 10¹⁷ will be greater than 4 x 10¹⁴ or 3 x 10¹⁷ > 4 x 10¹⁴