GSEB Solutions Class 7 Maths Chapter 13 Exponents and Powers InText Questions

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

Gujarat Board Textbook Solutions Class 7 Maths Chapter 13 Exponents and Powers InText Questions

Try These (Page 250)

Question 1.
Find five more such examples, where a number is expressed in exponential form. Also identify the base and the exponent in each case.
Solution:

Note:
1. x × x × x × x = x⁴ is read as ‘x raised to the power 4’ or ‘4^th power of x’.
2. x²y⁵ is read as ‘x squared into y raised to power 5’.
3. p⁶q³ is read as ‘p raised to the power 6 into q cubed’.

Try These (Page 251)

Question 1.
Express:
(i) 729 as a power of 3
(ii) 128 as a power of 2
(iii) 343 as a power of 7
Solution:
(i) 729
We have:
729 = 3 x 3 x 3 x 3 x 3 x 3 = 36
Thus, 729 = 3⁶

(ii) 128
We have:
128 = 2 x 2 x 2 x 2 x 2 x 2 x 2 = 2⁷
Thus, 128 = 2⁷

(iii) 343
We have:
343 = 7 x 7 x 7 = 7³
Thus 343 = 7³

Try These (Page 254)

Question 1.
Simplify and write in exponential form:
(i) 2⁵ x 2³
(ii) p³ x p²
(iii) 4³ x 4²
(iv) a³ x a² x a⁷
(v) 5³ x 5⁷ x 5¹²
(vi) (-4)¹⁰⁰ x (4)²⁰
Solution:
Since a^m x aⁿ = a^m+n, therefore:
(i) 2⁵ x 2³:
We have: 2⁵ x 2³ = 2⁵⁺³ = 2⁸

(ii) p³ x p²:
We have: p³ x p² = p³⁺² = p⁵

(iii) 4³ x 4²:
We have: 4³ x 4² = 4³⁺² = 4⁵

(iv) a³ x a² x a⁷
We have: a³ x a² x a⁷ = a³⁺²⁺⁷ = a¹²

(v) 5³ x 5⁷ x 5¹²
We have: 5³ x 5⁷ x 5¹² = 5³⁺⁷⁺¹² = 5²²

(vi) (-4)¹⁰⁰ x (-4)²⁰
We have: (-4)¹⁰⁰ x (-4)²⁰ = (-4)¹⁰⁰⁺²⁰ = (4)¹²⁰

Note: The above rule is possible only for same bases. It is not true for different bases. Thus, 2³ x 3² will not obey this rule.

Try These (Page 255)

Question 1.
Simplify and write in exponential form: (example 11⁶ – 11² = 11⁴ )
(i) 2⁹÷ 2³
(ii) 10⁸ ÷ 10⁴
(iii) 9¹¹ ÷ 9⁷
(iv) 20¹⁵ ÷ 20¹³
(v) 7¹³ ÷ 7¹⁰
Solution:
Since a^m ÷ aⁿ = a^m-n, therefore;
(i) 2⁹÷ 2³:
We have: 2⁹ ÷ 2³ = 2^9-3 = 2⁶

(ii) 10⁸ ÷ 10⁴:
We have: 10⁸ ÷ 10⁴ = 10^8-4 = 10⁴

(iii) 9¹¹ ÷ 9⁷
We have: 9¹¹ ÷ 9⁷ = 9^11-7 = 9⁴

(iv) 20¹⁵ ÷ 20¹³:
We have: 20¹⁵ ÷ 20¹³ = 20^15-13 = 20²

(v) 7¹³ ÷ 7¹⁰:
We have: 7¹³ ÷ 7¹⁰ = 7^13-10 = 7³

Try These (Page 255)

Question 1.
Simplify and write the answer in exponential form:
(i) (6²)⁴
(ii) (2²)¹⁰⁰
(iii) (7⁵⁰)²
(iv) (5³)⁷
Solution:
Since (a^m)ⁿ = a^m×n = amn, therefore;
(i) (6²)⁴:
We have: (6²)⁴ = 6^2×4 = 6⁸

(ii) (2²)¹⁰⁰:
We have: (2²)¹⁰⁰ = 2² x 100 = 2²⁰⁰

(iii) 7⁵⁰)²:
We have: (7⁵⁰)² = 7^50×2 = 7¹⁰⁰

(iv) (5³)7:
We have: (5³)7 = 5^3×7 = 5²¹

Try These (Page 256)

Question 1.
Put into another form using a^m x b^m = (ab)^m:
(i) 4³ x 2³
(ii) 2⁵ x b⁵
(iii) a² x t²
(iv) 5⁶ x (-2)⁶
(v) (-2)⁴ x (-3)⁴
Solution:
(i) 4³ x 2³:
We have: 4³ x 2³ = (4 x 2)³ = 8³

(ii) 2⁵ x b⁵:
We have: 2⁵ x b⁵ = (2 x b)⁵ = (2b)⁵

(iii) a² x t²:
We have: a² x t² = (a x t)² = (at)²

(iv) 5⁶ x (-2)⁶:
We have: 5⁶ x (-2)⁶ = [5 x (-2)]⁶ = (-10)⁶

(v) (-2)⁴ x (-3)⁴:
We have: (-2)⁴ x (-3)⁴ = [(-2) x (-3)]⁴ = [6]⁴

Try These (Page 257)

Question 1.
Put into another form using a^m ÷ bⁿ = ( \(\frac { a }{ b }\) )^m.
(i) 4⁵ ÷ 3⁵
(ii) 2⁵ ÷ b⁵
(iii) (-2)³ ÷ b³
(iv) p⁴ ÷ q⁴
(v) 5⁶ ÷ (-2)⁶
Solution:

Try These (Page 261)

Question 1.
Expand by expressing powers of 10 in the exponential form:
(i) 172
(ii) 5,643
(iii) 56,439
(iv) 1,76,428
Solution:
(i) 172:
We have:
172 = (1 x 100) + (7 x 10) + (2 x 1)
= 1 x 10² + 7 x 10¹ + 2 x 10⁰ (∵ 10° = 1)

(ii) 5,643:
We have:
5,643 = 5 x 1000 + 6 x 100 + 4 x 10 + 3 x 1
= 5 x 10³ + 6 x 10² + 4 x 10¹ + 3 x 10⁰

(iii) 56,439:
We have:
56,439 = 5 x 10000 + 6 x 1000 + 4 x 100 + 3 x 10 + 9 x 1
= 5 x 10⁴ + 6 x 10³ + 4 x 10² + 3 x 10¹ + 9 x 10⁰

(iv) 1,76,428:
We have:
1,76,428 = 1,00,000 + 7 x 10,000 + 6 x 1000 + 4 x 100 + 2 x 10 + 8 x 1
= 1 x 10⁵ + 7 x 10⁴ + 6 x 10³ + 4 x 10² + 2 x 10¹ + 8 x 10⁰