NCERT Solutions for Class 7 Maths Chapter 13 Exponents and Powers InText Questions

NCERT Solutions for Class 7 Maths Chapter 13 Exponents and Powers InText Questions

These NCERT Solutions for Class 7 Maths Chapter 13 Exponents and Powers InText Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 7 Maths Chapter 13 Exponents and Powers InText Questions

NCERT In-text Question Page No. 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.
Answer:

Note: 1. xxxxxxx = x4is 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 reas as ‘p raised to the power 6 into q cubed’.

NCERT In-text Question Page No. 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
Answer:
(i) 729
We have: 729 = 3 × 3 × 3 × 3 × 3 × 3 = 3⁶
Thus, 729 = 3⁶

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

(iii) 343
We have: 343 = 7 × 7 × 7 × 7= 7³
Thus, 343 = 7³

NCERT In-text Question Page No. 254

Question 1.
Simplify and write in exponential form:
(i) 2⁵ × 2³
(ii) p³ × p²
(iii) 4³ × 4²
(iv) a³ × a² × a⁷
(v) 5³ × 5⁷ × 5¹²
(vi) (-4)¹⁰⁰ × (-4)²⁰
Answer:
(i) 2⁵ × 2³
We have: 2⁵ × 2³ = 2^{5 + 3} = 28

(ii) p³ × p²
We have: p³ × p² = p^{3 + 2} = p⁵

(iii) 4³ × 4²
We have: 4³ × 4^{2<.sup> = 43+2 = p5}

(iv) a³ × a² × a⁷
We have: a³ × a² × a⁷ = a^{3 + 2 + 7} = a¹²

(v) 5³ × 5⁷ × 5¹²
We have: 5³ × 5⁷ × 5¹² = 5^{3 + 7 + 12} = 5²²

(vi) (-4)¹⁰⁰ × (-4)²⁰
We have: (-4)¹⁰⁰ × (-4)²⁰ = (-4)¹⁰⁰⁺² = (4)¹²⁰

Note: The above rule is possible only for same bases. It is not true for different bases. Thus, 2³ x 3² will no obev this rule.

NCERT In-text Question Page No. 255

Question 1.
Simplify and write in exponential form: (eg., 11⁶ ÷ 11² = 11⁴)
(i) 2⁹ ÷ 2³
(ii) 10⁸ ÷ 10⁴
(iii) 9¹¹ ÷ 9⁷
(iv) 20¹⁵ ÷ 20¹³
(v) 7¹³ ÷ 7¹⁰
Answer:
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

Question 15.
Simplify and write the answer in exponential form:
(i) (6²)⁴
(ii) (2²)¹⁰⁰
(iii) (7⁵⁰)²
(iv) ( 5³)⁷
Answer:
Since (a^m)ⁿ = a^mxn = a^mn, therefore;
(i) (6²)⁴
We have: (6²)⁴ = 6^{2 × 4} = 6⁸

(ii) (2²)¹⁰⁰
We have: ( 2²)¹⁰⁰ = 2^{2 × 100} = 2²⁰⁰

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

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

NCERT In-text Question Page No. 256

Question 1.
Put into another form using a^m × b^m= (ab)^m
(i) 4³ × 2³
(ii) 2⁵ × b⁵
(iii) a² × t²
(iv) 5⁶ × (-2)⁶
(v) (-2)⁴ × (-3)⁴
Answer:
Since a^m ÷ aⁿ = a^m-n, therefore;
(i) 4³ × 2³
We have: 4³ × 2³ = (4 × 2)³ = 8³

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

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

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

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

NCERT In-text Question Page No. 257

Question 2.
Put into another form using a^m ÷ b^m
(i) 4⁵ ÷ 3⁵
(ii) (-2)⁵ ÷ b⁵
(iii) (-2)³ ÷ b³
(iv) p⁴ ÷ q⁴
(v) 5⁶ ÷ (-2)⁶
Answer:

NCERT In-text Question Page No. 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
Answer:
(i) 172:
We have:
172 =(1 × 100) + ( 7 × 10) + (2 × 1)
= 1 × 10² + 7 × 10¹ + 2 × 1)
(∵10⁰ = 1)

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

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

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