# 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:

 Number Exponential form Base Exponent (i) 243 = 3 × 3 × 3 × 3 × 3 35 3 5 (ii) 625 = 5 × 5 × 5 × 5 54 5 4 (iii) 343 = 7 × 7 × 7 . 73 7 3 (iv) 1331 = 11 × 11 × 11 113 11 3 (v) 64 = 8 × 8 82 8 2

Note: 1. xxxxxxx = x4is read as ‘x raised to the power 4’ or ‘4th 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 = 25 + 3 = 28

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

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

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

(v) 53 × 57 × 512
We have: 53 × 57 × 512 = 53 + 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 am an = am-n, therefore;
(i) 29 ÷ 23
We have: 29 ÷ 23 = 29-3 = 26

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

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

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

(v) 713 ÷ 710
We have: 713 ÷ 710 = 713-10 = 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 (am)n = amxn = amn, therefore;
(i) (62)4
We have: (62)4 = 62 × 4 = 68

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

(iii) (750)2
We have: (750 )2 = 750 × 2 = 7100

(iv) ( 53)7
We have: (53)7 = 53 × 7 = 521

NCERT In-text Question Page No. 256

Question 1.
Put into another form using am × bm= (ab)m
(i) 43 × 23
(ii) 25 × b5
(iii) a2 × t2
(iv) 56 × (-2)6
(v) (-2)4 × (-3)4
Answer:
Since am ÷ an = am-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 am ÷ bm
(i) 45 ÷ 35
(ii) (-2)5 ÷ b5
(iii) (-2)3 ÷ b3
(iv) p4 ÷ q4
(v) 56 ÷ (-2)6
Answer:
(i) 45 ÷ 35
We have: 45 ÷ 35 = $$\left(\frac{4}{3}\right)^{5}$$

(ii) (-2)5 ÷ b5
We have: (-2)5 ÷ b5 = $$\left(\frac{2}{b}\right)^{5}$$

(iii) (-2)3 ÷ b3
We have: (-2)3 ÷ b3 = $$\left(\frac{-2}{b}\right)^{3}$$

(iv) p4 ÷ q4
We have: p4 ÷ q4 = $$\left(\frac{p}{q}\right)^{4}$$

(v) 56 ÷ (-2)6
We have: 56 ÷ (-2)6 = $$\left(\frac{5}{-2}\right)^{6}$$ = $$\left(-\frac{5}{2}\right)^{6}$$

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

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