RD Sharma Solutions Class 7 Chapter 6 Exponents

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Exercise 6.1

Question 1: Find the values of each of the following:

(i) 132

(ii) 73

(iii) 34

Solution 1:

(i) 132

= 13 × 13 =169

(ii) 73

= 7 × 7 × 7 = 343

(iii) 34

= 3 × 3 × 3 × 3 = 81

Question 2: Find the value of each of the following:

(i) (-7)2

(ii) (-3)4

(iii) (-5)5

Solution 2:

(i) (-7)2

= (-7) × (-7)

= 49

(ii) (-3)4

= (-3) × (-3) × (-3) × (-3)

= 81

(iii) (-5)5

= (-5) × (-5) × (-5) × (-5) × (-5)

= -3125

Question 3: Simplify:

(i) 3 × 102

(ii) 22 × 53

(iii) 33 × 52

Solution 3:

(i) 3 × 102

= 3 × 10 × 10

= 3 × 100

= 300

(ii) 22 × 53

= 2 × 2 × 5 × 5 × 5

= 4 × 125

= 500

(iii) 33 × 52

= 3 × 3 × 3 × 5 × 5

= 27 × 25

= 675

Question 4: Simply:

(i)  32 × 104

(ii)  24 × 32

(iii) 52 × 34

Solution 4:

(i)  3× 104

= 3 × 3 × 10 × 10 × 10 × 10

= 9 × 10000

= 90000

(ii) 24 × 32

= 2 × 2 × 2 × 2 × 3 × 3

= 16 × 9

= 144

(iii) 52 × 34

= 5 × 5 × 3 × 3 × 3 × 3

= 25 × 81

= 2025

Question 5: Simplify:

(i) (-2) × (-3)3

(ii) (-3)2 × (-5)3

(iii) (-2)5 × (-10)2

Solution 5:

(i) (-2) × (-3)3

= (-2) × (-3) × (-3) × (-3)

= (-2) × (-27)

= 54

(ii) (-3)2 × (-5)3

= (-3) × (-3) × (-5) × (-5) × (-5)

= 9 × (-125)

= -1125

(iii) (-2)5 × (-10)

= (-2) × (-2) × (-2) × (-2) × (-2) × (-10) × (-10)

= (-32) × 100

= -3200

Question 6: Simplify:

(i) (3/4)2

(ii) ((-2)/3)4

(iii) ((-4)/5)5

Solution 6:

(i) (3/4)2

= (3/4) × (3/4)

= (9/16)

(ii) ((-2)/3)4

= ((-2)/3) × ((-2)/3) × ((-2)/3) × ((-2)/3)

= (16/81)

(iii) ((-4)/5)5

= ((-4)/5) × ((-4)/5) × ((-4)/5) × ((-4)/5) × ((-4)/5)

= ((-1024)/3125)

Question 7: Identify the greater number in each of the following:

(i) 25 or 52

(ii) 34 or 43

(iii) 35 or 53

Solution 7:

(i) 25 or 52

25 = 2 × 2 × 2 × 2 × 2         = 32

52 = 5 × 5                              = 25

So, 25 > 52

(ii) 34 or 43

34 = 3 × 3 × 3 × 3                = 81

4= 4 × 4 × 4                       = 64

So, 34 > 43

(iii) 3or 53

35 = 3 × 3 × 3 × 3 × 3         = 243

53 = 5 × 5 × 5                       = 125

So, 35 > 53

Question 8: Express each of the following in exponential form:
(i) (-5) × (-5) × (-5)
(ii) ((-5)/7) × ((-5)/7) × ((-5)/7) × ((-5)/7)
(iii) (4/3) × (4/3) × (4/3) × (4/3) × (4/3)

Solution 8:

(i) (-5) × (-5) × (-5)
= (-5) × (-5) × (-5)
= (-5)3

(ii) ((-5)/7) × ((-5)/7) × ((-5)/7) × ((-5)/7)
= ((-5)/7) × ((-5)/7) × ((-5)/7) × ((-5)/7)
= ((-5)/7)4

(iii) (4/3) × (4/3) × (4/3) × (4/3) × (4/3)
= (4/3) × (4/3) × (4/3) × (4/3) × (4/3)
= (4/3)5

Question 9: Express each of the following in exponential form:
(i) x × x × x × x × a × a × b × b × b
(ii) (-2) × (-2) × (-2) × (-2) × a × a × a
(iii) (-2/3) × (-2/3) × x × x × x

Solution 9:

(i) x × x × x × x × a × a × b × b × b
= x × x × x × x × a × a × b × b × b
= x4a2b3

(ii) (-2) × (-2) × (-2) × (-2) × a × a × a
= (-2) × (-2) × (-2) × (-2) × a × a × a
= (-2)4a3

(iii) (-2/3) × (-2/3) × x × x × x
= (-2/3) × (-2/3) × x × x × x
= (-2/3)2 x3

Question 10: Express each of the following numbers in exponential form:
(i) 512
(ii) 625
(iii) 729

Solution 10:

(i) 512

By the prime factorization of 512

= 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2

= 29

(ii) 625

By the prime factorization of 625

= 5 × 5 × 5 × 5

= 54

(iii) 729

By the prime factorization of 729

= 3 × 3 × 3 × 3 × 3 × 3

= 36

Question 11: Express each of the following numbers as a product of powers of their prime factors:
(i) 36
(ii) 675
(iii) 392

Solution 11:

(i) 36

By the prime factorization of 36

= 2 × 2 × 3 × 3

= 22 × 32

(ii) 675

By the prime factorization of 675

= 3 × 3 × 3 × 5 × 5

= 33 × 52

(iii) 392

By the prime factorization of 392

= 2 × 2 × 2 × 7 × 7

= 23 × 72

Question 12: Express each of the following numbers as a product of powers of their prime factors:
(i) 450
(ii) 2800
(iii) 24000

Solution 12:

(i) 450

By the prime factorization of 450

= 2 × 3 × 3 × 5 × 5

= 2 × 32 × 52

(ii) 2800

By the prime factorization of 2800

= 2 × 2 × 2 × 2 × 5 × 5 × 7

= 24 × 52 × 7

(iii) 24000

By the prime factorization of 24000

= 2 × 2 × 2 × 2 × 2 × 2 × 3 × 5 × 5 × 5

= 26 × 3 × 53

Question 13: Express each of the following as a rational number of the form (p/q):
(i)  (3/7)2
(ii) (7/9)3
(iii) ((-2)/3)4

Solution 13:

(i) (3/7)2
= (3/7) × (3/7)
= (9/49)

(ii) (7/9)3
= (7/9) × (7/9) × (7/9)
= (343/729)

(iii) ((-2)/3)4
= ((-2)/3) × ((-2)/3) ×((-2)/3) ×((-2)/3)
= (16/81)

Question 14: Express each of the following rational numbers in power notation:
(i) (49/64)
(ii) ((-64)/125)
(iii) ((-1)/216)

Solution 14:

(i) (49/64)
As we all knows 72 (7 ×7) = 49 and 82 (8 ×8) = 64
So, (49/64) = (7/8)2

(ii)((-64)/125)
As we all knows 43 (4 ×4) = 64 and 53 (5 × 5) = 125
So, ((-64)/125) = ((-4)/5)3

(iii) ((-1)/216)
As we all knows 13 (1 ×1)= = 1 and 63 (6 ×6)= = 216
So,  ((-1)/216) =- (1/6)3

Question 15: Find the value of the following:
(i) ((-1)/2)2 × 23 × (3/4)2
(ii) ((-3)/5)4 × (4/9)4 × ((-15)/18)2

Solution 15:

(i) ((-1)/2)2 × 23 × (3/4)2
= 1/4 × 8 ×9/16
= 9/8

(ii)((-3)/5)4 × (4/9)4 × ((-15)/18)2
= (81/625) × (256/6561) × (225/324)
= (64/18225)

Question 16: If a = 2 and b= 3, the find the values of each of the following:
(i) (a + b)a
(ii) (a b)b
(iii) (b/a)b
(iv) 〖{(a/b)+(b/a)}〗a

Solution 16:

(i) (a + b)a
According to the question a= 2 and b= 3
= (a + b)
= (2 + 3)2
= (5)2
= 25

(ii) (a b)
According to the question a = 2 and b = 3
= (a × b)
= (2 × 3)3
= (6)3
= 216

(iii) (b/a)b
According to the question a =2 and b = 3
=(b/a)b
= (3/2)3
= (27/8)

(iv) 〖{(a/b)+(b/a)}〗a
According to the question a = 2 and b = 3
= {(a/b)+(b/a) }a
= 〖{(2/3)+(3/2)}〗2
= (4/9)+(9/4)
LCM of 9 and 4 is 36
= (169/36)

Exercise 6.2

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

(i) 23 × 24 × 25

(ii) 512 ÷ 53

(iii) (72)3

(iv) (32)5 ÷ 34

(v) 37 × 27

(vi) (521 ÷ 513) × 57

Solution 1:

(i) 23 × 24 × 25

According to the law of exponents: am × a× ap = a(m+n+p)

So, the equation is written as

= 2(3 + 4 + 5)

= 212

(ii) 512 ÷ 53

According to the law of exponents: a÷ a= am-n

So, the equation is written as

= 512 – 3

= 59

(iii) (72)3

According to the law of exponents: (am)n = am x n

So the equation is written as

= 72 x 3

= 76

(iv) (32)5 ÷ 34

According to the law of exponents: (am)n = am x n

So the equation is written as

= 32 x 5 ÷ 34

= 3 10 ÷ 34

According to the law of exponents: a÷ a= am-n

= 3(10 – 4)

= 36

(v) 37 × 27

According to the law of exponents: am × bm = (a × b)m

So the equation is written as

= (3 × 2)7

= 67

(vi) (521 ÷ 513) × 57

According to the law of exponents: a÷ a= am-n

= 5(21 -13) × 57

So the equation is written as

= 58 × 57

According to the law of exponents: am × an = a(m +n)

So the equation is written as

= 5(8+7)

= 515

Question 2: Simplify and express each of the following in exponential form:

(i) {(23)4 × 28} ÷ 212

(ii) (82 × 84) ÷ 83

(iii) (57/52)× 5

(iv) (54× x10y5)/ (54 × x7y4)

Solution 2:

(i) {(23)4 × 28} ÷ 212

= {212 × 28} ÷ 212

= 2(12 + 8) ÷ 212

= 220 ÷ 212

= 2 (20 – 12)

=  28

(ii) (82 × 84) ÷ 83

= 8(2 + 4) ÷ 83

= 86 ÷ 83

= 8(6-3)

= 83

= (23)3

= 29

(iii) (57/52) × 53

= 5(7-2) × 53

= 55 × 53

= 5(5 + 3)

= 58

(iv) (54× x10y5)/ (54 × x7y4)

= (54-4× x10-7y5-4)

= 50x3y[since 50 = 1]

= 1x3y

Question 3: Simplify and express each of the following in exponential form:

(i) {(32)3 × 26} × 56

(ii)(x/y) 12 × y24 × (23)4

(iii)(5/2)6 × (5/2)2

(iv) (2/3)5× (3/5)5

Solution 3:

(i) {(32)3 × 26} × 56

= {36 × 26} × 56

= 66 × 56

= 306

(ii) (x/y)12 × y24 × (23)4

= (x12/y12) × y24 × 212

= x12 × y24-12 × 212

= x12 × y12 × 212

= (2xy)12

(iii)(5/2)6 × (5/2)2

= (5/2)6+2

= (5/2)8

(iv) (2/3)5× (3/5)5

= (2/5)5

Question 4: Write 9 × 9 × 9 × 9 × 9 in exponential form with base 3.

Solution 4:

= 9 × 9 × 9 × 9 × 9

= (9)5 = (32)5

= 310

Question 5: Simplify and write each of the following in exponential form:

(i) (25)3 ÷ 53

(ii) (81)5 ÷ (32)5

(iii) 98 × (x2)5/ (27)4 × (x3)2

(iv) 32 × 78 × 136/ 212 × 913

Solution 5:

(i) (25)3 ÷ 53

= (52)3 ÷ 53

= 5÷ 53

= 56 – 3

= 53

(ii) (81)5 ÷ (32)5

= (81)5 ÷ 310                                     ->81 = 34

= (34)5 ÷ 310

= 320 ÷ 310

= 320-10

= 310

(iii) 98 × (x2)5/ (27)4 × (x3)2

= (32)8 × (x2)5/ (33)4× (x3)2

= 316 × x10/312 × x6

= 316-12 × x10-6

= 34 × x4

= (3x)4

(iv) (32 × 78 × 136)/ (212 × 913)

= (32 × 727× 136)/(212× 13× 73)

= (212 × 76 × 136)/(212× 13× 73)

= (7× 136)/(13× 73)

= 916/913

= 916-3

= 913

Question 6: Simplify:

(i) (35)11 × (315)4 – (35)18 × (35)5

(ii) (16 × 2n+1 – 4 × 2n)/(16 × 2n+2 – 2 × 2n+2)

(iii) (10 × 5n+1 + 25 × 5n)/(3 × 5n+2 + 10 × 5n+1)

(iv) (16)×(25)5× (81)3/(15)×(24)5× (80)3

Solution 6:

(i) (35)11 × (315)4 – (35)18 × (35)5

= (3)55 × (3)60 – (3)90 × (3)25

= 3 55+60 – 390+25

= 3115 – 3115

= 0

(ii) (16 × 2n+1 – 4 × 2n)/(16 × 2n+2 – 2 × 2n+2)

= (24 × 2(n+1) -22 × 2n)/(24 × 2(n+2) -2n+1 × 22)

= 22 × 2(n+3-2n)/)22× 2(n+4-2n+1)

= 2n × 23 – 2n/ 2n × 24 – 2× 2

= 2n(23 – 1)/ 2n(24 – 1)

= 8 -1 /16 -2

= 7/14

= (1/2)

(iii) (10 × 5n+1 + 25 × 5n)/(3 × 5n+2 + 10 × 5n+1)

= (10 × 5n+1 + 52 × 5n)/(3 × 5n+2 + (2 × 5) × 5n+1)

= (10 × 5n+1 + 5 × 5n+1)/(3 × 5n+2 + (2 × 5) × 5n+1)

= 5n+1 (10+5)/ 5n+1 (10+15)

= 15/25

= (3/5)

(iv) (16)×(25)5× (81)3/(15)×(24)5× (80)3

= (16)× (52)5× (34)3/(3 × 5 )×(3 × 8)5× (16 × 5)3

= (16)× (52)5× (34)3/37 × 5× 35 × 85× 163× 53

= (16)7/ 85 × 16 3

= (16)4/85

= (2 × 8)4/85

= 24/8

= (16/8)

= 2

Question 7: Find the values of n in each of the following:

(i) 52n × 53 = 511

(ii) 9 × 3n = 37

(iii) 8 × 2n+2 = 32

(iv) 72n+1 ÷ 49 = 73

(v) (3/2)4 ×  (3/2) = (3/2)2n+1

(vi) (2/3)10× {(3/2)2}5 = (2/3)2n – 2

Solution:

(i)  52n × 53 = 511

= 52n+3 = 511

On equating the coefficients, we find:

2n + 3 = 11

⇒2n = 11- 3

⇒2n = 8

⇒ n = (8/2)

⇒ n = 4

(ii)   9 × 3n = 37

= (3)2 × 3n = 37

= (3)2+n = 37

On equating the coefficients, we find:

2 + n = 7

⇒ n = 7 – 2 = 5

(iii) 8 × 2n+2 = 32

= (2)3 × 2n+2 = (2)5      [since 23 = 8 and 25 = 32]

= (2)3+n+2 = (2)5

On equating the coefficients, we find:

3 + n + 2 = 5

⇒ n + 5 = 5

⇒ n = 5 -5

⇒ n = 0

(iv) 72n+1 ÷ 49 = 73

= 72n+1 ÷ 72 = 73  [since 49 = 72]

= 72n+1-2 = 73

= 72n-1=73

On equating the coefficients, we find:

2n – 1 = 3

⇒ 2n = 3 + 1

⇒ 2n = 4

⇒ n =4/2 =2

(v) (3/2)4 ×  (3/2) = (3/2)2n+1

= (3/2)4+5 = (3/2)2n+1

= (3/2)9 = (3/2)2n+1

On equating the coefficients, we find:

2n + 1 = 9

⇒ 2n = 9 – 1

⇒ 2n = 8

⇒ n =  8/2

=4

(vi) (2/3)10× {(3/2)2}5 = (2/3)2n – 2

= (2/3)10 × (3/2)10 = (2/3)2n – 2

= 2 10 × 310/310 × 210 = (2/3)2n – 2

= 1 = (2/3)2n – 2

= (2/3)0 = (2/3)2n – 2

On equating the coefficients, we find:

0 =2n -2

2n -2 =0

2n =2

n = 1

Question 8: If (9n × 32 × 3n – (27)n)/ (33)5 × 23 = (1/27), find the value of n.

Solution 8:

= (9n × 32 × 3n – (27)n)/ (33)5 × 23 = (1/27)

= (32)n × 33 × 3– (33)n/ (315 × 23) = (1/27)

= 3(2n+2+n) – (33)n/ (315 × 23) = (1/27)

= 3(3n+2)– (33)n/ (315 × 23) = (1/27)

= 33n × 32 – 33n/ (315 × 23) = (1/27)

= 33n × (32 – 1)/ (315 × 23) = (1/27)

= 33n × (9 – 1)/ (315 × 23) = (1/27)

= 33n × (8)/ (315 × 23) = (1/27)

= 33n × 23/ (315 × 23) = (1/27)

= 33n/315 = (1/27)

= 33n-15 = (1/27)

= 33n-15 = (1/33)

= 33n-15 = 3-3

On equating the coefficients, we find:

3n -15 = -3

⇒ 3n = -3 + 15

⇒ 3n = 12

⇒ n = 12/3 = 4

Exercise 6.3

Question 1: Express the following numbers in the standard form:
(i) 3908.78
(ii) 5,00,00,000
(iii) 3,18,65,00,000
(iv) 846 × 107
(v)723 × 109

Solution 1:

(i) 3908.78

= 3.90878 × 103 [Here the decimal point is moved left up to 3 places]

(ii) 5,00,00,000

= 5,00,00,000.00 = 5 × 107 [Here, the decimal point is moved left up to 7 places]

(iii) 3,18,65,00,000

= 3,18,65,00,000.00

= 3.1865 × 109 [here, the decimal point is moved left up to 9 places]

(iv) 846 × 107

= 8.46 × 102 × 10 [here the decimal point is moved left up to 2 places]

= 8.46 × 109 [since am × an = am+n]

(v) 723 × 109

= 7.23 × 102 × 109 [here the decimal point is moved left up to 2 places]

= 7.23 × 1011 [ since am × an = am+n]

Question 2: Write the following numbers in the usual form:
(i) 4.83 × 107
(ii) 3.21 × 105
(iii) 3.5 × 103

Solution 2:

(i) 4.83 × 107

= 483 × 107-2 [hence the decimal point is moved right up to 2 places]

= 483 × 105

= 4, 83, 00,000

(ii) 3.21 × 10

= 321 × 105-2 [hence the decimal point is moved right up to 2 places]

= 321 × 103

= 3, 21,000

(iii) 3.5 × 103

= 35 × 103-1 [hence the decimal point is moved right up to 1 places]

= 35 × 102

= 3,500

Question :3. Express the numbers appearing in the following statements in the standard form:

(i) The distance between the Earth and the Moon is 384,000,000 meters.
(ii) Diameter of the Earth is 1, 27, 56,000 meters.
(iii) Diameter of the Sun is 1,400,000,000 meters.
(iv) The universe is estimated to be about 12,000,000,000 years old.

Solution:

(i) The distance between the Earth and the Moon is 384,000,000 meters.

Distance between the Earth and the Moon is 3.84 × 108 meters.

[Here the decimal point is moved left up to 8 places.]

(ii) Diameter of the Earth is 1, 27, 56,000 meters.

Diameter of the Earth is 1.2756 × 107 meters.

[Here the decimal point is moved left up to 7 places]

(iii) Diameter of the Sun is 1,400,000,000 meters.

Diameter of the Sun is 1.4 × 109 meters.

[Here the decimal point is moved left up to 9 places]

(iv) The universe is estimated to be about 12,000,000,000 years old.

The universe is estimated to be about 1.2× 1010 years old.

[Here the decimal point is moved left up to 10 places]

Exercise 6.4

Question 1: Write the following numbers in the expanded exponential forms:
(i) 20068
(ii) 420719
(iii) 7805192
(iv) 5004132
(v) 927303

Solution 1:

(i) 20068

= 2 × 104 + 0 × 103 + 0 × 102 + 6 × 101 + 8 × 100

(ii) 420719

= 4 × 105 + 2 × 104 + 0 × 103 + 7 × 102 + 1 × 101 + 9 × 100

(iii) 7805192

= 7 × 106 + 8 × 105 + 0 × 104 + 5 × 103 + 1 × 102 + 9 × 101 + 2 × 100

(iv) 5004132

= 5 × 106 + 0 × 105 + 0 × 104 + 4 × 103 + 1 × 102 + 3 × 101 + 2 × 100

(v) 927303

= 9 × 105 + 2 × 104 + 7 × 103 + 3 × 102 + 0 × 101 + 3 × 100

Question 2: Find the number from each of the following expanded forms:
(i) 7 × 104 + 6 × 103 + 0 × 102 + 4 × 101 + 5 × 100
(ii) 5 × 105 + 4 × 104 + 2 × 103 + 3 × 100
(iii) 9 × 105 + 5 × 102 + 3 × 101
(iv) 3 × 10+ 4 × 102 + 5 × 100

Solution 2:

(i) 7 × 104 + 6 × 103 + 0 × 102 + 4 × 101 + 5 × 100

= 7 × 10000 + 6 × 1000 + 0 × 100 + 4 × 10 + 5 × 1

= 70000 + 6000 + 0 + 40 + 5

= 76045

(ii) 5 × 105 + 4 × 104 + 2 × 103 + 3 × 100

= 5 × 100000 + 4 × 10000 + 2 × 1000 + 3 × 1

= 500000 + 40000 + 2000 + 3

= 542003

(iii) 9 × 105 + 5 × 102 + 3 × 101

= 9 × 100000 + 5 × 100 + 3 × 10

= 900000 + 500 + 30

= 900530

(iv) 3 × 10+ 4 × 102 + 5 × 100

= 3 × 10000 + 4 × 100 + 5 × 1

= 30000 + 400 + 5

= 30405