Class 7 - Maths - Exponents and Powers

Exercise 13.1

Question 1:

Find the value of:

(i) 26            (ii) 93              (iii) 112               (iv) 54

(i) 26 = 2 * 2 * 2 * 2 * 2 * 2 = 64

(ii) 93 = 9 * 9 * 9 = 729

(iii) 112 = 11 * 11 = 121

(iv) 54 = 5 * 5 * 5 * 5 = 625

Question 2:

Express the following in exponential form:

(i) 6 * 6 * 6 * 6 (ii) t * t  (iii) b * b * b * b (iv) 5 * 5 * 7 * 7 * 7 (v) 2 * 2 * a * a

(vi) a * a * a * c * c * c * c  * d

(i) 6 * 6 * 6 * 6 = 64

(ii) t * t = t2

(iii) b * b * b * b = b4

(iv) 5 * 5 * 7 * 7 * 7 = 52 * 73

(v) 2 * 2 * a * a = 22 * a2

(vi) a * a * a * c * c * c * c  * d = a3 * c4 * d

Question 3:

Express each of the following numbers using exponential notations:

(i) 512          (ii) 343            (iii) 729             (iv) 3125

(i) 512 = 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 = 29

(ii) 343 = 7 * 7 * 7 = 73

(iii) 729 = 3 * 3 * 3 * 3 * 3 * 3 = 36

(iv) 3125 = 5 * 5 * 5 * 5 * 5 = 55

Question 4:

Identify the greater number, wherever possible, in each of the following:

(i) 43 and 34       (ii) 53 or 35    (iii) 28 or 82     (iv) 1002 or 2100       (v) 210 or 102

(i) 43 = 4 * 4 * 4 = 64

34 = 3 * 3 * 3 * 3 = 81

Since 64 < 81

So, 34 is greater than 43

(ii) 53 = 5 * 5 * 5 = 125

35 = 3 * 3 * 3 * 3 * 3 = 243

Since 125 < 243

So, 35 is greater than 53

(iii) 28 = 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 = 256

82 = 8 * 8 = 64

Since, 256 > 64

Thus, 28 is greater than 82

(iv) 1002 = 100 * 100 = 10,000

2100 = 2 * 2 * 2 * 2 * 2 * …..14 times * ……… * 2 = 16,384 * ….. * 2

Since, 10,000 < 16,384 * ……. * 2

Thus, 2100 is greater than 1002.

(v) 210 = 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 = 1,024

102 = 10 * 10 = 100

Since, 1,024 > 100

Thus, 210 is greater than 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

(i) 648 = 23 * 34

(ii) 405 = 5 * 34

(iii) 540 = 22 * 33 * 5

(iv) 3,600 = 24 * 32 * 52

Question 6:

Simplify:

(i) 2 * 103           (ii) 72 * 22             (iii) 23 * 5            (iv) 3 * 44                 (v) 0 * 102

(vi) 52 * 33              (vii) 24 * 32       (viii) 32 * 104

(i) 2 * 103 = 2 * 10 * 10 * 10 = 2,000

(ii) 72 * 22 = 7 * 7 * 2 * 2 = 196

(iii) 23 * 5 = 2 * 2 * 2 * 5 = 40

(iv) 3 * 44 = 3 * 4 * 4 * 4 * 4 = 768

(v) 0 * 102 = 0 * 10 * 10 = 0

(vi) 52 * 33 = 5 * 5 * 3 * 3 * 3 = 675

(vii) 24 * 32 = 2 * 2 * 2 * 2 * 3 * 3 = 144

(viii) 32 * 104 = 3 * 3 * 10 * 10 * 10 * 10 = 90,000

Question 7:

Simplify:

(i) (-4)3          (ii) (-3) * (-2)3           (iii) (-3)2 * (-5)2           (iv) (-2)3 * (-10)3

(i) (-4)3 = (-4) * (-4) * (-4) = -64

(ii) (-3) * (-2)3 = (-3) * (-2) * (-2) * (-2) = 24

(iii) (-3)2 * (-5)2 = (-3) * (-3) * (-5) * (-5) = 225

(iv) (-2)3 * (-10)3 = (-2) * (-2) *(-2) *(-10) *(-10) *(-10) = 8000

Question 8:

Compare the following numbers:

(i) 2.7 * 1012; 1.5 * 108               (ii) 4 * 1014; 3 * 1017

(i) 2.7 * 1012 and 1.5 * 108

On comparing the exponents of base 10,

2.7 * 1012 > 1.5 * 108

(ii) 4 * 1014  and 3 * 1017

On comparing the exponents of base 10,

4 * 1014 < 3 * 1017

Exercise 13.2

Question 1:

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

(i) 32 * 34 * 38               (ii) 615/610                (iii) a3 * a2             (iv) 7x * 72          (v) (52)3 /53

(vi) 25 * 55                   (vii) a4 * b4                (viii) (34)3              (ix) (220/215) * 23        (x) 8t/82

(i) 32 * 34 * 38 = 32+4+8 = 314                                                     [Since am * an = am+n ]

(ii) 615/610 = 615-10 = 65                                                             [Since am / an = am-n ]

(iii) a3 * a2 = a3+2 = a5                                                               [Since am * an = am+n ]

(iv) 7x * 72 = 7x+2                                                                       [Since am * an = am+n ]

(v) (52)3 /53 = 56 /53 =  56-3 = 53                                               [Since (am)n = amn and am * an = am+n ]

(vi) 25 * 55 = (2 * 5)5 = 105                                                      [Since am * bm = (a * b)m ]

(vii) a4 * b4 = (a * b)4                                                               [Since am * bm = (a * b)m ]

(viii) (34)3 = 312                                                                          [Since (am)n = amn ]

(ix) (220/215) * 23 = 220-15 * 23 = 25 * 23

= 25+3 = 28                                [Since am / an = am-n and am * an = am+n ]

(x) 8t/82 = 8t-2                                                                                                                       [Since am / an = am-n  ]

Question 2:

Simplify and express each of the following in exponential form:

(i) (23 * 34 * 4)/(3 * 32)          (ii) [(52)3 * 54]/57         (iii) 254 / 53            (iv) (3 * 72 * 118)/(21 * 113)

(v) 37/(34 * 33)      (vi) 20 + 30 + 40     (vii) 20 * 30 * 40   (viii) (30 + 20) * 50    (ix) (28 * a5)/(43 * a3)

(x) (a5/a3) * a8         (xi) (45 * a8 b3)/ (45 * a5 b2)    (xii) (23 * 2)2

(i) (23 * 34 * 4)/(3 * 32) = (23 * 34 * 22)/(3 * 25)

= (23+2 * 34)/(3 * 25)

= (25 * 34)/(3 * 25)

= 25-5 * 34-1

= 20 * 33

= 1 * 33

= 33

(ii) [(52)3 * 54]/57 = [56 * 54]/57

= 56+4/57

= 510/57

= 510-7

= 53

(iii) 254 / 53 = (52)4 / 53

= 58 / 53

= 58-3

= 55

(iv) (3 * 72 * 118)/(21 * 11) = (3 * 72 * 118)/(3 * 7 * 113)

= 31-1 * 72-1 * 118-3

= 30 * 7 * 115

= 1 * 7 * 115

= 7 * 115

(v) 37/(34 * 33) = 37/34+3

= 37/37

= 30

= 1

(vi) 20 + 30 + 40 = 1 + 1 + 1 = 3

(vii) 20 * 30 * 40 = 1 * 1 * 1 = 1

(viii) (30 + 20) * 50 = (1 + 1) * 1 = 2 * 1 = 2

(ix) (28 * a5)/(43 * a3) = (28 * a5)/{(22)3 * a3)}

= (28 * a5)/(26 * a3)}

= 28-6 * a5-3

= 22 * a2

= 4a2

= (2a)2

(x) (a5/a3) * a8 = a5-3 * a8

= a2 * a8

= a2+8

= a10

(xi) (45 * a8 b3)/ (45 * a5 b2) = 45-5 * a8-5 b3-2

= 40 * a3 b

= 1 * a3 b

= a3 b

(xii) (23 * 2)2 = (23+1)2 = (24)2 = 24*2 = 28

Question 3:

(i) 10 * 1011 = 10011       (ii) 23 > 52       (iii) 23 * 32 = 65        (iv) 30 = (1000)0

(i) 10 * 1011 = 10011

LHS: 10 * 1011 = 1011+1 = 1012 =

RHS: 10011 = (102)11 = 102*11 = 1022

Since LHS ≠ RHS

So, it is false.

(ii) 23 > 52

LHS: 23 = 8

RHS: 52 = 25

Since LHS is not greater than RHS

Therefore, it is false.

(iii) 23 * 32 = 65

LHS: 23 * 32 = 8 * 9 = 72

RHS: 65 = 7,776

Since LHS ≠ RHS

Therefore, it is false.

(iv) 30 = (1000)0

LHS: 30 = 1 and RHS: (1000)0 = 1

Since LHS = RHS

Therefore, it is true.

Question 4:

Express each of the following as a product of prime factors only in exponential form:

(i) 108 * 192        (ii) 270          (iii) 729 * 64         (iv) 768

(i) 108 x 192 = (22 * 33)*(26 * 3)

= 22+6 * 33+1

= 28 * 34

(ii) 270 = 2 * 35 * 5

(iii) 729 x 64 = 36 * 26

(iv) 768 = 28 * 3

Question 5:

Simplify:

(i) {(25)2 * 73}/(83 * 7)         (ii) (25 * 52 * t8)/(103 * t4)         (iii) (35 * 105 * 25)/(57 * 65)

(i) {(25)2 * 73}/(83 * 7) = (25*2 * 73)/{(23)3 * 7)}

= (210 * 73)/(23*3 * 7)

= (210 * 73)/(29 * 7)

= 210-9 * 73-1

= 2 * 72

= 2 * 49

= 98

(ii) (25 * 52 * t8)/(103 * t4) =(52 * 52 * t8)/{(5 * 2)3 * t4}

= (52+2 * t8)/(53 * 23 * t4)

= (54 * t8)/(53 * 23 * t4)

= (54-3 * t8-4)/8

= 5t4/8

(iii) (35 * 105 * 25)/(57 * 65) = {(35 * (2 * 5)5 * 52)}/{57 * (2 * 3)5}

= (35 * 25 * 55+2)/(57 * 25 * 35)

= (35 * 25 * 57)/(57 * 25 * 35)

= 35-5 * 25-5 * 57-7

= 30 * 20 * 50

= 1 * 1 * 1

= 1