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RS Aggarwal 2019,2020 solution class 7 chapter 5 Exponents Exercise 5A

Exercise 5A

Page-90

Question 1:

Write each of the following in power notation:

(i) 57×57×57×57
(ii) (-43)×(-43)×(-43)×(-43)×(-43)
(iii) (-16)×(-16)×(-16)
(iv) (−8) × (−8) × (−8) × (−8) × (−8)

Answer 1:

(i) 57×57×57×57 = (57)4

(ii) (-43)×(-43)×(-43)×(-43)×(-43)=(-43)5

(iii) (-16)×(-16)×(-16)=(-16)3

(iv) (-8)×(-8)×(-8)×(-8)×(-8)=(-8)5

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Question 2:

Express each of the following in power notation:

(i) 2536
(ii) -2764
(iii) -32243
(iv) -1128

Answer 2:

(i) 2536=5262                               [since 25 = 52 and 36 = 62]
          =(56)2

(ii) -2764=(-3)343                     [since −27 = (−3)3 and 64 = 43]

              =(-34)3

(iii) -32243=(-2)535                   [since −32 = (−2)5 and 243 = 35]
               =(-23)5

(iv) -1128=(-1)727                    [since (−1)7 = −1 and 128 = 27]
              =(-12)7

Question 3:

Express each of the following as a rational number:

(i) (23)5
(ii) (-85)3
(iii) (-1311)2
(iv) (16)3
(v) (-12)5
(vi) (-32)4
(vii) (-47)3
(viii) (−1)9

Answer 3:

(i) (23)5=(2)5(3)5=2×2×2×2×23×3×3×3×3=32243

(ii) (-85)3=(-8)3(5)3=(-8)×(-8)×(-8)5×5×5=-512125

(iii) (-1311)2=(-13)2(11)2=(-13)×(-13)11×11=169121

(iv) (16)3=(1)3(6)3=1×1×16×6×6=1216

(v) (-12)5=(-1)5(2)5=(-1)×(-1)×(-1)×(-1)×(-1)2×2×2×2×2=-132

(vi) (-32)4=(-3)4(2)4=(-3)×(-3)×(-3)×(-3)2×2×2×2=8116

(vii) (-47)3=(-4)3(7)3=(-4)×(-4)×(-4)7×7×7=-64343

(viii) (-1)9=-1       [Since (-1) an odd natural number = -1]

Question 4:

Express each of the following as a rational number:

(i) (4)−1
(ii) (−6)−1
(iii) (13)-1
(iv) (-23)-1

Answer 4:

(i) (4)-1=(41)-1=(14)1=14                                 [since (ab)-n=(ba)n]

(ii) (-6)-1=(-61)-1=(1-6)1=-16                    [since (ab)-n=(ba)n]

(iii) (13)-1=(31)1=31                                           [since (ab)-n=(ba)n]

(iv) (-23)-1=(3-2)1=-32                                  [since (ab)-n=(ba)n]
                     

Question 5:

Find the reciprocal of each of the following:

(i) (38)4
(ii) (-56)11
(iii) 67
(iv) (−4)3

Answer 5:

We know that the reciprocal of (ab)m is (ba)m.

(i) Reciprocal of (38)4=(83)4

(ii) Reciprocal of (-56)11=(-65)11

(iii) Reciprocal of 67 = Reciprocal of (61)7= (16)7

(iv) Reciprocal of (− 4)3 = Reciprocal of (-41)3 = (-14)3

Question 6:

Find the value of each of the following:

(i) 80
(ii) (−3)0
(iii) 40 + 50
(iv) 60 × 70

Answer 6:

(i) 80 = 1
(ii) (−3)0 = 1
(iii) 40 + 50 = 1 + 1 = 2
(iv) 60 × 70 = 1 × 1 = 1

Note: a0 = 1

Question 7:

Simplify each of the following and express each as a rational number:

(i) (32)4×(15)2
(ii) (-23)5×(-37)3
(iii) (-12)5×23×(35)2
(iv) (23)2×(-35)3×(72)2
(v) {(-34)3-(-52)3}×42

Answer 7:

(i) (32)4×(15)2=3424×1252=81×116×25=81400

(ii) (-23)5×(-37)3=(-2)5(3)5×(-3)3(7)3
                                  = =(-2)5(7)3×(-1)(3)3(3)5=-32×-1×33-5343=-32×-1×3-2343=-32×-1×1343×9=323087               [since 3-2=19]

                                
(iii) (-12)5×23×(34)2=(-1)525×23×3242
                                       =(-1)525×23×32(22)2
                                       =-1×23×3225×24
                                       =-1×23×3229 =-1×23-9×32  = -9×2-6=-926=-964                  [since (ab)-1=(ba)1]

(iv) (23)2×(-35)3×(72)2=2232×(-3)353×7222
                                             -1×33-2×7253=-1×31×7253=-1×3×49125=-147125

(v) {(-34)3-(-52)3}×42={(-3343)-(-5323)}×42
                                             ={(-2764)-(-1258)}×16
                                             ={-2764+1258}×16
                                             =(-27+100064)×16
                                              =(97364×16)=9734

Question 8:

Simplify and express each as a rational number:

(i) (49)6×(49)-4
(ii) (-78)-3×(-78)2
(iii) (43)-3×(43)-2

Answer 8:

(i) (49)6×(49)-4=(49)6+(-4)                                             [since an×am=an+m]
                        =  (49)2(4)2(9)2=4×49×9=1681

(ii) (-78)-3×(-78)2=(-78)(-3)+2                                  [since an×am=an+m]
                             = (-78)-1            
                              =  (8-7)1                                          [since (ab)-1= (ba)1]
                              = (8×-1-7×-1)=-87

(iii) (43)-3×(43)-2=(43)(-3)+(-2)                                    [since an×am= an+m]
                           = (43)-5
                           = (34)5                                                [since (ab)-1=(ba)1]
                          = (3)5(4)5=3×3×3×3×34×4×4×4×4=2431024

Question 9:

Express each of the following as a rational number:

(i) 5−3
(ii) (−2)−5
(iii) (14)-4
(iv) (-34)-3
(v) (-3)-1×(13)-1
(vi) (57)-1×(74)-1
(vii) (5−1−7−1)−1
(viii) {(43)-1-(14)-1}-1
(ix) {(32)-1÷(-25)-1}
(x) (2325)0

Answer 9:

Note: [(ab)-1=(ba)1]


(i) 5−3 = (51)-3=(15)3=(1)3(5)3=1125


(ii) (−2)−5 = (-21)-5=(1-2)5=(1)5(-2)5=1×-1-32×-1=-132


(iii) (14)-4=(41)4=(4)4(1)4=2561=256


(iv) (-34)-3=(4-3)3 = (4)3(-3)3=64-27=64×-1-27×-1=-6427


(v) (-3)-1×(13)-1=(1-3)1×(31)1=(1×3-3×1)1=(3-3)1=1-1=1×-1-1×-1=-11=-1


(vi) (57)-1×(74)-1=(75)1×(47)1=(7×45×7)1=45


(vii) (5-1-7-1)-1=(15-17)-1=(7-535)-1
                                                     = (235)-1=(352)1=352

(viii) {(43)-1-(14)-1}-1 = {(34)1-(41)1}-1=(34-41)-1
                                                               = (3-164)-1=(-134)-1= (4-13)1=(4×-1-13×-1)= -413

(ix) {(32)-1÷(-25)-1}={(23)1÷(5-2)1}
                                 =(23×-25)=-415

(x) (2325)0=1          [since a0 = 1 for every integer a]

Question 10:

Simplify:

(i) [{(-14)2}-2]-1
(ii) {(-23)2}3
(iii) (-32)3÷(-32)6
(iv) (-23)7÷(-23)4

Answer 10:

(i)

[{(-14)2}-2]-1=[(-14)2×-2]-1                 [since {(ab)m}n=(ab)mn]
                          =[(-14)-4]-1
                           =(-14)(-4)×(-1)=(-14)4=(-1)4(4)4=1256

(ii)

{(-23)2}3=(-23)2×3                               [since {(ab)m}n=(ab)mn]
                  = (-23)6
                  = (-2)6(3)6=64729                      [since (−2)6 = 64 and (3)6 = 729]

(iii)

(-32)3÷(-32)6=(-32)3-6                    [since am÷an=am-n]          
                             = (-32)-3 
                            =(2-3)3                      [since (ab)-1=(ba)1]
                            =(2×-1-3×-1)3=(-23)3=(-2)3(3)3=-827

(iv)

(-23)7÷(-23)4=(-23)7-4                   [since am÷an=am-n]
                            =(-23)3
                             = (-2)3(3)3=-827

Question 11:

By what number should (−5)−1 be multiplied so that the product is (8)−1?

Answer 11:

Let the required number be x.
(−5)-1 × x = (8)-1
1-5× x=18
x = 18×(-5) = -58
Hence, the required number is -58.

Question 12:

By what number should 3−3 be multiplied to obtain 4?

Answer 12:

Let the required number be x.
(3)-3 x x = 4
133×x=4
127×x=4
x = 4 x 27 = 108
Hence, the required number is 108.

Question 13:

By what number should (−30)−1 be divided to get 6−1?

Answer 13:

Let the required number be x.
(-30)-1 ÷ x = 6-1
1(-30)×1x=16
1(-30x)=16
x = 6(-30)=1-5
                       =-15
Hence, the required number is -15.

Question 14:

Find x such that (35)3×(35)-6=(35)2x-1.

Answer 14:

(35)3×(35)-6=(35)2x-1
(35)3+(-6)=(35)2x-1      [since am×an=am+n]
(35)-3=(35)2x-1
On equating the exponents:
−3 = 2x − 1
⇒ 2x = −3 + 1
⇒ 2x   = −2
x = (-22)=-1

Question 15:

Simplify: 35×105×2557×65.

Answer 15:

35×105×2557×65=35×(2×5)5×5257×(2×3)5
                        =35×25×55×5257×25×35
                        =35×25×5735×25×57=35-5×25-5×57-7=30×20×50=1×1×1=1

Page-92

Question 16:

Simplify: 16×2n+1-4×2n16×2n+2-2×2n+2.

Answer 16:

    16×2n+1-4×2n16×2n+2-2×2n+2
24×2n+1-22×2n24×2n+2-2n+1×22
22×(2n+3-2n)22×(2n+4-2n+1)
2n×23-2n2n×24-2n×2
2n(23-1)2n(24-2)=8-116-2=714=12

Question 17:

Find the value of n when:

(i) 52n × 53 = 59
(ii) 8 × 2n+2 = 32
(iii) 62n+1 ÷ 36 = 63

Answer 17:

(i) 52n × 53 = 59
     52n+3 = 59          [since an × am = am+n]

     On equating the coefficients:
     2n + 3 = 9
     ⇒ 2n = 9 − 3
     ⇒ 2n = 6
     ∴ n =62=3

(ii) 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:
      3 + n + 2 = 5
      ⇒ n + 5 = 5
      ⇒ n = 5 − 5
      ∴ n = 0

(iii) 62n+1 ÷ 36 = 63
       ⇒ 62n+1 ÷ 62 = 63       [since 36 = 62]
       ⇒ 62n+162=63
       ⇒ 62n+1-2=63          [since aman=am-n ]
       ⇒ 62n-1 = 63
        On equating the coefficients:
        2n - 1 = 3
        ⇒ 2n = 3 + 1
        ⇒ 2n = 4
         ∴ n = 42=2

Question 18:

If 2n−7 × 5n−4 = 1250, find the value on n.

Answer 18:

    2n-7×5n-4=1250
2n27×5n54=2×54                       [since 1250 = 2 × 54]
2n×5n27×54=2×54
⇒  2n×5n=2×54×27×54           [using cross multiplication]
⇒  2n×5n=21+7×54+4               [since am × an = am+n ]
2n×5n=28×58
(2×5)n=(2×5)8                      [since an × bn = (a × b)n ]
10n=108
n = 8

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