RD Sharma solution class 7 chapter 6 Exponents Exercise 6.2

Exercise 6.2

Page-6.28

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

Answer 1:

 We have

(i)  23 x 24 x 25 = 2(3 + 4 + 5) = 212             [since am + an + ap = a(m+n+p)]

(ii) 512 ÷ 53 = 51253 = 512 - 3 = 59                [ since am ÷ an = am-n ]
(iii) (72)3 = 76                                                       [since (am)n = amn ]

(iv)(32)5 ÷ 34 = 310 ÷ 34                             [since (am)n = amn ]
                       = 3(10 - 4) = 36                      [since am ÷ an = am-n ]

(v) 37 ​× 27 = (3 x 2)7 = 67                          [since am x bm = (a x b)m ]

 (vi) (521 ÷  513) x 57 = 5(21 -13) x 57         [since am ÷ an = am-n ]
                                   = 58 x 57                  [since am x bn =a(m +n)]
                                   = 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) 5752×53
(iv) 54×x10y554× x7y4

Answer 2:

We have
(i)  {(23)4 x 28} ÷ 212
    = {212 x 28} ÷ 212
    = 2(12 + 8) ÷ 212
 
   = 220 ÷ 212
   
= 2 (20 - 12) =  28

(ii) (82 x 84)  ÷ 83
    = 8(2 + 4) ÷ 83
    = 86 ÷ 83
   = 8(6-3) = 83 = (23)3 = 29

(iii) 5752 x 53 = 5(7-2) x 53
                        = 55 x 53
                        = 5(5 + 3 ) = 58

(iv) 54 ×x10y554×x7y4 = 5(4-4)×x(10-7)×y(5-4)
                         = 50×x3×y               [since 50 = 1]
                         = 1×x3y =x3y

Question 3:

Simplify and express each of the following in exponential form:
(i) {(32)3×26}×56
(ii) xy12×y24×(23)4
(iii) 526×522
(iv) 235×355

Answer 3:

We have
(i)   {(32)3 x 26} x 56
    = {36 x 26} x 56             [since (am)n = amn]
    = 66 x 56                        [since am x bm = (a x b)m ]
    = 306

(ii)  
xy12×y24×(23)4
=x12y12×y24×212
=x12×y24y12×212
=x12×y24-12×212
=x12×y12×212
=(2xy)12 [since am×bm×cm=(a×b×c)m]
(iii)
526×522
=528 [since am×an=am+n]

(iv)
235×355
=23×355 [since am×bm=(a×b)m]
=255

Question 4:

Write 9 × 9 × 9 × 9 × 9 in exponential form with base 3.

Answer 4:

We have
9 x 9 x 9 x 9 x 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× 136212× 913

Answer 5:

We have
(i) (25)3 ÷ 53
= (52)3÷ 53
= 56 ÷ 53
= 5653=56-3=53

(ii) (81)5 ÷ (32)5
= (34)5 ÷ (32)5
= (3)20 ÷ (3)10
= 320310=320-10=310

(iii)

98×(x2)5(27)4×(x3)2
=(32)8×(x2)5(33)4×(x3)2
=316×(x)10312×(x)6
=316-12× (x)10-6 = 34× x4= (3x)4

(iv)

32×78×136212×913
=32×72×76×136212×(13×7)3
=(21)2×76×136212×133×73
=76×136133×73
=916913=916-3=913

Question 6:

Simplify:
(i) (35)11×(315)4-(35)18×(35)5
(ii) 16×2n+1-4×2n16×2n+2-2×2n+2
(iii) 10×5n+1+25×5n3×5n+2+10×5n+1
(iv) (16)7×(25)5× (81)3(15)7× (24)5 × (80)3

Answer 6:

We have
(i) (35)11× (315)4- (35)18×(35)5
   = 355 x 360 - 390 x 325
   = 3(55 + 60) - 3(90 + 25)
   = 3115 - 3115
   = 0

(ii) 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

 (iii)

10×5n+1+25×5n3×5n+2+10×5n+1
=10×5n+1+(5)2×5n3×5n+2+2×5×5n+1
=10×5n+1+5×5n+13×5n+2+2×5×5n+1
=5n+1(10+5)3×5×5n+1+10×5n+1
=5n+1(15)5n+1(15+10)=5n+1×155n+1×25=1525=35


(iv)
(16)7×(25)5×(81)3(15)7×(24)5×(80)3
=(16)7×(52)5×(34)3(3×5)7×(3×8)5×16×53
=(16)7×(5)10×(3)1237×57×35×85×163×53
=(16)7×(5)10×(3)1237×35×57×53×85×163
=(16)7×(5)10×(3)12312×510×85×163
=(16)785×163
=(16)7-385=(16)485=(2×8)485=24×8485=248=168=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) 324×325=322n+1
(vi) 2310+3225=252n-2

Answer 7:

We have

(i) 52n x 53 = 511
= 52n+3 = 511
On equating the coefficients, we get
2n + 3 = 11
⇒2n = 11- 3
⇒2n = 8
⇒ n =82=4
(ii) 9 x 3n = 37
= (3)2 x 3n = 37
= (3)2+n = 37
On equating the coefficients, we get
2 + n = 7
⇒ n = 7 - 2  = 5

(iii) 8 x 2n+2 = 32
= (2)3 x 2n+2 = (2)5      [since 23 = 8 and 25 = 32]
= (2)3+n+2 = (2)5 
On equating the coefficients, we get
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+172=73
=72n+1-2=73 [since aman=am-n]
= 72n-1 =73
On equating the coefficients, we get
2n - 1 = 3
⇒ 2n = 3 + 1
⇒ 2n = 4
⇒ n = 42=2
(v) 324×325=322n+1
=32(4+5)=32(2n+1)
=329=322n+1
On equating the coefficients, we get
2n + 1 = 9
⇒ 2n = 9 - 1
⇒ 2n = 8
⇒ n =82=4
 (vi) 2310×3225=252n-2
=2310×3210=252n-2
=210×310310×210=252n-2
=1=252n-2
=250=252n-2 [since 250=1]
On equating the coefficients, we get
⇒ 0 = 2n - 2
⇒ 2n = 2
⇒ n = 22=1

Page-6.29

Question 8:

If 9n×32× 3n-(27)n(33)×23=127, find the value of n.

Answer 8:

We have

9n×32×3n-(27)n(33)5×23=127
=(32)n×32×3n-(33)n(3)15×23=127
=(3)2n+2+n-(3)3n(3)15×23=127
=(3)3n+2-(3)3n(3)15×23=127
=(3)3n×(3)2-(3)3n(3)15×23=127
=(3)3n(32-1)(3)15×23=127
=(3)3n ×8(3)15×23=127
=(3)3n ×23(3)15×23=127
=33n315=127
=33n-15=133
=33n-15=3-3
On equating the coefficients, we get
3n -15 = -3
⇒ 3n = -3 + 15
⇒ 3n = 12
⇒ n =123=4

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