Exercise 5A
Page-90Question 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
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
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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