ML Aggarwal Solution Class 9 Chapter 8 Indices Exercise 8
Exercise 8
Question 1
(i) (8116)−34
⇒(3424)−3/4
⇒[(32)4]−3/4
⇒(32)4×−34
⇒(32)−3
⇒(23)3
⇒2333
⇒2333
⇒2×2×23×3×3
⇒827
⇒(12564)−2/3
⇒(5343)−2/3
⇒[(54)3]−2/3
⇒(54)3×−23
⇒(54)−2
⇒(45)2
⇒4252
⇒1625
Question 2
(i) (2a-3b2)3
⇒23a-3×3b2×3
⇒8a-9b6
(ii) a−1+b−1(ab)−1
⇒1a+1b1ab
⇒b+aab1ab
⇒b+aab×ab1
⇒a+b
Question 3
(i) x−1y−1x−1+y−1
⇒(xy)−11x+1y
⇒1xyy+xxy
⇒1x+y
(ii) (4×107)(6×10−5)8×1010
⇒4×107×6×10−58×1010
⇒3×107−51010
⇒3×1021010
⇒31010−2
⇒3108
Question 4
(i) 3ab−1+2ba−1
Sol :
⇒3a(1b)+2b(1a)
⇒3ab+2ab
⇒5ab
(ii) 50×4-1+81/3
⇒1×14+(23)13
⇒14+2
⇒1+84
⇒94
Question 5
(i) (8125)−1/3
⇒(2353)−1/3
⇒(25)3×−13
⇒(25)−1
⇒52
(ii) (0.027)−13
⇒(271000)−1/3
⇒(33103)−1/3
⇒(310)8×−1/8
⇒(310)3×−13
⇒103
Question 6
(i) (−127)−2/3
⇒−(133)−2/3
⇒−1(33)−2/3
⇒−133×−23
⇒−13−2
⇒-32
⇒-9
(ii) (64)-2/3 ÷ 9-3/2
⇒(43)-2/3÷(92)-3/2
⇒43×−23÷32×−37
⇒4-2÷3-3
⇒4−23−3
⇒142133
⇒3342
⇒2716
Question 7
(i) (27)2n/3×8−n/6(18)−n/2
⇒(33)2n3×(23)−n/6(2×9)−n/2
⇒33×2n3×23×−n62−n2×32×−n2
⇒32n×2−n/22−n/2×3−n
⇒32n2−n/2×12n/2×3n
⇒32n+n2−n2×2n2
⇒33n2−n/2⋅2n/2
⇒33n×2n/22n/2
⇒33n
(ii) 5⋅(25)n+1−25(5)2n5⋅(5)2n+3−(25)n+1
⇒51⋅(52)n+1−52⋅52n51⋅52n+3−(52)n+1
⇒51+2n+2−52+2n51+2n+3−52n+2
⇒52n+3−52+2n52n+4−52+2n
⇒52n+1⋅53−52⋅52n52n⋅54−52⋅52n
⇒52h[53−52]53n[54−52]
⇒125−25625−25
⇒100600
⇒16
Question 8
(i) (8−43÷2−2)12
⇒((23)−43÷2−2)1/2
⇒(2−42−2)1/2
⇒(2−4+2)1/2
⇒(2−2)1/2
⇒2-1
⇒12
(ii) (278)2/3−(14)−2+5∘
⇒(3323)23−(122)−2+1
⇒(32)3×23−122×−2+1
⇒(32)2−12−4+1
⇒94−24+1
⇒94−16+1
⇒94−15
⇒9−604
⇒−514
Question 9
(i) (3x2)−3×(x9)23
Sol :
⇒1(3x2)3×x9×23
⇒133⋅x2×3×x3×2
⇒127⋅x6×x6
⇒127
(ii) (8x4)1/3÷x1/3
⇒(23⋅x4)1/3÷x1/3
⇒23⋅13⋅x4⋅13x13
⇒2⋅x4/3x1/3
⇒2⋅x4×13−x13
⇒x13[2x4−1]
Question 10
(i) (32)0+3−4×36+(13)−2
⇒30+3−4+6+13−2
⇒30+3-4+6+32
⇒1+132+32
⇒1+9+9
⇒19
(ii) 952−3⋅(5)0−(181)−12
⇒952−3(1)−(192)−12
⇒952−3−192×−12
⇒32×52−3−19−1
⇒35-3-9
⇒35-3-32
⇒3(34-1-3)
⇒3(81-1-3)
⇒3(77)
⇒231
Question 11
(i) 1634+2(12)−1⋅30
⇒(24)34+2(12−1)⋅1
⇒24×34+2⋅2
⇒23+4
⇒8+4
⇒12
(ii) (81)34−(132)−25+(8)13(12)−1⋅(2)0
⇒(34)34−(125)−2/5+(23)13(12−1)
⇒33−125×−25+2(2)
⇒33−12−2+2(2)
⇒27-22+4
⇒27-4+4
⇒27
Question 12
(i) (64125)−23÷1(256625)14+(√253√64)0
⇒(4353)−2/3÷1(4454)14+1
⇒(45)3×−23÷1(45)4×14+1
⇒(45)−2÷1(45)+1
⇒(54)2(15)+1
⇒(54)2×(45)+1
⇒54+1
⇒5+44
⇒94
(ii) 5n+3−6×5n+19×5n−22×5n
⇒5n⋅53−6×5n⋅59×5n−22×5n
⇒5n[53−6×5]5n[9−4]
⇒125−305
⇒955
⇒19
Question 13
(i) [(64)23⋅2−2÷80]−12
⇒((43)23⋅122÷1)−1/2
⇒(4222)−1/2
⇒(42)2×−12
⇒2-1
⇒12
(ii) 3n×9n+1 ÷ 3n-1×9n-1
⇒3n×32(n+1) ÷ 3n-1×32(n-1)
⇒3n×32n+2 ÷ 3n-1×32n-2
⇒3n+2n+23n−1+2n−2
⇒33n+233n−3
⇒33n⋅3233n⋅3−3
⇒32×3+3
⇒32+3
⇒35
⇒243
Question 14
(i) √22×4√2563√64−(12)−2
⇒(22)1/2×(44)1/4(43)13−12−2
⇒2×44−22
⇒2-4
⇒-2
(ii) 3−67×4−37×937×26722+20+2−2
⇒3−67×32⋅37×22×−37×2674+1+122
⇒3−67×367×2−67×2−674+1+14
⇒3−67+67×2−67+6716+4+14
⇒30×20(214)
⇒1(214)
⇒421
Question 15
(i) (32)25×(4)−12×(8)132−2÷(64)−13
⇒(25)−25×(27)−12×(23)13122÷(43)−13
⇒2−2×2−1×21122÷(4−1)
⇒2−1−2+1(122)(122)
⇒2-2
⇒122
⇒14
(ii) 52(x+6)×25−7+2x(125)2x
⇒52x+12×52(−7+2x)(53)2x
⇒52x+12×5−14+4x56x
⇒52x+12−14+4x56x
⇒56x−256x
⇒56x⋅5−256x
⇒5-2
⇒152
⇒125
Question 16
(i) 72n+3−49n+2((343)n+1)2/3
⇒72n+3−72(n+2)(73(n+1))2/3
⇒72n+3−72n+472(n+1)
⇒72n+3−72n+472n+2
⇒72n⋅73−72n⋅7472n⋅n2
⇒22x[73−74]72n⋅72
⇒343−240149
⇒−205849
⇒-42
(ii) (27)4/3+(32)0.8+(0.8)−1
⇒(33)43+(25)810+(810)−1
⇒34+25×45+(45)−1
⇒34+24+54
⇒81+16+54
⇒97+54
⇒388+54
⇒3934
Question 17
(i) (√32−√5)13⋅(√32+√5)13
⇒[(√25−√5)⋅(√25+√5)]13
⇒[(√25)2−(√5)2]13
⇒[(√25)2−(√5)2]13
⇒(25−5)13
⇒(32−5)13
⇒(27)13
⇒(33)13
(ii) (x13−x−13)(x23+1+x−23)
⇒(x1/3−1x1/3)(x2/3+1+1x2/3)
∵It is in the form of (a-b)(a2+ab+b2)=a3-b3
∴Here a=x13;b=1x13
∴(x1/3)3−(1x1/3)3
⇒x3/3−1x3/3
⇒x−1x
Question 18
(i) (xmxn)l−(xnxl)m⋅(xlxm)n
⇒(xm-n)l.(xn-l)2.(xl-n)n
⇒xml-nl.xmn-nl.xnl-nm
⇒xml-nl+mn-ml+nl-nm
⇒x0
⇒1
(ii) (xa+bxc)a−b⋅(xb+cxa)b−c⋅(xc+axb)c−a
⇒x(a+b)(a−b)xc(a−b)⋅x(b+c)(b−c)xa(b−c)⋅x(c+a)(c−a)xb(c−a)
⇒xa2−b2xac−bc⋅xb2−c2xab−ac⋅xc2−a2xbc−ab
⇒xa2−b2+b2−c2+c2−a2xac−bc+ab−ac+bc−ab
⇒x0x0
⇒1
Question 19
(i) lm√xlxm×mn√xmxn.nl√xnxl
⇒1m√x1−m⋅mn√xm−n⋅nl√xn−t
⇒(xl−m)1lm⋅(xm−n)1nm(xn−l)1nl.
⇒xl−mlm⋅xm−nnm⋅xn−lnl
⇒xl−mlm+m−nnm+n−lnl
⇒xn(l−m)+l(m−n)+m(n−l)lmn
⇒xnl−nm+lm−ln+mn−lnlmn
⇒x0
⇒1
(ii) (xaxb)a2+ab+b2⋅(xbxc)b2+bc+c2⋅(xcxa)c2+ac+a2
⇒x(a−b)(a2+ab+b2).x(b−c)(b2+bc+c2).x(c−a)(c2+ac+a2)
⇒xa3−b3⋅xb3−c3⋅xc3−a3⋅
⇒xa3−b3+b3−c3+c3−a3
⇒x0
⇒1
(iii) (xax−b)a2−ab+b2⋅(xbx−c)b2−bc+c2⋅(xcx−a)c2−ac+a2
⇒x(a−c−b))(a2−ab+b2).x(b−c−c))(b2−bc+c2).x(c−(−a))(c2−ac+a2)
⇒x(a+b)(a2−ab+b2).x(b+c)(b2−cb+c2).x(c+a)(c2−ac+a2)
⇒xa3+b3⋅xb3+c3⋅xc3+a3
⇒xa3+b3+b3+c3+c3+a3
⇒x2a3+2b3+2c3
⇒x2(a3+b3+c3)
Question 20
(i) (a-1+b-1)÷(a-2-b-2)
⇒(1a+1b)÷(1a2−1b2)
⇒(b+aab)÷(b2−a2a2b2)
⇒(b+a)ab(b2−a2)a2b2
⇒(b+a)ab×(ab)2(b2−a2)
⇒b+aab⋅(ab)2(b+a)(b−a)
⇒abb−a
(ii) 11+am−n+1an−m+1
⇒11+am−n+11+a−(m−n)
⇒11+am−n+11+1am−n
⇒11+am−n+1am−n+1am−n
⇒11+am−n+am−nam−n+1
⇒1+am−n1+am−n
⇒1
Question 21
(i) (a+b)−1(a−1+b−1)=1ab
L.H.S⇒(a+b)-1(a-1+b-1)
⇒1a+b(1a+1b)
⇒1a+b(b+aab)
⇒1a+b(a+bab)
⇒1ab
⇒R.H.S
(ii) x+y+zx−1y−1+y−1z−1+z−1x−1=xyz
Sol :
L.H.S⇒x+y+zx−1y−1+y−1z−1+z−1x−1
⇒x+y+z1xy+1yz+1xz
⇒x+y+zz+x+yxyz
⇒x+y+z(x+y+z)xyz
⇒xyz
⇒R.H.S
Question 22
Sol :
Given :
a=cz ; b=ax ; c=by
⇒a=cz
⇒a=(by)z (∵c=by)
⇒a=(ax)yz
⇒a1=axyz
∴Bases are equal ; so exponents are also equal
∴xyz=1
Hence proved
Question 23
Sol :
Given :
a=xyp-1 ; b=xyq-1 ; c=xyr-1
L.H.S⇒aq-r.br-q .cp-q
⇒(xyp-1)q-r.(xy)q-1(r-p).(xyr-1)p-q
⇒xypq-pr-q+r.xyqr-qp-r+p.xyrp-rq-p+q
⇒xypq-pr-q+r+qr-qp-r+p+rp-rq-p+q
⇒xy0
⇒1
⇒R.H.S
∴Hence proved
Question 24
Sol :
Given : 2x=3y=6-z
Let 2x=3y=6-z=k
⇒2=k1/x
⇒3=k1/y
⇒6=k−1/2
⇒16=k1/2
⇒12×3=k1/2
⇒1k1x.⋅k1y=k1/2
⇒1=k1z⋅k1x⋅1y
⇒k=k1x+1y+1z
∴1x+1y+12=0
Question 25
Sol :
Given : 2x=3y=6z
Let 2x=3y=6z=k
⇒2=k1/x
⇒3=k1/y
⇒12=k1/z
⇒22⋅3=k1/z
⇒(k1/x)2⋅k1/y=k1/z
⇒k2x⋅k1y=k1z
⇒2x+1y=1z
⇒2y+xxy=1z
⇒2x=1z−1y
⇒2x=y−zyz
⇒x=2yzy−z
∴Hence proved
Question 26
(i) (3x2)0
Sol :
⇒1
(ii) (xy)-2
⇒1(xy)2
⇒1x2y2
(iii) (-27aa)2/3
⇒-(33aa)2/3
⇒-(3a3)3 × 2/3
⇒-(3a3)2
⇒-32a3×2
⇒9a6
Question 27
Given : a=3 ; b=-2
(i) aa+bb
⇒33+(-2)-2
⇒33+1(−2)2
⇒27+14
⇒108+14
⇒1094
⇒3-2+(-2)3
⇒132−8
⇒19−8
⇒1−729
⇒−719
Question 28
Given : x=103×0.0099 ;
y=10-2×110
⇒√xy⇒√103×0.009910−2×110
⇒=√103+2×0.0099110
⇒⇒√105×0.0099110
⇒√3
⇒3
Question 29
Given : x=9 ; y=2 ; z=8
⇒x12⋅y−1⋅z23
⇒912⋅2−1⋅823
⇒(32)12⋅(12)(23)23
⇒312⋅22
⇒3⋅12⋅4
⇒6
Question 30
Given : x4y2z3=49392
2493922246962123482617433087310297343749771
⇒x4y2z3=243273
∵x,y,z are different primes
∴x=2 ; y=3 ; z=7
Question 31
Given : 3√a6b−4=ax⋅b2y
⇒(a6b−4)13=axb2y
⇒a63⋅b−43=axb2y
∴x=63
⇒x=2
⇒2y=−43
⇒y=−43×2
⇒y=−23
Question 32
Given :
⇒(p+q)-1(p-1+q-1)=paqb
⇒1p+q(1p+1q)=pa⋅qb
⇒1p+q(q+pqp)=pa⋅qb
⇒1qp=paqb
⇒(qp)-1=pa.qb
⇒p-1.q-1=pa.qb
a=-1
b=-1
∴L.H.S⇒a+b+z
⇒-1+2-1
⇒0
=R.H.S
Question 33
Sol :
Given :
⇒(p−1q2p2q−4)7÷(p3q−5p−2q3)−5=pxqy
⇒(p−7⋅q2×7p2×7q−4×7)÷(p3×5⋅q−5x−5p−2×−5q3×−5)=pxqy
⇒(p−7q14p14q−28)÷(p15⋅q25p10q−15)=px⋅qy
⇒(p-7-14q14+28)÷(p15-10.q25+15)=pxqy
⇒(p-21.q42)÷(p5.q40)=px.qy
⇒(p−21⋅q42p5⋅q40)=px⋅qy⇒
⇒(p-21-5.q42-20)=px.qy
⇒(p-26.q2)=px.qy
∴x=-26 ; y=2
∴x+y=-26+2
=-24
Question 34
(i) 52x+3=1
⇒52x+3=50 (∵50=1)
∴2x+3=0
⇒2x=-3
⇒x=−32
(ii) (13)√x=44−34−6
⇒(13)√x=256−81−6
⇒(13)√x=169
⇒(13)√x=(13)2
⇒(13)√x=132
⇒∴√x=2
⇒x=212
(iii) (√35)x+1=12527
⇒(35)x+12=5333
⇒(35)x+12=(53)3
⇒(35)x+12=(35)−3
∴x+12=−3
⇒x+1=-6
⇒x=-6-1
⇒x=-7
(iv) (3√4)2x+12=132
⇒[(22)13]4x+12=132
⇒(223)4x+12=125
⇒24x+13=2−5
⇒4x+13=−5
⇒4x+1=-15
⇒4x=-15-1
⇒4x=-16
⇒x=−164
⇒x=-4
Question 35
(i) √pq=(qp)1−2x
⇒(pq)12=(pq)−(1−2x)
⇒(pq)12=(pq)−1+2x
⇒12=−1+2x
⇒−1+2x=12
⇒2x=12+1
⇒2x=1+22
⇒2x=32
⇒x=32×2
⇒x=34
(ii) 4x−1×(0.5)3−2x=(18)x
⇒22(x−1)×(12)3−2x=(123)x
⇒22x−2×123−2x=123x
⇒22x−2×22x−3=2−3x
∴2x-2+2x-3=-3x
⇒4x-5=-3x
⇒4x+3x=5
⇒7x=5
⇒x=57
Question 36
Given : 53x=125
⇒104=0.001
⇒53x=125
⇒53x=53
⇒3x=3
⇒x=33
⇒x=1
∵10y=0.001
⇒10y=(11000)
⇒10y=(1103)
⇒10y=10-3
⇒y=-3
∴x=1 , y=-3
Question 37
Given : 9n32⋅3n−27n33m⋅23=127
⇒32n323n−33n33m⋅23=133
⇒32n+2+n−33n33m⋅8=133
⇒33n+2−33n33m⋅8=133
⇒33n(32−1)33m⋅8=133
⇒33n(9−1)33m⋅8=133
⇒33n⋅833m⋅8=133
⇒33n.33=33m
⇒33(n+1)=33m
⇒m=n+1
Question 38
Given : 34x=(81)-1
⇒34x=(34)-1
⇒34x=3-4
⇒4x=-4
⇒x=−44
⇒x=-1
and
⇒101y=0.0001
⇒101y=110000
⇒101y=1104
⇒101y=10−4
⇒1y=−4
⇒y=−14
Question 39
Given : 3x+1=9x-2
⇒3x+1=32(x-2)
⇒3x+1=32x-4
⇒x+1=2x-4
⇒2x-x=1+4
⇒x=5
⇒21+x=21+5
=26
=64
Question 40
(i) 3(2x+1)-2x+2+5=0
⇒3.2x+3-2x.22+5=0
⇒3.2x+3-2x.4+5=0
⇒3.2x-4.2x+8=0
⇒-2x+8=0
⇒2x=8
⇒2x=23
⇒x=3
(ii) 3x=9.3y
⇒3x=32.3y
⇒3x=32+y
⇒x=2+y
and 8.2y=4x
⇒232y=22x
⇒3+y=2x
⇒x=3+y2
Comments
Post a Comment