DAV Class 8 Maths Solution Chapter 3 Exponents and Radicals
DAV School Books Class 8 Maths Solution Chapter 3 exponents and radicals all Question Answer. DAV Class 8 3rd chapter exponents and radicals full Chapter explanation by expert teacher.
DAV School Books Class 8 Maths Solution Chapter 3 Exponents and Radicals:
Worksheet 1
1) Express each of the following in exponential form:
i)5√3
ii)11√(272)
iii)7√(11/3)
iv)∛(2/5)-3
v)13√(111)3
vi)7√(29)2
vii)∛(2)-6
viii)7√(15/341)-3
Ans.
i)5√3 = 3(1/5)
ii)11√(272) = 11√36=3(6/11)
iii)7√(11/3) = (11/3)(1/7)
iv)∛(2/5)-3=(2/5)-(3/3)=(2/5)-(1)
v)13√(111)3=(111)3/13
vi)7√(29)2=(29)2/7
vii)∛(2)-6 = 2-(6/3) = 2-2
viii)7√(15/341)-3 = (15/341)(-3/7)
2) Express each of the following as radicals:
i)(21)(1/8)
ii)(25)(3/4)
iii)(2/9)(1/9)
iv)(100)(-5/6)
v)(-221)(1/5)
vi)(17/231)(-5/6)
vii)(15/21)(2/5)
Ans.
i)(21)(1/8) = 8√21
ii)(25)(3/4)= 4√253
iii)(2/9)(1/9)= 9√(2/9)
iv)(100)(-5/6)= 6√(100)(-5)
v)(-221)(1/5)= 5√(-221)
vi)(17/231)(-5/6)= 6√(17/231)(-5)
vii)(15/21)(2/5)= 5√(15/21)2
3) Express each of the following with positive indices:
i) x(-1/2)
ii)x(-2/5)
iii)7/{x(-5/6)}
iv)(x-3)4
Ans.
i) x(-1/2)= (1/x)(1/2)
ii)x(-2/5)= (1/x)(2/5)
iii)7/{x(-5/6)}= 7x(5/6)
iv)(x-3)4=(1/x)12
Worksheet 2
1) Simplify:
i) x(1/2) × x(5/2)
ii) [x(6/5)]/[x(1/5)]
iii) (x7)0
iv) [5x(5/6)] × [6x(1/6)]
v) [x(-7/2)] × [2x(-1/2)]
Ans.
i) x(1/2) × x(5/2)
= x(1/2)+(5/2)
= x3
ii) [x(6/5)]/[x(1/5)]
= x(6/5)+(1/5)
= x(7/5)
iii) (x7)0 = 1
iv) [5x(5/6)] × [6x(1/6)]
= 30x(5/6)+(1/6)
= 30x(6/6)=30x
v) [x(-7/2)] × [2x(-1/2)]
= 2x(-7/2)+(-1/2)
= 2x-(4)
= 2/x4
2) Find the value:
i) (512)(-2/9)
ii) [(216)(2/3)](1/2)
iii) 10÷8(-1/3)
iv) 16(3/4)
v) [27(1/3)] × [16(-1/4)]
vi) 1/[(34)2]-2
vii) [27(-2/3)] × [81(5/4)]/(1/3)-3
viii) 64(1/2) × (64(1/2)+1)
ix) [36(7/2) – 36(9/2)]/36(5/2)
x) 4 × 81(-1/2) ×(81(1/2)+81(3/2))
Ans.
i) (512)(-2/9)
= 9√512(-2)
= 2(-2)
= 1/22
= 1/4
ii)[(216)(2/3)](1/2)
= [62](1/2)
= [36](1/2)
= 6
iii)10÷8(-1/3)
= 10/8(-1/3)
=10×8(1/3)
= 10×2
= 20
iv)16(3/4)
= (4√16)3
= 23
= 8
v)[27(1/3)] × [16(-1/4)]
= ∛27 × (1/4√16)
= 3 × (1/2)
= 3/2
vi)1/[(34)2]-2
= 1/3-16
= 1 × 43046721
= 43046721
vii)[27(-2/3)] × [81(5/4)]/(1/3)-3
= [(∛27)(-2) × (4√81)5]/33
= (35/32)/33
=35-2-3
=30
=1
viii)64(1/2) × (64(1/2)+1)
= √64×(√64 + 1)
= 8×9
= 72
ix)[36(7/2) – 36(9/2)]/36(5/2)
= [(√36)7– (√36)9]/ (√36)5
= [67/65]- [69/65]
= 62– 64
=36 – 1296
= 1260
x)4 × 81(-1/2) ×(81(1/2)+81(3/2))
= 4 × (1/√81) × [√81+(√81)3]
=4 × 1/9 × (9+729)
= 4×738/9
= 4×82
= 328
3) Evaluate:
i) (0.04)(3/2)
ii) (6.25)(3/2)
iii) (0.03125)(-2/5)
iv) (0.008)(2/3)
Ans.
i) (0.04)(3/2)
= (√0.04)3
= (0.2)3
= 0.008
ii) (6.25)(3/2)
= (√6.25)3
= 2.53
= 15.625
iii) (0.03125)(-2/5)
= (5√0.03125)(-2)
= (0.5)(-2)
= 1/0.25
= 1/(1/4)
= 4
iv) (0.008)(2/3)
= (∛0.008)2
= 0.22
= 0.04
4) Evaluate:
i) (62+82)(1/2)
ii)[5(8(1/3)+27(1/3))3](1/4)
iii) (172 – 82)(1/2)
iv)(13+23+33)(-5/2)
Ans.
i) (62+82)(1/2)
= √(62+82)
= √(36+64)
= √100
= 10
ii) [5(8(1/3)+27(1/3))3](1/4)
= [5×(∛8 + ∛27)3] (1/4)
= [5×(2 + 3)3] (1/4)
= [5 × 53] (1/4)
= 54×(1/4)
= 5
iii) (172 – 82)(1/2)
= √(289 – 64)
= √(225)
= 15
iv) (13+23+33)(-5/2)
= (1+8+27)(-5/2)
= 36(-5/2)
= (1/6)5
= 1/7776
5) Simplify and express the answers with positive indices:
i)2x(1/6)×2x(-7/6)
ii)[4√{(1/x)(-12)}](-2/3)
iii) a(4/7)+a(10/7)
Ans.
i)2x(1/6)×2x(-7/6)
= 4x(1/6)+(-7/6)
= 4x(-6/6)
= 4x(-1)
= 4/x
ii)[4√{(1/x)(-12)}](-2/3)
= 4√(x12)(-2/3)
= 4√x(-24/3)
= x(-24/12)
= x-2
= 1/x2
iii) a(4/7)+a(10/7)
It cannot be simplified any further.
6) Verify that
[729(-5/3)](-1/2) = (729)(-5/3)×(-1/2)
Ans.
[729(-5/3)](-1/2)
=[∛(729)-5](-1/2)
= [9(-5)](-1/2)
= [1/95](-1/2)
= (1/59049)(-1/2)
= √59049
=243
Now let’s simplify (729)(-5/3)×(-1/2)
(729)(-5/3)×(-1/2)
=(729)(5/6)
= (36)(5/6)
= 35
=243
So we can say that [729(-5/3)](-1/2) = (729)(-5/3)×(-1/2).
Solve the given exponential equations.
i) (√6)x-2 = 1
ii) 34x=(1/81)
iii) (√2)x= 28
iv)2(2x+1) = 4(2x-1)
Ans.
i) (√6)x-2 = 1
or, (√6)x-2 = (√6)0
or, x-2 = 0
or, x = 2
ii) 34x=(1/81)
or, 34x=(3)(-4)
or, 4x = -4
so, x = -1
iii) (√2)x= 28
or, 2(x/2)=28
or, x/2 = 8
or, x = 16
iv)2(2x+1) = 4(2x-1)
or, 2(2x+1) = 22(2x-1)
or, 2x+1 = 2(2x-1)
or, 2x+1 = 4x-2
or, 2x = 3
so, x = 3/2
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