Class 8 ICSE Maths Solutions Chapter 2

Class 8 ICSE Maths Solutions Chapter 2 Exponents (Powers) (Selina Concise)

Welcome to NCTB Solutions. Here with this post we are going to help 8th class students for the Solutions of Selina Class 8 ICSE Math Book, Chapter 2, Exponents (Powers). Here students can easily find step by step solutions of all the problems for Exponents (Powers), Exercise 2A and 2B Also here our mathematics teacher’s are solved all the problems with easily understandable methods with proper guidance so that all the students can understand easily. Here in this post students will get chapter 2 solutions. Here in this post all the solutions are based on ICSE latest Syllabus.

Exponents (Powers) Exercise 2(A) Solution :

Question no – (1)

Solution :

(i) (3–1 × 9–1) ÷ 3–2

= (1/3 × 1/9) ÷ 1/32

= 1/27 × 9

= 1/3

(ii) (3-1× 4-1) ÷ 6-1

= (1/3 × 1/4) ÷ 1/6

= 1/12 × 6

= 1/2

(iii) (2-1 + 3-1

= (1/2 + 1/3)³

= (3 + 2/6)³

= (5/6)³

= 125/216

(iv) (3-1 ÷ 4-1

= (1/3 ÷ 1/4)²

= (1/3 × 4)²

= 16/9

(v) (2² + 3²) × (1/2)²

= (4 + 9) × 1/4

= 13/4

(vi) (5² – 3²) × (2/3)-3

= (25 – 9) × (3/2)³

= 16 × 27/8

= 54

(vii) [(1/4)-3 – (1/3)-3] ÷ (1/6)-3

= (4 × 4 × 4 – 3 × 3 × 3) ÷ 216

= 37 × 1/216

= 37/216

(viii) [(- 3/4)-2

= (- 3/4)-4

= (4/3) ⁴

= 256/81

(ix) {(3/5)-2}-2

= (3/5)⁴

= 81/625

(x) (5-1 × 3-1) ÷ 6-1

= (1/5 × 1/3) ÷ 1/6

= 1/15 × 6

= 2/5

Question no – (2)

Solution :

As per the question,

1125 = 3m × 5n

∴ 3² × 5³ = 3m × 5n

Now, confirming,

m = 2 and n = 3.

Question no – (3)

Solution :

According to the question,

9 × 3x = (27)2x-3

= 3² × 3x = (3³)2x-3

= 3x = 2 = 36x -9

Now, x + 2 = 6x – 9

=  x – 6x = – 9 – 2

= – 5x = – 11

= x = 11/5

Exponents (Powers) Exercise 2(B) Solution :

Question no – (1)

Solution :

(i) Given,  1⁸ × 30 × 5³ × 2²

= 1 × 1 × 125 × 4

= 500

(ii) (4⁷)² × (4-3)⁴

= 4^14 × 4^- 12

= 4^14 – 12

= 4²

= 16

(iii) (2^-9 ÷ 2^-11)^3

= (2^ – 9/2- 11)^3

= (2²)³

= 2⁶

= 64

(iv) (2/3)-4 × (27/8)-2

= 2-4/3-4 × 5-6/2-6

= 2²/3²

= 4/9

(v) (56/28)0 ÷ (2/5)3 × 16/25

= 1 × 53/23× 24/52

= 10

(vi) (12)-2 × 3³

= (2² × 3)-2 × 3³

= 2-4 × 3-2 + 3

= 2-4 × 3-1

= 3/2⁴

= 3/16

(vii) (-5)⁴ × (-5) ⁶ ÷ (-5)⁹

= – 5⁴ + 6 – 9

= (-5)^1

= -5

(viii) (-1/3)4 ÷ (- 1/3)8  × (-1/3)5

= (- 1/3)^4 + 5 – 8

= (- 1/3)^1

= -1/3

(ix) 9^0 × 4^-1 ÷ 2^-4

= 1 × 1/4 × 2⁴

= 4

(x) (625)^-3/4

= (5)^4×-3/4

= (5)^-3

= 1/53

= 1/125

(xi) (27/64)^-2/3

= (3)^3×-2/3/(4)^3×-2/3

= 4²/3²

= 16/9

(xii) (1/32)^-2/5

= (1/25)^-2/5

= 1/2-2

= 4

(xiii) (125)^-2/3 ÷ (8)^2/3

= (5)^3×-2/3 ÷ (2)^3×2/3

= 5^-2 ÷ 22

= 1/25 × 1/4

= 1/100

(xiv) (243)^2/5 ÷ (32)^-2/5

= (3)^5×2/5 ÷ (2)^5×-2/5

= 3² ÷ 2^-2

= 9 × 2²

= 36

(xv) (-3)^4 – (4√3)^0 × (-2)^5 ÷ (64)^2/3

= 3⁴ + 2⁵ ÷ 2⁴^6 × 2/3

= 3⁴ + 2⁵ ÷ 2⁴

= 3⁴ + 2⁵/2⁴

= 81 + 2^5 – 4

= 81 + 2

= 83

(xvi) (27)^2/3 ÷ (81/16)^-1/4

= (3)^ 3 × 2/3 ÷ (3/2)^ 4 × – 1/4

= 3² ÷ 3^-1/2^-1

= 27/2

Question no – (2)

Solution :

(i) 8^4/3 + 25^3/2 – (1/27)^-2/3

= (2)^3 × 4/3 + (5)^ 2 × 3/2 – (1/3^3)^-2/3

= 2⁴ + 5³ – 3²

= 16 + 125 – 9

= 132

(ii) [(64)^-2]^-3 ÷ [{(- 8)²}³]²

= (2⁶)⁶ ÷ (- 8)^1

= 2^36 ÷ (- 8)^12

= 2^36/(-2)^36

= 2^36/2^36

= 1

Question no – (3)

Solution :

(i) (- 5)0

= 1

(ii) 80 + 40 + 20

= 1 + 1 + 1

= 3

(iii) (8 +4 + 2)0

= 1

(iv) 4x0

= 4× 1

= 4

(v) (4x)0

= 1

(vi) [(10³)0]⁵

= 1

(vii) (7×0)²

= 49

(viii) 9^0 + 9^- 1 – 9^- 2 + 9^1/2 – 9^- ½

= 1 + 1/9 – 1/9² + (3²)^1/2 – (3)² × (- 1/2)

= 1 + 1/9 – 1/18 + 3 – 3 – 1

= 1 + 1/9 – 1/81 + 3/1 – 1/3

= 81 + 9 – 1 + 243 – 27/81

= 305/81

Question no – (4)

Solution :

(i) a⁵ b²/a² b^- 3

= a³ b⁵…(Simplified)

(ii) 15y⁸ ÷ 3y³

= 5y⁵…(Simplified)

(iii) x^10y⁶ ÷ x³y^- 2

= x^10 y⁶/ x³y^-2

= x⁷y⁸…(Simplified)

(iv) 5z^16 ÷ 5z^16

= 5/15  z^16/z^16

= 1/3z27…(Simplified)

(v) (36x^2)^1/2

= (6x)^2 × 1/2

= 6x …(Simplified)

(vi) (125x^-3)^1/3

= (5³)^1/3

= 5x-1…(Simplified)

(vii) (2x²y^-3)^-2

= 2^1/2x^-4 y⁶

= 1/4 × y⁶/x⁴

= y⁶/4x⁴…(Simplified)

(viii) (27x^-3y^6)^2/3

= (3³)^2/3 x^- 2 y⁴

= 3² x^-2y⁴

= 9x^-2y⁴

= 9y⁴/x²…(Simplified)

(ix) (- 2x^2/3 y^- 3/2)⁶

= (-2)⁶ x^2/3 × 6 y^ – 3/2 × 6

= 64x⁴ y-9

= 64x⁴/y⁹…(Simplified)

Question no – (5)

Solution :

Given, (x^a+b)^a-b. (x^b+c)^b-c. (x^c+a)^c-a

= x^a² – b², x^ b² – c², x^c² – a²

= x^a² – b²+ b² – c² + c² – a²

= x 0

= 1…(Simplified)

Question no – (6)

Solution :

(i) 5√x^20y^- 10 z⁵ ÷ x³/y³

= (x^20 y^- 10 z⁵)^1/5 ÷ x³/y³

= x^-3 . y^-2 + 3. 2^1

= xy2…(Simplified)

(ii) [256a^16/81b^4]^-3/4

= 4^4× – 3/4.a^16 × – 3/4 /3^4× – 3/4 . b ^4× – 3/4 a

= 3³ b³/4³ a^12

= 27b³/64 a^12

= 27/64 b³.a^-12…(Simplified)

Question no – (7)

Solution :

(i) (a^-2)^-2(ab)^-3

= a⁴ b^- 2 a ^- 3 b^ – 3

= a^4 – 3, b^- 2- 3

= a^1, b- 5…(Simplified)

(ii) (x^n y^-m) ^4 × (x^3 y^-2)^-n

= x ^4n, y – ^4m,…(Simplified)

(iii) (125a^- 3/y⁶)^- 1/3

= (5³ a^- 3/y⁶)^- 1/3

= 5^- 1 a^1/y – 2…(Simplified)

(iv) (32x^- 5/243y^- 5)^- 1/5

= (2⁵ x^- 5/3⁵ – y^- 5)^- 1/5

= 2^- 1. x^+ 1/3^- 1. y^+1

= 3x/2y…(Simplified)

(v) (a^-2b)^1/2 × (ab^-3)^1/3

= (a^-1. b^1/2. × a^1/3. b^-1)

= a ^-2/3. b ^ -1/2

= 1/a^2/3. b^1/2 …(Simplified)

(vi) (xy)^m-n. (yz)^n-1. (zy)^1-m

= x^m-n. y^m-n. y^n-1. z^n-1.x^1- m. z^1 – m

= x^m – n + 1 – m y^m – n + n – 1 z ^n – 1 + 1 – m

= x^- n. y^m – 1. z^n-m …(Simplified)

Question no – (8)

Solution :

Given in the question,

(x^a/x^-b)^a-b.  (x^b/x^-c)^b-c. (x^c/x^-a)^c-a = 1

L.H.S, (x^a +b)^a – b. (x^b+c)^b – c. (x^c + a) ^c – a

= x^a² – b². x^ b² – c². x^ c² – a²

= x^ a² – b² + b² – c² + c² – a²

= x 0

= 1

= L.H.S = R.H.S …(Proved)

Question no – (9)

Solution :

In the given question, x^5+n × (x^2)^3n+1 /x^7n-2

= x^5+n x^6n + 2/x^7n-2

= x^5 + n + 6n + 2 – 7n + 2

= x⁹

Question no – (10)

Solution :

Given in the question,

a^2n+1 × a^(2n+1)(2n-1)/a^n(4n-1) ×(a²)^2n+3

= a^2n + 1 × a^(2n + 1)(2n – 1)/a^n(4n – 1) ×(a²)^2n + 3

= a^2n + 1 a^(2n)² – (1)²/a^4n² – n . a^4n + 6

= a^2n + 1 + 4n² – 1 – 4n² + n – 4n – 6

= a^-n-6

Question no – (11)

Solution :

L.H.S (m + n)^-1 (m^-1 + n^-1) = (mn)^-1

= 1/m + n(1/m + 1/n)

= 1/m + n(n + m/mn)

= 1/mn

= (mn)^- 1

L.H.S = R.H.S …[Proved]

Question no – (13)

Solution :

(i) Given in the question, 12^-5 × 12^2n+1 = 12^13 ÷ 12^7

= 126-5 + 2n + 1 = 12^13/12^7

= 12^2n – 4 = 12^13 – 7 = 12^6

Comparing,

= 2n – 4 = 6

= n = 10/2

= 5

Thus, the value of n will be 5.

(ii) Given in the question, a^2n – 3 × (a²)^n + 1/(a⁴)^ – 3 = (a³)³ ÷ (a⁶)^- 3

= a^2n – 3 × a^2n + 2/a^- 12 = a⁹ ÷ a^18

= a^2n -3 + 2n + 2 – (-12) = a^9 – (- 18)

= a^4n + 11 = a27

Comparing,

= 4n + 11 = 27

= n = 16/4

= 4

Therefore, the value of n will be  4.

Question no – (14)

Solution :

(i) a^2n + 3. a^(2n+1)(n+2)/(a³)^2n+1. a^n(2n+1)

= a^2n – 3. a^2n + 4n + n + 2/a^6n + 3. a^2nc + n

= a ^2n + 3 + 2n² + 5n + 2/a^6n + 3 + 2n² + n

= a^2n² + 7x + 5/a^2n² + 7x + 3

= a²…(Simplified)

(ii) x^2n +7. (x^2)^3n+2/x^4(2n+3)

= x^2n + 7 + 6n + 4/8n + 12

= x^8n + 11 – 8n – 12

= x^-1

= 1/x…(Simplified)

Next Chapter Solution :

Updated: June 20, 2023 — 10:02 am