# Selina Concise Solutions for Chapter 2 Exponents (Power) Class 8 ICSE Mathematics

### Exercise 2 A

1. Evaluate:
(i) (3-1 × 9-1) ÷ 3-2
(ii) (3-1 × 4-1) ÷ 6-1
(iii) (2-1 + 3-1)3
(iv) (3-1 ÷ 4-1)2
(v) (22 + 32) × (1/2)2
(vi) (52– 32) × (2/3)-3
(vii) [(1/4)-3– (1/3)-3] ÷ (1/6)-3
(viii) [(- ¾)-2]2
(ix) {(3/5)-2}-2
(x) (5-1× 3-1) ÷ 6-1
Solution
(i) (3-1 × 9-1) ÷ 3-2
(1/3 × 1/9) ÷ 1/3 × 1/3
= 1/27 ÷ 1/9
= 1/27 ×  9/1
= 1/3

(ii) (3-1× 4-1) ÷ 6-1
= (1/3 × 1/4) ÷ 1/6
= 1/12 ÷ 1/6
= 1/12 × 6/1
= ½

(iii) (2-1 + 3-1)3
= (1/2 + 1/3)3
= {(1 × 3)/(2 × 3) + (1 × 2)/(3 × 2)}3
= {(3 + 2)3/6}
= (5/6)3
= (5 × 5 × 5)/(6 × 6 × 6)
= 125/216

(iv) (3-1 ÷ 4-1)2
= (1/3 ÷ ¼)2
= (1/3 ×  4/1)2
= (4/3)2
= 16/9
= 1 7/9

(v) (22 + 32) × (1/2)2
=  (2 × 2) + (3 × 3) × (1/2 × 1/2)
= 4 + 9 × 1/4
= 13/4
= 3 1/4

(vi) (52 - 32) × (2/3)-3
= (5 × 5) – (3 × 3) × (3/2)3
= 25 – 9 × (3/2 × 3/2 × 3/2)
= 16 × 27/8
= 54

(vii) [(1/4)-3  – (1/3)-3] ÷  (1/6)-3
= [(4/1)3 - (3/1)3] ÷ (6/1)3
= (4/1 × 4/1 × 4/1 -  3/1 × 3/1 × 3/1) ÷ (6/1)3
= 64 - 27 × (1/6 × 1/6 × 1/6)
= 37 × 1/216
= 37/216

(viii) [(- ¾)-2]2
= (- ¾)-2×2
= (- ¾)-4
= (4/3)4
= (4 × 4 × 4 × 4/3 × 3 × 3 × 3)
= 256/81
= 3 13/81

(ix) {(3/5)-2}-2
= (3/5)-2×(-2)
= (3/5)4
= (3 × 3 × 3 × 3)/(5 × 5 × 5 × 5)
= 81/625

(x) (5-1 × 3-1) ÷ 6-1
= (1/5 × 1/3) ÷ 1/6
= 1/15 ÷ 1/6
= 1/15 × 6/1
= 2/5

2. If 1125 = 3m × 5n; find m and n.
Solution
1125 = 32 × 53
The factors of 1125 are 3 × 3 × 5 × 5 × 5
∴ 1125 = 3 × 3 × 5 × 5 × 5
Now comparing, 32 × 53 = 3m × 5n
∴ m = 2 , n = 3

3. Find x, if 9 × 3x = (27)2x- 3

Solution 3:
9 × 3x = (27)2x - 3
⇒ 3× 3x = (3 × 3 × 3)2x - 3
⇒ 3x + 2 = (3)3(2x - 3)
⇒ 3x + 2 = (3)6x - 9
Since, bases are same, compare them,
x + 2 = 6x - 9
6x – x = 9 + 2
⇒ 5x = 11
⇒ x = 11/2
⇒ x = 2 1/5

### Exercise 2 B

1. Compute:
(i) 18 × 30 × 53 × 22
(ii) (47)2 × (4-3)4
(iii) (2-9 ÷ 2-11)3
(iv) (2/3)-4 × (27/8)-2
(v) (56/28)0 ÷ (2/5)3 × 16/25
(vi) (12)-2 × 33
(vii) (-5)4 × (-5)6 ÷ (-5)9
(viii) (-1/3)÷ (-1/3)8 × (-1/3)5
(ix) 90 × 4-1 ÷ 2-4
(x) (625)-3/4
(xi) (27)/64)-2/3
(xii) (1/32)-2/5
(xiii) (125)-2/3 ÷ (8)2/3
(xiv) (243)2/5 ÷ (32)-2/5
(xv)
(xvi) (27)2/3 ÷ (81/16)-1/4
Solution
(i) 18 × 30 × 53 × 22
= 1 × 1 × 5 × 5 × 5 × 2 × 2
= 125 × 4
= 500

(ii) (47)2  × (4-3)4
= 414 × 4-12
= 414-12
= 42
= 4 × 4
= 16

(iii) (2-9 ÷ 2-11)3
= (2-9/2-11)3
= (2-9+11)3
= (22)3
= 26
= 2 × 2 × 2 × 2 × 2 × 2
= 64

(iv) (2/3)-4 × (27/8)-2
= (2/3)-4 × (33/23)-2
= 2-4/3-4 × 3-6/2-6
= 2-4/2-6 × 3-6/3-4
= 2-4+6 × 1/3-4+6
= 22/32
= 4/9

(v) (56/28)0 ÷  (2/5)3 × 16/25
= 1 ÷ 23/53 × (2 × 2 × 2 × 2)/(5 ×  5) [∵ (56/28)0 = 1]
= 1 × 53/23 × 24/52
= 53-2 × 24-3
= 51 × 21
= 10

(vi) (12)-2 × 33
= (2 × 2 × 3)-2 × 33
= (22 × 3)-2 × 33
= 2-2×2 × 3-2 × 33
= 2-4 × 3-2+3
= 2-4 × 31
= 3/24
= 3/(2 × 2 × 2 × 2)
= 3/16

(vii) (-5)4 × (-5)6 ÷ (-5)9
= (5)4 × (-5)6 × 1/(-5)9
= (-5)4+6-9
= (-5)10-9
= -5

(viii) (- 1/3)4 ÷ (- 1/3)8 × (- 1/3)5
= (- 1/3)4 × 1/(- 1/3)8 × (- 1/3)5
= (- 1/3)4+5-8
= (- 1/3)9-8
= - 1/3

(ix) 90 × 4-1 ÷ 2-4
= 1 × 1/41 × 1/2-4
= 1 × ¼ × 24
= 1 × 1/22 × 24
= 24-2
= 22
= 4

(x) (625)-3/4
= (5 × 5 × 5 × 5)-3/4
= (54)-3/4
= 54× -3/4
= 5-3
= 1/53
= 1/(5 × 5 × 5)
= 1/125

(xi) (27/64)-2/3
= [(33)/(43)]-2/3
= (33× -2/3)/(43× -2/3)
= 3-2/4-2
= 42/32
= (4 × 4)/(3 × 3)
= 16/9
= 1 7/9

(xii) (1/32)-2/5
= (1/2 × 2 × 2 × 2 × 2)2/5
= (1/25)-2/5
= 1/(25× -2/5)
= 1/(2-2)
= 22
= 4

(xiii) (125)-2/3 ÷ (8)2/3
= (53)-2/3 ÷ (23)2/3
= 5-2/3×3 ÷ 23× 2/3
= 5-2 ÷ 22
= 1/52 × 1/22
= 1/25 × ¼
= 1/100

(xiv) (243)2/5 ÷ (32)-2/5
= (3 × 3 × 3 × 3 × 3)2/5 ÷ (2 × 2 × 2 × 2 × 2)-2/5
= (35)2/5 ÷ (25)-2/5
= 35× 2/5 ÷ 22/5 ×5
= 32 ÷ 2-2
= 32 × 1/(2-2)
= 32 × 2+2
= 3 × 3 × 2 × 2
= 36

(xv) (-3)4 – (∜3)× (-2)5 ÷ (64)2/3
= (-3 × -3 × -3 × -3)
= -1 × -2 × -2 × -2 × -2 ÷ (26)2/3
Note:  (∜(3)0  = 1
= 34 + (2)5 ÷  26× 2/3
= 34 + 25 ÷ 24
= 34 + 25/24
= 34 + 25-4
= 34 + 2
= 3 × 3 × 3 × 3 + 2
= 81 + 2
= 83

(xvi) (27)2/3 ÷ (81/16)-1/4
= (33)2/3 ÷ (34/24)-1/4
= 33× 2/3 ÷ (31/4 ×4)/(21/4 ×4)
=  32 ÷ 3-1/2-1
= 32 × 2-1/3-1
= 32+1 × 2-1
= 32× 1/(2+1)
= (3 × 3 × 3)/2
= 27/2
= 13 ½

2. Simplify:
(i) 84/3 + 253/2 – (1/27)-2/3
(ii) [(64)-2]-3 ÷ [{(-8)2}3]2
(iii) (2-3 – 2-4) (2-3+ 2-4)
Solution
(i) 84/3 + 253/2 – (1/27)2/3
= (23)4/3 + (52)3/2 – (1/33)-2/3
= (23× 4/3) + (52× 2/3) – 1/(33× -2/3)
= 2+ 5- 1/3-2
= 16 + 125 - 32
= 141 - 9
= 132

(ii) [(64)-2]-3  ÷ [{(-8)2}3]2
= (26)-2×3 ÷ (-8)2
= 26x(6)  ÷ (-8)12
= (2)+36 ÷ (-8)12
(2)+36 ÷ [(-2)3]12
= (2)36 ÷ (-2)36
(2)36/(-2)36
= (2)36/(2)36 (∵ 36 is  even)
= (2)36 - 36
= (2)0 = 1 (∵ a= 1)

(iii) (2-3 - 2-4)(2-3 + 2-4)
= (2-3)- (2-4)2
{∵ (a - b)(a + b) = a2- b2}
= 2-6 - 2-8
= 1/2- 1/28
= 1/64 - 1/256
= (4 - 1)/256
= 3/256

3. Evaluate :
(i) -50
(ii) 8+ 40 + 20
(iii) (8 + 4 + 2)0
(iv) (4x)0
(v) [(103)0]5
(vi) (7x0)2
(vii) 9+ 9-1 - 9-2 + 91/2 - 9-1/2
Solution
(i) (-5)= 1 (∵ a0 = 1)

(ii) 8+ 4+ 20
= 1 + 1 + 1
= 3 (∵  a0 = 1)

(iii) (8 + 4 + 2)0
= (14)0
= 1 (∵ a0 = 1)

(iv) 4x0
= 4 × 1
= 4

(v) [(103)0]5
= (103×0×5
= 100
= 1

(vi) (7x0)2
= 7× x2
= 49 × 1
= 49

(vii) 9- 9-1 - 9-2 + 91/2 - 9-1/2
1 + 1/9 - 1/9+ (32)1/2 - (32)-1/2
1 + 1/9 - 1/81 + 31/2 - 3(-1/2)
= 1 + 1/9 - 1/81 + 3 - 3-1
= 1 + 1/9 - 1/81 + 3/1 - 1/3
= (81 + 9 - 1 + 243 - 27)/81
= (333 - 28)/81
= 305/81
= 3 62/81

4. Simplify:
(i) (a5b2)/(a2b-3)
(ii) 15y8  ÷ 3y3
(iii) x10y6 ÷ x3y-2
(iv) 5z16 ÷ 15z-11
(v) (36x2)1/2
(vi) (125x-3)1/3
(vii) (2x2y-3)-2
(viii) (27x-3y6)2/3
(ix) (-2x2/3y-3/2)6
Solution
(i) (a5b2)/(a2b-3)
= a5-2. b2+3
= a3b5

(ii) 15y8 ÷ 3y3
= 15y8/3y3
= 5y8-3
= 5y5

(iii) x10y6 ÷ x3y-2
= x10y6/x3y-2
= x10-3.y6+2
= x7y8

(iv) 5z16 ÷ 15z-11
= 5z16/15z-11
= 5/15z16+11
= 1/3z27

(v) (36x2)1/2
= (36)1/2. x2× 1/2
= (6 × 6)1/2.x
= (62)1/2.x
= 6x

(vi) (125-3)1/3
= (125)1/3. x-3× 1/3
= (5 × 5 × 5)1/3. x-1
= (53)1/3.x-1
= 5x-1

(vii) (2x2y-3)-2
= 2-2. x2×-2.y-3×-2
= 1/22x-4.y6
= ¼ × y6/x4
= y6/4xy4
= ¼.y6.x-4

(viii) (27x-3y6)2/3
= (27)2/3. x-3× 2/3. y6× 2/3
= (3 × 3 × 3)2/3. x-2. y4
= [(3 × 3 × 3)1/3]2. x-2. y4
= 32.x-2y4
= 9x-2y4
= 9y4/x2
= 9x-2 y4

(ix) (-2x2/3y-3/2)6
= (-2)6. x2/3 ×6. y-3/2 ×6
= 64x4y-9
= 64x4/y9
= 64x4y-9

5. Simplify:
(xa+b)a-b.(xb+c)b-c.(xc+a)c-a
Solution
(xa+b)a-b.(xb+c)b-c.(xc+a)c-a
= x(a+b)(a-b). x(b+c)(b-c). x(c+a)(c-a)
= xa2-b2. xb2-c2. xc2-a2
= xa2 - b2+b2-c2+c2-a2
= x0
= 1

6. Simplify:
(i)
(ii) [256a16/81b4]-3/4
Solution

(i)
= (x20y-10z5)1/5 ÷ x3/y3
= x20× 1/5. y-10× 1/5. z5× 1/5 ÷ x3/y3
= x4. y-2. z-1 × y3/x3
= x4-3. y-2+3. z1
= xyz

(ii) [256a16/81b4]-3/4
= [44a16/34b4]-3/4
= (44×-3/4).(a16×-3/4)/(34×-3/4).(b4×-3/4)
= (4-3.a-12)/(3-3.b-3)
= (33b3)/(43a12)
= 27b3/64a12
= 27/64. a-12b3

7. (i) (a-2)-2.(ab)-3
(ii) (xny-m)4 × (x3y-2)-n
(iii) (125a-3)/(y6)-1/3
(iv)(32x-5)/(243y-5)-1/5
(v) (a-2b)1/2 × (ab-3)1/3
(vi) (xy)m-n.(yz)n-1.(zx)1-m
Solution
(i) (a-2b)-2.(ab)-3
= (a2×-2.b-2).(a-3.b-3)
= a+4. b-2. a-3. b-3
= a4-3. b-2-3
= ab-5
= a/b5

(ii) (xny-m)4 × (x3y-2)-n
= x4ny-4m× x-3ny2n
= x4n-3n.y-4m+2n
= xny-4m+2n

(iii) [125a-3/y6]-1/3
= [53a-3/y6]-1/3
= (53×-1/3.a-3×-1/3)/(y6x-1/3)
= 5-1.a1/y-2

(iv) [32x-5/243y-5]-1/5 = [25x-5/35y-5]-1/5
=  {(25×-1/5).(x-5×-1/5)}/{(35×-1/5)(5-5×-1/5)}
= (2-1x+1)/(3-1y+1)
= 3x/2y

(v) (a-2b)1/2 × (ab-3)1/3
= (a-2×1/2.b1/2) × (a1/3.b-3×1/3)
= a-1b1/2 × a1/3b-1
= a-1+1/3.b1/2-1
= a-2/3b-1/2
= 1/(a2/3/b1/2)

(vi) (xy)m-n.(yz)n-1(xz)l-m
= xm-n.ym-n.yn-1.zn-1.xl-m.zl-m
= xm-n+l-m.ym-n+n-l.zn-l+l-m
= xl-n.ym-l.zn-m

8. Show that:
(xa/x-b)a-b.(xb/x-c)b-c.(xc/x-a)c-a = 1
Solution
L.H.S. = (xa/x-b)a-b. (xb/x-c)b-c.(xc/x-a)c-a
= (xa+b)a-b. (xb+c)b-c.(xc+a)c-a
= x(a+b)(a-b). x(b+c)(b-c). x(c+a)(c-a)
= xa2-b2. xb2-c2. xc2-a2
= xa2-b2+b2-c2+c2-a2
= x0
= 1 = R.H.S

9. Evaluate:
(x5+n)(x2)3n+1/(x7n-2)
Solution
(x5+n)× x2(3n+1)/(x7n-2)
= (x5+n)×(x2)(3n+1)/(x7n-2)
= (x5+n)×(x6n+2)/(x7n-2)
= x5+n+6n+2-7n+2
=  x9

10. Evaluate:
(a2n+1)×a(2n+1)(2n-1)/(an(4n-1)) × (a2)(2n+3)
Solution
(a2n+1) × a(2n+1)(2n-1)/(an(4n-1)) × (a2)(2n+3)
= [(a2n+1) × a(2n)2-(1)2]/[(a4n2-n) × (a2(2n+3))]
= (a2n+1) × (a4n2-1)/(a4n2-n) × (a4n+6)
= a2n+1+4n2-1-4n2+n-4n-6
= a-n-6
= a-(n+6)
= 1/(an+6)

11. (m + n)-1(m-1+ n-1) = (mn)-1
Solution
L.H.S = (m + n)-1(m-1+ n-1)
= 1/(m + n)(1/m + 1/n)
= 1/(m + n).(n + m)/mn
= 1/mn
= (mn)-1
= R.H.S
Hence, proved.

12. Prove that:
(i) (xa/xb)1/ab (xb/xc)1/bc(xc/xa)1/ca = 1
(ii) 1/(1 + xa-b) + 1/(1 + xb-a) = 1
Solution
(i)  (xa/xb)1/ab (xb/xc)1/bc (xc/xa)1/ca = 1
L.H.S = (xa/xb)1/ab (xb/xc)1/bc (xc/xa)1/ca
= (xa-b)1/ab  (xb-c)1/bc  (xc-b)1/ca
=  x(a-b)/ab x(b-c)/bc x(c-a)/ca {(xa)b= xab}
= x{(a-b)/ab}+{(b-c)/bc}+{(c-a)/ca}
= x{(ac-bc+ab-ac+bc-ab)/abc}
= x0 = 1 = RHS  (∵ x0 = 1)

(ii) 1/(1 +  xa-b) + 1/(1 + xb-a) = 1
L.H.S. = 1/(1 +  xa-b) + 1/(1 + xb-a)
= 1/(xa-a +  xa-b) + 1/(xb-b + xb-a)
= 1/(xa. x-a  +  x. x-b) + 1/(xb. x-b + xb. x-a)
= 1/xa(x-a + x-b) + 1/(xb(x-b + x-a)
= 1/(x-a + x-b)[1/xa + 1/xb]
= 1/(x-a + x-b)[x-a + x-b] = 1 = R.H.S.

13. Find the values of n, when:
(i) 12-5 × 122n+1 = 1213 ÷ 127
(ii) {(a2n-3) × (a2)n+1}/(a4)-3 = (a3)3 ÷ (a6)-3
Solution
(i) 12-5 × 122n+1 = 1213 ÷ 127
= 12-5+2n+1 = 1213/127
= 122n-4 = 1213-7
= 122n-4 = 126
Comparing both sides, we get
2n – 4 = 6
⇒ 2n = 6 + 4
⇒ 2n = 10
⇒ n = 5

(ii) {(a2n-3)× (a2)n+1}/(a4)-3 = (a3)3 ÷ (a6)-3
= (a2n-3) × (22n+2)/(a-12) = a9 ÷ a-18
(a2n-3 × 22n+2)/(a-12) = (a9)/(a-18)
= a-2n-3+2n+2-(-12) = a9-(-18)
= a4n+11 = a27
Comparing both sides, we get
4n+ 11 = 27
⇒ 4n = 27 -11
⇒ n = 16/4 = 4

14. Simplify:
(i) {(a2n-3). a(2n+1)(n+2)}/{(a3)2n+1. an(2n+1)}
(ii) {(x2n+7).(x2)3n+2}/x4(2n+3)
Solution
(i) {(a2n-3). a(2n+1)(n+2)}/{(a3)2n+1. an(2n+1)}
Given expression = {(a2n+3). a(2n2+4n+n+2)}/{(a6n+3).(a2n2+n)}
= {(a2n+3+2n2+5n+2)/(a6n+3+2n2+n)}
= {(a2n2+7n+5)/(a2n2+7n+3)
= {a(2n2+7n+3)+2}/{(a2n2+7n+3)} = a2

(ii) {(x2n+7).(x2)3n+2}/x4(2n+3)
Given expression = {(x2n+7).(x6n+4))}/(x8n+12)
= {(x2n+7+6n+4))/(x8n+12)
= (x8n+11)/((x8n+12)
=(x8n+11-8n-12) = x-1
= 1/x