# ICSE Solutions for Selina Concise Chapter 7 Indices Exponents Class 9 Maths

### Exercise 7(A)

1.1 Evaluate : 33 ×(243)2/3 × 91/3

1.2. Evaluate : 5−4 ×(125)5/3 ÷ (25)1/2

1.3. Evaluate : (27/125)2/3 × (9/25)−3/2

1.4. Evaluate : 70 × (25)-3/2 −5−3

1.5. Evaluate : (16/81)−3/4 × (49/9)3/2 + (343/216)2/3

2.1. Simplify : (8x3 + 125y3)2/3

2.2. Simplify : (a+b)−1 . (a−1 + b−1)

2.3. Simplify : (5n+3 −6×5n+1)/(9×5n × 22)

2.4. Simplify : (3x2)−3 × (x9)2/3

3.1. Evaluate : √(1/4) + (0.01)−1/2 – (27)2/3

3.2. Evaluate : (27/8)2/3 – (1/4)−2 + 5°

= −51/4

(3−4/2−6)1/4

(27−3/9−3)1/5

#### 4.3. Simplify the following and express with positive index :

(32)−2/5 + (125)−2/3

#### 4.4. Simplify the following and express with positive index :

[1−{1−(1−n)−1}−1]−1

#### 6. If 1960 = 2a. 5b. 7c, calculate the value of 2-a. 7b. 5-c.

1960 = 2a×5b ×7c
⇒ 2×2×2×5×7 ×7 = 2×5b×7c
⇒ 23×51 ×7= 2a ×5b×7c
⇒ 2a×5b×7c = 2×51×72
Comparing powers of 2,5 and 7 on the both sides of equation, We have
a = 3; b = 1 and c = 2
Hence,
Value of 2-a×7b× 5-c

= 2-3×71×5-2

#### 7.1. Simplify :

(83a × 25 × 22a)/(4 × 211a ×2−2a)

#### 9. If a = xm+n. yl ; b = xn+l. ym and c = xl+m. yn,

Prove that : am–n. bn–l. cl–m = 1

### Exercise 7(B)

#### 1. Solve for x:

(i) 22x+1 = 8

(ii) 25x-1 = 4 23x+1

(iii) 34x+1 = (27)x+1

(iv) (49)x+4 = 72 (343)x+1

#### 2. Find x, if:

(i) 42x = 1/32

(ii) √(2x+3=16

(iii) √(3/5)x+1 = 125/27

(iv) ∛(2/3)x−1 = 27/8

#### 3. Solve:

(i) 4x–2 – 2x+1 = 0

(ii) 3x2 : 3x = 9:1

#### 4. Solve :

(i) 8×22x + 4×2x+1 = 1 + 2x

(ii)22x + 2x+2 – 4×23 = 0

(iii) (√3)x-3 = (∜3)x+1

#### 6. Solve x and y if

(√32)x  ÷2y+1  = 1 and 8y – 164−x/2 = 0

7. Prove that:

(i) (xa/xb)a+b-c × (xb/xc)b+c-a × (xc/xa)c+a-b

(ii) xa(b-c)/xb(a-c) ÷ (xb/xa)c = 1

x + y + 2 = 0

#### 12. If 5x+1 = 25x–2, find the value of

3x–3 × 23–x

13. If 4x+3 = 112 + 8×4x, find the value of (18x)3x.

14. (i) 9x+2 = 720 + 9x

(ii) Solve for x: (a3x + 5)2. (ax)4 = a8x + 12

(iii) Solve for x: (81)3/4 –(1/32)−2/5 + x(1/2)−1. 20 = 27

(iv) Solve for x: 23x + 3 = 23x + 1 + 48

(v) Solve for x: 3(2x + 1) – 2x + 2 + 5 = 0

### Exercise 7(C)

#### 1. Evaluate

(i) 95/2 – 3×80 –(1/81)−1/2

(ii) (64)2/3 − ∛125 −1/2−5 + (27)−2/3 × (25/9)−1/2

(iii)[(-2/3)-2]3 × (1/3)-4 × 3-1 ×1/6

#### 2. Simplify: (3×9n+1 −9×32n)/(3×32n+3 −9n+1)

3. Solve: 3x-1× 52y-3 = 225.

#### 14. Evaluate:  (xq/x1)1/q1 × (x1/xp)1/1p ×(xp/xq)1/pq

15 (i). Prove that : a-1/(a-1 + b-1)  + a-1/(a-1b-1) = 2b2/(b2 – a2)