# Frank Learning Maths Class 8 Chapter 2 Exponents Exercise 2.1 Solutions

## Frank Learning Maths Class 8 Chapter 2 Exponents Exercise 2.1 Solutions

This post is created to help all the CBSE Class 8 students for the Solutions of Frank Learning Maths Class 8 Mathematics Book, Chapter 2 Exponents Exercise 2.1. Here students can easily find step by step solutions of all the problems for Exponents Exercise 2.1. Exercise wise proper solutions for every problems. All the problem are solved with easily understandable methods so that all the students can understand easily.

Chapter 2 – Exponents

Exponents – Exercise 2.1 all Questions Solution

(1) Simplify

Solution :

(a) {(1/3)-3 – (1/3)-3} ÷ (1/4)-3

= {1/3-3 – 1/2-3} ÷ (1/4-3)

= {33 – 23} ÷ 43

= (27 – 8) × 1/43

= 19 × 1/64

= 19/64…(Simplified)

(b) {(1/3)-1 × (-9)-1}-1

= {(1/3-1) × (-9-1)}-1

= {31 × 1/-91}-1

= (- 1/3)-1

= -3…(Simplified)

(c) (52 – 32) × (2/3)2

= (25 – 9) × (4/9)

= 16 ×  4/9

= 64/9…(Simplified)

(d) {(2/3)2}3 × (1/3)-4 × (3)-2 × (2)-1

= (4/9)3 × (1/3-4) × 1/32 × 1/21

= 4 × 4 × 4 /9 × 9 × 9 ×  34 × 1/32 × 1/2

= 32/9…(Simplified)

(e) (4-1 ÷ 6-1)3

= (1/4 ÷ 1/6)3

= (1/4 × 6)3

= (3/2)3

= 33/23

= 27/8…(Simplified)

(f) {6-1 – 5-1} ÷ 3-1

= (1/6 – 1/5) × 2/9 ÷ 1/3

= 5 – 6/30 ÷ 1/3

= 1/30 × 3

= 1/10…(Simplified)

(2) Evaluate

Solution :

(a) (1/3)-2 + (1/2)-2 + (1/4)-2

= 1/3-2 + 1/2-2 + 1/4-2

= 32 + 22 + 42

= 9 + 4 + 16

= 29

(b) (1/5)45 × (1/5)-60 – (1/5)+28 × (1/5)-43

= 1/545 × 560 – 1/528 × 543

= (5)60 – 45 – (5)43 – 28

= 515 – 515

= 0

(c) {(2+ 3-1) × 92}

= (1 + 1/3) × 92

= 4/3 × 9 ×

= 108

(d) 3-5 × 10-5 × 125/5-7 × 6-5

= 3-5 × 2-5 × 5+5 × 125/5-7 × 2-5 × 3-5

= 125/(5-2)

= 125 × 52

= 3125

(e) (4/3)-2 × (32/4)(2)

= (4/3) -2 × (32/4) -2

= 4 -2/3 -2 × 3 -4/4 -2

= 3 -2 × 4

= 1/9

(f) (13 + 23 + 33)-5/2

= (1 + 8 + 27)-5/2

= (36)-5/2

= (62)-5/2

= 6-5

= 1/65

= 1/7776

(g) (-7)-3 × 492 × 11-4 × 1212

= – 1/73 × 73 × 1/114 × 114

= 1

(h) 3-5 × 10-4 × 625/5-3 × 6-7

= 3-5 × 2-4 × 5-4 × 625/5-3 × 2-7 × 3-7

= 32 × 23 × 625/5

= 9000

(i) 16 × 102 × 64/24 × 42

= 42 × 22 × 52 × 22 × 16/24 × 42

= 25 × 16

= 400

(j) (3-22 × (52-3 × (t-32/(3-25 × (53-2 × (t-43

= 3-4 × 5-6 × t-6/3-10 × 5-6 × t-12

= 36 × t6

= 729t6

(3) Find x.

Solution :

(a) (- 1/7)-5 ÷ (- 1/7)-7 = (-7)x

= 75 ÷ 77 = (-7) x

= 75/77 = (-7) x

= 7-2 = (-7) x

= (-7) -2 = (-7) x

∴ x = -2

Hence, the x will be -2

(b) (2/5) 2x + 6 × (2/5) 3 = (2/5) x + 2

= 2x + 6 + 3 = x + 2

= x = 2 – 9

∴ x = -7

Thus, the x will be -7

(c) 2x + 2x + 2x = 192

= 3 × 2x = 192

= 2x = 192/3

= 2x = (2) 6

∴ x = 6

Therefore, the x will be 6

(d) (-6/7) x – 7 = 1

= (- 6/7) x – 7 = (- 6/7) 0

= x – 7 = 0

∴ x = 7

Hence, the x will be 7

(e) 23x = 82x + 1

= 8x = 82x + 1

= x = 2x + 1

= -x = 1

∴ x = -1

Thus, the x will be -1

(f) 5x + 5x – 1 = 750

= 5x (1 + 1/5) = 750

= 5x 6/5 = 750

= 5x = 125 × 5

= 5x = (5) 4

∴ x = 4

Therefore, the x will be 4

Question no – (4)

Solution :

Let the multiplied number be x

∴ (-5)-1 × x = (11)-1

X = 11-1/-5-1

= – 1/11 × 5

= – 5/11

Hence, -5/11 should be multiplied.

Question no – (5)

Solution :

Let the number should be x

∴ (-1/2)-1 × x = (- 5/9)-1

X = (- 5/9)-1 × (2)-1

= – 9/5 × (1/2)

= – 9/10

Therefore, -9/10 should be multiplied.

Question no – (6)

Solution :

Let the number should be x

∴ (-12)-1/x = (-4)-1

x = -(12)-1/(-4)-1

= -3-1

= 1/3

Thus, 1/3 should be divided.

Question no – (7)

Solution :

Let the number should be x

∴ (2/7)-2 × x = (5/7)-1

= (7/2)2 × x = (7/5)

X = 7/5 × (/2/7)2

= 4/35

Therefore, 4/35 should be multiplied.

Updated: February 2, 2023 — 2:02 pm