ML Aggarwal Exponents and Powers Exe-2 Class 8 Ch-2 ICSE Maths Solutions. We Provide Step by Step Answer of  Exe-2 Questions for Exponents and Powers as council prescribe guideline for upcoming board exam. Visit official Website CISCE for detail information about ICSE Board Class-8.

## ML Aggarwal Exponents and Powers Exe-2 Class 8 ICSE Maths Solutions

 Board ICSE Publications Avichal Publishig Company (APC) Subject Maths Class 8th Chapter-2 Exponents and Powers Writer ML Aggarwal Book Name Understanding Topics Solution of Exe-2 Questions Edition 2023-2024

### Exponents and Powers Exe-2

ML Aggarwal Class 8 ICSE Maths Solutions

Page-40

#### Question 1.Evaluate:

(i) (3/5)-2

(ii) (-3)-3

(iii) (2/7)-4

##### (i)(3/5)-2

By using the formula, (a-n = 1/an)

(3/5)-2 = (5/3)2

= (5/3) × (5/3)

= 25/9

##### (ii)(-3)-3

By using the formula, (a-n = 1/an)

(-3)-3 = (-1/3)3

= (-1/3) × (-1/3) × (-1/3)

= – 1/27

##### (iii)(2/7)-4

By using the formula, (a-n = 1/an)

(2/7)-4 = (7/2)4

= (7/2) × (7/2) × (7/2) × (7/2)

= 2401/16

#### Question 2. Simplify:

(i) [(2)-1 + (4)-1 + (3)-1]-1

(ii) [(4)-1 – (5)-1]2 × (5/8)-1

(iii) [40 + 42 – 23] × 3-2

(iv) [(5)2 – (1/4)-2] × (3/4)-2

= [(6+3+4)/12]-1

= [13/12]-1

= 12/13

#### [(4)-1 – (5)-1]2 × (5/8)-1 = [(1/4) – (1/5)]2 × (8/5)1

= [(5-4)/20]2 × (8/5)

= [1/20]2 × (8/5)

= (1/20) × (1/20) × (8/5)

= 1/250

= 9 × (1/9)

= 1

#### [(5)2 – (1/4)-2] × (3/4)-2 = [25 – (4)2] × (4/3)2

= [25 – 16] × (16/9)

= 9 × (16/9)

= 16

#### Question 3. Find the multiplicative inverse of the following:

(i) (81/16)-3/4

(ii) {(-3/2)-4}1/2

(iii) (5/7)-2 × (5/7)4 ÷ (5/7)3

##### (i) (81/16)-3/4

(81/16)-3/4 = (16/81)3/4

= (24/34)3/4

= (2/3)4×3/4

= (2/3)3

= (2/3) × (2/3) × (2/3)

= 8/27

So, the multiplicative inverse of 8/27 is 27/8.

##### (ii) {(-3/2)-4}1/2

{(-3/2)-4}1/2 = (-3/2)-4×1/2

= (-3/2)-2

= (2/-3)2

= (2/-3) × (2/-3)

= 4/9

So, the multiplicative inverse of 4/9 is 9/4.

##### (iii) (5/7)-2 × (5/7)4 ÷ (5/7)3

(5/7)-2 × (5/7)4 ÷ (5/7)3 = (7/5)2 × (5/7)4 ÷ (5/7)3

= (7/5)2 × (5/7)4-3

= (7/5)2 × (5/7)

= (7/5) × (7/5) × (5/7)

= 7/5

So, the multiplicative inverse of 7/5 is 5/7.

#### Question 4.

(i) Express 16-2 as a power with base 2.
(ii) Express 125-4 as a power with base 5.

(16)-2 = (24)-2

= 2-8

= 1/28

(125)-4 = (53)-4

= 5-12

= 1/512

#### Question 5. Simplify and write in exponential form with positive exponent:

(i) [{(5/7)2}-1]-3

(ii) (2/7)2 × (7/2)-3 ÷ {(7/5)-2}-4

(iii) (4/5)2 × 54 × (2/5)-2 ÷ (5/2)-3

(iv) [8-1 × 53]/2-4

#### [{(5/7)2}-1]-3 = {(5/7)2}-1×-3

= (5/7)2×3

= (5/7)6

##### (ii) (2/7)2 × (7/2)-3 ÷ {(7/5)-2}-4

(2/7)2 × (7/2)-3 ÷ {(7/5)-2}-4 = (2/7)2 × (2/7)3 ÷ (7/5)8

= (2/7)2 × (2/7)3 × (5/7)8

= (22/72) × (23/73) × (58/78)

= (22+3/72+3) × (58/78)

= (25/75) × (58/78)

= (25 × 58)/75+8

= (25 × 58)/713

##### (iii) (4/5)2 × 54 × (2/5)-2 ÷ (5/2)-3

(4/5)2 × 54 × (2/5)-2 ÷ (5/2)-3 = ((22)2/52) × 54 × (2-2/5-2) × (2-3/5-3)

= (24 × 54 × 2-2 × 2-3) / (52 × 5-2 × 5-3)

= 24-2-3 × 54-2+2+3

= 2-1 × 57

= 57/21

= [2-3 × 53]/2-4

= 2-3+4 × 53

= 21 × 53

= 2 × 5 × 5 × 5

= 250

#### Question 6. Simplify and write the following in exponential form:

(i) ((-2)3)2 + 5-3 ÷ 5-5 – (-1/2)0

(ii) 3-5 × 32 ÷ 3-6 + (22 × 3)2 + (2/3)-1 + 2-1 + (1/19)-1

##### (i) ((-2)3)2 + 5-3 ÷ 5-5 – (-1/2)0

((-2)3)2 + 5-3 ÷ 5-5 – (-1/2)0 = (-2)6 + 1/53 ÷ 1/55 – 1

= 64 + 1/53 × 55 – 1

= 64 + 55-3 – 1

= 64 + 52 – 1

= 64 + 25 – 1

= 88

##### (ii) 3-5 × 32 ÷ 3-6 + (22 × 3)2 + (2/3)-1 + 2-1 + (1/19)-1

3-5 × 32 ÷ 3-6 + (22 × 3)2 + (2/3)-1 + 2-1 + (1/19)-1 = 3-5+2+6 + (24 × 32) + 3/2 + ½ + 191

= 33 + (16×9) + 4/2 + 19

= 27 + 144 + 2 + 19

= 192

#### Question 7. Simplify and write in exponential form with negative exponent:

(i) 53 × (4/5)3

(ii) [(3/7)-2]-3

(iii) (5/9)-2 × (5/3)2 ÷ (1/5)-2

(iv) 2-1 [(5/3)4 + (3/5)-2] ÷ (17/9)

(v) (-7)3 × (1/-7)-9 ÷ (-7)10

##### (i) 53 × (4/5)3

53 × (4/5)3 = 53 × (43/53)

= 53-3 × 43

= 50 × 43

= 1 × 43

= (1/4)-3

#### [(3/7)-2]-3 = (3/7)-2×-3

= (3/7)6

= (7/3)-6

##### (iii) (5/9)-2 × (5/3)2 ÷ (1/5)-2

(5/9)-2 × (5/3)2 ÷ (1/5)-2 = (5-2/9-2) × (52/32) ÷ (1/5-2)

= (5-2 × 52 × 5-2)/ [(32)-2 × 32]

= (5-2+2-2)/(3-4+2)

= 5-2/3-2

= (5/3)-2

##### (iv) 2-1 [(5/3)4 + (3/5)-2] ÷ (17/9)

2-1 [(5/3)4 + (3/5)-2] ÷ (17/9) = 2-1 [(54/34) + (3-2/5-2)] ÷ (17/9)

= 2-1 [(54/34) + (52/32)] ÷ (17/9)

= 2-1 [(625/81) + (25/9)] ÷ (17/9)

= ½ [(625 + 225)/81] × (9/17)

= ½ × (850/81) × 9/17

= 25/9

= (5/3)2

= (3/5)-2

##### (v) (-7)3 × (1/-7)-9 ÷ (-7)10

(-7)3 × (1/-7)-9 ÷ (-7)10 = (-7)3 × (-7)9 ÷ (-7)10

= (-7)3+9-10

= (-7)2

= (1/-7)-2

#### Question 8. Simplify:

(i) (49 × z-3) / (7-3 × 10 × z-5) (z ≠ 0)

(ii) (93 × 27 × t4) / (32 × 34 × t2)

(iii) [(3-2)2 × (52)-3 × (-t-3)2] / [(3-2)5 × (53)-2 × (t-4)3]

(iv) (2-5 × 15-5 × 500) / (5-6 × 6-5)

##### (i) (49 × z-3) / (7-3 × 10 × z-5) (z ≠ 0)

(49 × z-3) / (7-3 × 10 × z-5) = (72 × z-3) / (7-3 × 10 × z-5)

= (72+3 × z-3+5) / 10

= (75 × z2)/10

##### (ii) (93 × 27 × t4) / (32 × 34 × t2)

(93 × 27 × t4) / (32 × 34 × t2) = [(32)3 × (3)3 × t4] / (32 × 34 × t2)

= (36 × 33 × t4) / (32 × 34 × t2)

= 36+3-2-4 × t4-2

= 33 × t2

= 27t2

#### [(3-2)2 × (52)-3 × (t-3)2] / [(3-2)5 × (53)-2 × (t-4)3] = [3-4 × 5-6 × t-6] / [3-10 × 5-6 × t-12]

= 3-4+10 × 5-6+6 × t-6+12

= 36 × 50 × t6

= 36 × 1 × t6

= 729t6

##### (iv) (2-5 × 15-5 × 500) / (5-6 × 6-5)

(2-5 × 15-5 × 500) / (5-6 × 6-5) = (2-5 × (3×5)-5 × 22 × 53) / (5-6 × (2×3)-5)

= (2-5 × 3-5 × 5-5 × 22 × 53) / (5-6 × 2-5 × 3-5)

= 2-5+2+5 × 3-5+5 × 5-5+3+6

= 22 × 30 × 54

= 4 × 1 × 625

= 2500

### Exponents and Powers Exe-2

ML Aggarwal Class 8 ICSE Maths Solutions

Page-41

#### Question 9. By what number should (3/-2)-3 be divided to get (2/3)2?

The required number is (3/-2)-3 ÷ (2/3)2

(3/-2)-3 ÷ (2/3)2 = (3-3)/(-2)-3 × (32)/(22)

= [3-3 × 32]/[-2-3 × -22]

= [3-3+2]/(-2)-3+2

= 3-1/-2-1

= -21/31

= -2/3

#### Question 10. Find the value of m for which 9m ÷ 3-2 = 94.

9m ÷ 3-2 = 94

(32)m ÷ 3-2 = (32)4

32m ÷ 3-2 = 38

32m+2 = 38

Now by comparing the powers,

2m + 2 = 8

2m = 8 – 2

= 6

m = 6/2

= 3

#### Question 11. If (-5/7)-4 × (-5/7)12 = {(-5/7)3} × (-5/7)-1, find the value of x.

(-5/7)-4 × (-5/7)12 = {(-5/7)3} × (-5/7)-1

(-5/7)-4 × (-5/7)12 = (-5/7)3x × (-5/7)-1

(-5/7)-4+12 = (-5/7)3x-1

(-5/7)8 = (-5/7)3x-1

Now by comparing the powers, we get

8 = 3x – 1

3x = 8 + 1

3x = 9

x = 9/3

= 3

#### Question 12. Find x, if (-2/3)-13 × (3/-2)8 = (-2/3)-2x+1

(-2/3)-13 × (3/-2)8 = (-2/3)-2x+1

(-2/3)-13 × (-2/3)-8 = (-2/3)-2x+1

(-2/3)-13-8 = (-2/3)-2x+1

(-2/3)-21 = (-2/3)-2x+1

Now by comparing the powers,

-21 = -2x +1

2x = 1 + 21

2x = 22

x = 22/2

= 11

(ML Aggarwal Exponents and Powers Exe-2 Class 8)

#### Question 13.

(i) If 52x-1 = 1/(125)x-3, find x.

(ii) If (9n × 35 × 273)/(3×814) = 27, find n.

##### (i) If 52x-1 = 1/(125)x-3, find x.

52x-1 = 1/(53)x-3

52x-1 = 1/53x-9

52x-1 = 5-3x+9

Now by comparing the powers,

2x – 1 = -3x + 9

2x + 3x = 9 + 1

5x = 10

x = 10/5

= 2

#### [(32)n × 35 × (33)3]/(3×(34)4) = 33 [32n × 35 × 39] = (3×316) = 33

32n+5+9-1-16 = 33

Now by comparing the powers,

2n + 5 + 9 – 1 – 16 = 3

2n – 3 = 3

2n = 3 + 3

2n = 6

n = 6/2

= 3

—  : End of ML Aggarwal Exponents and Powers Exe-2 Class 8 ICSE Maths Solutions :–