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

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

Answer:

(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

Answer :

(i) [(2)^-1 + (4)^-1 + (3)^-1]^-1

[(2)^-1 + (4)^-1 + (3)^-1]^-1 = [(1/2) + (1/4) + (1/3)]^-1

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

= [13/12]^-1

= 12/13

(ii) [(4)^-1 – (5)^-1]² × (5/8)^-1

[(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

(iii) [4⁰ + 4² – 2³] × 3^-2

[4⁰ + 4² – 2³] × 3^-2 = [1 + 16 – 8] × (1/3²)

= 9 × (1/9)

= 1

(iv) [(5)² – (1/4)^-2] × (3/4)^-2

[(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

Answer:

(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.

Answer:

(i) Express 16^-2 as a power with base 2.

(16)^-2 = (2⁴)^-2

= 2^-8

= 1/2⁸

(ii) Express 125^-4 as a power with base 5.

(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

Answer:

(i) [{(5/7)²}^-1]^-3

[{(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)

= 2^4-2-3 × 5^4-2+2+3

= 2^-1 × 5⁷

= 5⁷/2¹

(iv) [8^-1 × 5³]/2^-4

[8^-1 × 5³]/2^-4 = [(2³)^-1 × 5³]/2^-4

= [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

Answer:

(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 + 5^5-3 – 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

Answer:

(i) 5³ × (4/5)³

5³ × (4/5)³ = 5³ × (4³/5³)

= 5^3-3 × 4³

= 5⁰ × 4³

= 1 × 4³

= (1/4)^-3

(ii) [(3/7)^-2]^-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)

Answer:

(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)

= 3^6+3-2-4 × t^4-2

= 3³ × t²

= 27t²

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

[(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?

Answer:

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 9^m ÷ 3^-2 = 9⁴.

Answer:

9^m ÷ 3^-2 = 9⁴

(3²)^m ÷ 3^-2 = (3²)⁴

3^2m ÷ 3^-2 = 3⁸

3^2m+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.

Answer:

(-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

Answer:

(-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 5^2x-1 = 1/(125)^x-3, find x.

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

Answer:

(i) If 5^2x-1 = 1/(125)^x-3, find x.

5^2x-1 = 1/(5³)^x-3

5^2x-1 = 1/5^3x-9

5^2x-1 = 5^-3x+9

Now by comparing the powers,

2x – 1 = -3x + 9

2x + 3x = 9 + 1

5x = 10

x = 10/5

= 2

(ii) If (9ⁿ × 3⁵ × 27³)/(3×81⁴) = 27

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

3^2n+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 :–

Return to –  ML Aggarwal Maths Solutions for ICSE Class -8

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