Maths Ace Class 8 Solutions Chapter 2

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Maths Ace Class 8 Solutions Chapter 2 Exponents

Welcome to NCTB Solutions. Here with this post we are going to help 8th class students for the Solutions of Maths Ace Prime Class 8 Math Book, Chapter 2, Exponents. Here students can easily find step by step solutions of all the problems for Exponents, Exercise 2.1 and 2.2 Also here our mathematics teacher’s are solved all the problems with easily understandable methods with proper guidance so that all the students can understand easily. Here in this post students will get chapter 2 solutions.

Exponents Exercise 2.1 Solution :

Question no – (1) 

Solution :

(a) (4/7)² × (4/7)²

= (4/7)^{2 + 2}

= (4/7)⁴

= 256/2401

(b) (3/2)³ × (2/3)²

= (3/2)³ × (3/2)^-2

= (3/2)^3-2

= 3/2

(c) (3/4)³ ÷ (3/4)^-2

= (3/4)³ × (3/4)²

= (3/4)⁵

= 243/1024

(d) (3/2)³ × (2/3)²

= (3/2)³ × (3/2)^-2

= (3/2)^3-2

= 3/2

(e) (2/3)² ÷ (4/5)³ × (3/5)²

= (2/3)² × (5/4)³ × (3/5)²

= (2/3 × 5/4 × 3/5)⁷

(f) ((-3/4)^-3)^-2 × ((3/4)²)^-4

= (+27/64)⁺² × (9/16)^-4

= (+3/4)⁺⁶ × (3/4)^-8

= (3/4)^-2

= 16/9

Question no – (2) 

Solution : 

(a) (3/7)^-2 × (3/7)^-3

= (3/7)^-2-3

= (3/7)^-5

= (7/3)⁵

(b) (1/2)⁵ ÷ (1/2)³

= (1/2)^5-3

= (1/2)²

(c) ((4/3)^-2)^-2

= (4/3)⁴

(d) (2/3)² × (-3/2)³

= 4/9 × -27/8

= -3/2

(e) ((2/5)³)²

= (8/125)²

= (2³/5³)²

= 2⁶/5⁶

(f) ((3/4)²)^-2 × 1/4 ÷ (4/3)³

= (3/4)^-4 × 1/4 × (3/4)³

= (3/4)^-4+3 × 1/4

= (3/4)^-1 × 1/4

= 4/3 × 1/4

= 1/3

(g) (1/2)² × ((1/2)³)^-4 ÷ ((1/2)²)^-5

= (1/2)² × (1/2)^-12 ÷ (1/2)^-10

= (1/2)^2-12 × (2)^-10

= (1/2)^-10 × 2¹⁰

= (2)¹⁰ × 2^-10

= 2⁰

= 1

Question no – (3) 

Solution : 

(a) ((-3/2)^-3)^-2

= (2/3)^-6

(b) 36t^-3/6^-3 × t^-5(t ≠ 0)

= 6² × t^-3/6^-3 × t^-5

= 6^{2 + 3} × t^{-3 + 5}

= 6⁵ × t²

= 1/6^-5 . t^-2

(c) ((-2/3)^-2)³ × (1/3)^-4 × 3^-1 × 1/12

= {(- 3/2)²}³ × (3)⁴

= -3⁶/2⁶ × 3⁴

= -3⁶ – 4/2⁶

= – 32/26

= 3^-2/2⁶

= (2/3)^-6-2

= (2/3)^-8

(d) (1/4)^-2 × (1/2)^-3 × (8)⁴

= (1/2)^-4 × (1/2)^-3 × (1/2)^-12

= (1/2)^-4-3-12

= (1/2)^-19 (Negative exponent)

Question no – (4) 

Solution : 

(a) (2/3)¹² = (3/2)^-3x

= (2/3)¹² = (2/3)^3x

= 3x = 12

= x = 12/3

= x = 4

Thus, the value of x is 4.

(b) ((-8)²)³ = (-8)^2x

= (-8)⁶ = (-8)^2x

= 2x = 6

= x = 6/2

= 3

Hence, the value of x is 3.

(c) (3/4)²¹ × (3/4)³ = (3/4)^3x

= (3/4)²⁴ = (3/4)^3x

= 3x = 24

= x = 24/3

= 8

Thus, the value of x is 8.

(d) (2/3)³ × (2/6)^-6 = (2/3)^x-4

= (2/3)^{3 – 6} = (2/3)^{x – 4}

= x – 4 = -3

= x = -3 + 4

= 1

Therefore, the value of x is 1.

(e) ((3/4)^-6 ÷ (3/4)³)^x = (3/4)^-9

= ((3/4)^-6 × (3/4)^-3)^x = (3/4)^-9

= (3/4)^-9x = (3/4)^-9

= -9x = -9

= x = 9/9

= 1

Thus, the value of x is 1.

Question no – (5) 

Solution : 

(a) 2^m/2^-2 = 2³

= 2^m = 2³ × 2^-2

= 2 ^3-2

∴ 2^m = 2¹

∴ m = 1

(b) 3^m × 3⁶/3^-3 = 3¹⁸

= 3^m × 3⁶ = 3^{18 – 3}

= 3¹⁶

= 3^m = 3^{15 – 6}

= 3⁹

∴ m = 9

(c) 6^m × 6³ × 6^-2/6^-5 = 6¹²

= 6^m × 6³ × 6^-2 = 6^{12 – 5}

= 6⁷

= 6^m × 6¹ = 6⁷

= 6^7-1

∴ 6^m = 6⁶

= m = 6

Question no – (6) 

Solution : 

As per the asked question,

Given number = (-2/3)^-3

Quotient is = (9/8)^-2

The required number is,

= (-3/2)³/(8/9)²

= (3/-2)³ × (9/8)²

= 3³/-2³ × (3²/2³)²

= -3³ × 3⁴/2³ × 2⁶

= -3⁷/2⁹

Therefore, the required number will be -3⁷/2⁹

Question no – (7) 

Solution : 

1^st, 144 × 2^-3

= 144 × ½

= 12

2^nd, 144 × 12^-1

= 144 × 1/12

= 12

3^rd, 144 × 3^-2

= 144 × 1/9

= 16

Now, 144 × 2^-3 × 12^-1 × 3^-2

= 144 × 1/8 × 1/12 × 1/9

= 3/18

= 1/6

Therefore, the final product will be 1/6.

Question no – (9) 

Solution : 

(a) 1/9(3^3x) = 3^3x-2

= L.H.S, 1/9 × 3^3x

= 1/3² × 3^3x

= 3^{3x – 2}

= R.H.S

(b) 5^x+1 = (25 × 5^x)1/5

= R.H.S, (25 × 5^x) 1/5

= (5^x × 5^x) × 1/5

= (5^x × 5¹)

= 5^{x + 1}

= L.H.S

(c) (3/5)^x-1 = (3^x ×5/5^x ×3)

= R.H.S, (3^x × 5/5^x × 3¹)

= 3x ^{– 1}/5^{x – 1}

= (3/5)^{x – 1}

= L.H.S (Proved)

Question no – (10) 

Solution : 

Given, x = (3/2)² × (2/3)^-4

= (3/2)² × (3/2)⁴

= (3/2)⁶

∴ x^-2 = {(3/2)⁶}^-2

= (3/2)^-12

Therefore, the value of x^-2 is (3/2)^-12

Exponents Exercise 2.2 Solution :

Question no – (1) 

Solution : 

(a) 0.000000067 in standard form,

= 6.7 × 10^-8

(b) 12,54,36,000 in standard form,

= 1.25436 × 10²

(c) 345 in standard form,

= 3.45 × 10²

(d) 0.00028968 in standard form,

= 2.8968 × 10^-4

Question no – (2)

Solution : 

(a) 5.23 × 10⁴

= 52300

The usual form of 5.23 × 10⁴ is 52300

(b) 2.05 × 10^-3

= 0.00205

The usual form of 2.05 × 10^-3 is 0.00205

(c) 2 × 10^-5

= 0.00002

The usual form of 2 × 10^-5 is 0.00002

(d) 3.45 × 10⁸

= 345000000

The usual form of 3.45 × 10⁸ is 345000000

(e) 2.49 × 10¹⁵

=  2490000000000000

The usual form of 2.49 × 10¹⁵ is 2490000000000000

(f) 8.99 × 10^-7

= 0.000000899

The usual form of 8.99 × 10^-7 is 0.000000899

Question no – (3) 

Solution : 

Given, One angstrom is the ten billionth part of a metre

In scientific notation = 10^-9 m

Question no – (4) 

Solution : 

According to the question,

Single calculation in about = 0.0000034 seconds

Calculate 4,000 such calculations = ?

∴ Total time,

= (4000 × 0.0000034)

= (1.36 × 10^-2)

Therefore, calculator will it take (1.36 × 10^-2)

Question no – (5) 

Solution : 

Hour = 1h = 3600 s = 3.6 × 10³ s

Day = 1 day = 8.64 × 10⁴ seconds

Week = 1 week = 6.048 × 10⁵ s

Month = 1 m = 30 day × 8.64 × 10⁴ = 2.592 × 10⁶ s

Year = 1 year = (3.156 × 10⁷) seconds

Question no – (6) 

Solution : 

As we know that, 1 hectare = 10, 000 sq. m

∴ (10000 × 10000) cm

= 10⁸ sq.cm

Therefore, this will be 10⁸ sq. cm.

Next Chapter Solution : 

👉 Chapter 3 👈