RD Sharma Solutions For Class 8 Chapter 2 - Powers Exercise 2.2 ...

RD Sharma Solutions for Class 8 Maths Exercise 22 Chapter 2 Powers RD Sharma Solutions for Class 8 Maths Exercise 22 Chapter 2 Powers

In Exercise 2.2 of RD Sharma Class 8 Maths Chapter 2 Powers, we shall discuss problems based on the decimal number system and laws of integral exponents. RD Sharma Class 8 Solutions for Maths are solved by our subject experts in order to help students effortlessly learn the concepts given in the Maths textbook. Students can easily download the PDF of Exercise 2.2 Solutions from the links provided below.

Chapter 2 Powers

• RD Sharma Solutions for Class 8 Maths Chapter 1 Rational Numbers
• RD Sharma Solutions for Class 8 Maths Chapter 2 Powers
• RD Sharma Solutions for Class 8 Maths Chapter 3 Squares and Square Roots
• RD Sharma Solutions for Class 8 Maths Chapter 4 Cubes and Cube Roots
• RD Sharma Solutions for Class 8 Maths Chapter 5 Playing with Numbers
• RD Sharma Solutions for Class 8 Maths Chapter 6 Algebraic Expressions and Identities
• RD Sharma Solutions for Class 8 Maths Chapter 7 Factorization
• RD Sharma Solutions for Class 8 Maths Chapter 8 Division of Algebraic Expressions
• RD Sharma Solutions for Class 8 Maths Chapter 9 Linear Equations in One Variable
• RD Sharma Solutions for Class 8 Maths Chapter 10 Direct and Inverse Variations
• RD Sharma Solutions for Class 8 Maths Chapter 11 Time and Work
• RD Sharma Solutions for Class 8 Maths Chapter 12 Percentage
• RD Sharma Solutions for Class 8 Maths Chapter 13 Profit, Loss, Discount and Value Added Tax (VAT)
• RD Sharma Solutions for Class 8 Maths Chapter 14 Compound Interest
• RD Sharma Solutions for Class 8 Maths Chapter 15 Understanding Shapes – I (Polygons)
• RD Sharma Solutions for Class 8 Maths Chapter 16 Understanding Shapes – II (Quadrilaterals)
• RD Sharma Solutions for Class 8 Maths Chapter 17 Understanding Shapes – II (Special Types of Quadrilaterals)
• RD Sharma Solutions for Class 8 Maths Chapter 18 Practical Geometry (Constructions)
• RD Sharma Solutions for Class 8 Maths Chapter 19 Visualising Shapes
• RD Sharma Solutions for Class 8 Maths Chapter 20 Mensuration – I (Area of a Trapezium and a Polygon)
• RD Sharma Solutions for Class 8 Maths Chapter 21 Mensuration – II (Volumes and Surface Areas of a Cuboid and a Cube)
• RD Sharma Solutions for Class 8 Maths Chapter 22 Mensuration – III (Surface Area and Volume of a Right Circular Cylinder)
• RD Sharma Solutions for Class 8 Maths Chapter 23 Data Handling – I (Classification and Tabulation of Data)
• RD Sharma Solutions for Class 8 Maths Chapter 24 Data Handling – II (Graphical Representation of Data as Histograms)
• RD Sharma Solutions for Class 8 Maths Chapter 25 Data Handling – III (Pictorial Representation of Data as Pie Charts)
• RD Sharma Solutions for Class 8 Maths Chapter 26 Data Handling – IV (Probability)
• RD Sharma Solutions for Class 8 Maths Chapter 27 Introduction to Graphs

Exercise 2.2

• Exercise 2.1 Chapter 2 Powers
• Exercise 2.2 Chapter 2 Powers
• Exercise 2.3 Chapter 2 Powers

RD Sharma Solutions for Class 8 Maths Exercise 2.2 Chapter 2 Powers

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Access Answers to RD Sharma Solutions for Class 8 Maths Exercise 2.2 Chapter 2 Powers

1. Write each of the following in exponential form:

(i) (3/2)-1 × (3/2)-1 × (3/2)-1 × (3/2)-1

(ii) (2/5)-2 × (2/5)-2 × (2/5)-2

Solution:

(i) (3/2)-1 × (3/2)-1 × (3/2)-1 × (3/2)-1

(3/2)-4 (we know that a-n = 1/an, an = a×a…n times)

(ii) (2/5)-2 × (2/5)-2 × (2/5)-2

(2/5)-6 (we know that a-n = 1/an, an = a×a…n times)

2. Evaluate:

(i) 5-2

(ii) (-3)-2

(iii) (1/3)-4

(iv) (-1/2)-1

Solution:

(i) 5-2

1/52 = 1/25 (we know that a-n = 1/an)

(ii) (-3)-2

(1/-3)2 = 1/9 (we know that a-n = 1/an)

(iii) (1/3)-4

34 = 81 (we know that 1/a-n = an)

(iv) (-1/2)-1

-21 = -2 (we know that 1/a-n = an)

3. Express each of the following as a rational number in the form p/q:

(i) 6-1

(ii) (-7)-1

(iii) (1/4)-1

(iv) (-4)-1 × (-3/2)-1

(v) (3/5)-1 × (5/2)-1

Solution:

(i) 6-1

1/61 = 1/6 (we know that a-n = 1/an)

(ii) (-7)-1

1/-71 = -1/7 (we know that a-n = 1/an)

(iii) (1/4)-1

41 = 4 (we know that 1/a-n = an)

(iv) (-4)-1 × (-3/2)-1

1/-41 × (2/-3)1 (we know that a-n = 1/an, 1/a-n = an)

1/-2 × -1/3

1/6

(v) (3/5)-1 × (5/2)-1

(5/3)1 × (2/5)1

5/3 × 2/5

2/3

4. Simplify:

(i) (4-1 × 3-1)2

(ii) (5-1 ÷ 6-1)3

(iii) (2-1 + 3-1)-1

(iv) (3-1 × 4-1)-1 × 5-1

(v) (4-1 – 5-1) ÷ 3-1

Solution:

(i) (4-1 × 3-1)2

(1/4 × 1/3)2 (we know that a-n = 1/an)

(1/12)2

1/144

(ii) (5-1 ÷ 6-1)3

(1/5 ÷ 1/6)3 (we know that a-n = 1/an)

(1/5 × 6)3 (we know that 1/a ÷ 1/b = 1/a × b/1)

(6/5)3

216/125

(iii) (2-1 + 3-1)-1

(1/2 + 1/3)-1 (we know that a-n = 1/an)

LCM of 2 and 3 is 6

((3+2)/6)-1

(5/6)-1 (we know that 1/a-n = an)

6/5

(iv) (3-1 × 4-1)-1 × 5-1

(1/3 × 1/4)-1 × 1/5 (we know that a-n = 1/an)

(1/12)-1 × 1/5 (we know that 1/a-n = an)

12 × 1/5

12/5

(v) (4-1 – 5-1) ÷ 3-1

(1/4 – 1/5) ÷ 1/3 (we know that a-n = 1/an)

LCM of 4 and 5 is 20

(5-4)/20 × 3/1 (we know that 1/a ÷ 1/b = 1/a × b/1)

1/20 × 3

3/20

5. Express each of the following rational numbers with a negative exponent:

(i) (1/4)3

(ii) 35

(iii) (3/5)4

(iv) ((3/2)4)-3

(v) ((7/3)4)-3

Solution:

(i) (1/4)3

(4)-3 (we know that 1/an = a-n)

(ii) 35

(1/3)-5 (we know that 1/an = a-n)

(iii) (3/5)4

(5/3)-4 (we know that (a/b)-n = (b/a)n)

(iv) ((3/2)4)-3

(3/2)-12 (we know that (an)m = anm)

(v) ((7/3)4)-3

(7/3)-12 (we know that (an)m = anm)

6. Express each of the following rational numbers with a positive exponent:

(i) (3/4)-2

(ii) (5/4)-3

(iii) 43 × 4-9

(iv) ((4/3)-3)-4

(v) ((3/2)4)-2

Solution:

(i) (3/4)-2

(4/3)2 (we know that (a/b)-n = (b/a)n)

(ii) (5/4)-3

(4/5)3 (we know that (a/b)-n = (b/a)n)

(iii) 43 × 4-9

(4)3-9 (we know that an × am = an+m)

4-6

(1/4)6 (we know that 1/an = a-n)

(iv) ((4/3)-3)-4

(4/3)12 (we know that (an)m = anm)

(v) ((3/2)4)-2

(3/2)-8 (we know that (an)m = anm)

(2/3)8 (we know that 1/an = a-n)

7. Simplify:

(i) ((1/3)-3 – (1/2)-3) ÷ (1/4)-3

(ii) (32 – 22) × (2/3)-3

(iii) ((1/2)-1 × (-4)-1)-1

(iv) (((-1/4)2)-2)-1

(v) ((2/3)2)3 × (1/3)-4 × 3-1 × 6-1

Solution:

(i) ((1/3)-3 – (1/2)-3) ÷ (1/4)-3

(33 – 23) ÷ 43 (we know that 1/an = a-n)

(27-8) ÷ 64

19 ÷ 64

19 × 1/64 (we know that 1/a ÷ 1/b = 1/a × b/1)

19/64

(ii) (32 – 22) × (2/3)-3

(9 – 4) × (3/2)3 (we know that 1/an = a-n)

5 × (27/8)

135/8

(iii) ((1/2)-1 × (-4)-1)-1

(21 × (1/-4))-1 (we know that 1/an = a-n)

(1/-2)-1 (we know that 1/an = a-n)

-21

-2

(iv) (((-1/4)2)-2)-1

((-1/16)-2)-1 (we know that 1/an = a-n)

((-16)2)-1 (we know that 1/an = a-n)

(256)-1 (we know that 1/an = a-n)

1/256

(v) ((2/3)2)3 × (1/3)-4 × 3-1 × 6-1

(4/9)3 × 34 × 1/3 × 1/6 (we know that 1/an = a-n)

(64/729) × 81 × 1/3 × 1/6

(64/729) × 27 × 1/6

32/729 × 27 × 1/3

32/729 × 9

32/81

8. By what number should 5-1 be multiplied so that the product may be equal to (-7)-1?

Solution:

Let us consider a number x

So, 5-1 × x = (-7)-1

1/5 × x = 1/-7 (we know that 1/an = a-n)

x = (-1/7) / (1/5)

= (-1/7) × (5/1) (we know that 1/a ÷ 1/b = 1/a × b/1)

= -5/7

9. By what number should (1/2)-1 be multiplied so that the product may be equal to (-4/7)-1?

Solution:

Let us consider a number x

So, (1/2)-1 × x = (-4/7)-1

1/(1/2) × x = 1/(-4/7) (we know that 1/an = a-n)

x = (-7/4) / (2/1)

= (-7/4) × (1/2) (we know that 1/a ÷ 1/b = 1/a × b/1)

= -7/8

10. By what number should (-15)-1 be divided so that the quotient may be equal to (-5)-1?

Solution:

Let us consider a number x

So, (-15)-1 ÷ x = (-5)-1 (we know that 1/a ÷ 1/b = 1/a × b/1)

1/-15 × 1/x = 1/-5 (we know that 1/an = a-n)

1/x = (1×-15)/-5

1/x = 3

x = 1/3

11. By what number should (5/3)-2 be multiplied so that the product may be (7/3)-1?

Solution:

Let us consider a number x

So, (5/3)-2 × x = (7/3)-1

1/(5/3)2 × x = 1/(7/3) (we know that 1/an = a-n)

x = (3/7) / (3/5)2

= (3/7) / (9/25)

= (3/7) × (25/9) (we know that 1/a ÷ 1/b = 1/a × b/1)

= (1/7) × (25/3)

= 25/21

12. Find x, if

(i) (1/4)-4 × (1/4)-8 = (1/4)-4x

Solution:

(1/4)-4 × (1/4)-8 = (1/4)-4x

(1/4)-4-8 = (1/4)-4x (we know that an × am = an+m)

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

When the bases are same we can directly equate the coefficients

-12 = -4x

x = -12/-4

= 3

(ii) (-1/2)-19 ÷ (-1/2)8 = (-1/2)-2x+1

Solution:

(-1/2)-19 ÷ (-1/2)8 = (-1/2)-2x+1

(1/2)-19-8 = (1/2)-2x+1 (we know that an ÷ am = an-m)

(1/2)-27 = (1/2)-2x+1

When the bases are same we can directly equate the coefficients

-27 = -2x+1

-2x = -27-1

x = -28/-2

= 14

(iii) (3/2)-3 × (3/2)5 = (3/2)2x+1

Solution:

(3/2)-3 × (3/2)5 = (3/2)2x+1

(3/2)-3+5 = (3/2)2x+1 (we know that an × am = an+m)

(3/2)2 = (3/2)2x+1

When the bases are same we can directly equate the coefficients

2 = 2x+1

2x = 2-1

x = 1/2

(iv) (2/5)-3 × (2/5)15 = (2/5)2+3x

Solution:

(2/5)-3 × (2/5)15 = (2/5)2+3x

(2/5)-3+15 = (2/5)2+3x (we know that an × am = an+m)

(2/5)12 = (2/5)2+3x

When the bases are same we can directly equate the coefficients

12 = 2+3x

3x = 12-2

x = 10/3

(v) (5/4)-x ÷ (5/4)-4 = (5/4)5

Solution:

(5/4)-x ÷ (5/4)-4 = (5/4)5

(5/4)-x+4 = (5/4)5 (we know that an ÷ am = an-m)

When the bases are same we can directly equate the coefficients

-x+4 = 5

-x = 5-4

-x = 1

x = -1

(vi) (8/3)2x+1 × (8/3)5 = (8/3) x+2

Solution:

(8/3)2x+1 × (8/3)5 = (8/3)x+2

(8/3)2x+1+5 = (8/3) x+2 (we know that an × am = an+m)

(8/3)2x+6 = (8/3) x+2

When the bases are same we can directly equate the coefficients

2x+6 = x+2

2x-x = -6+2

x = -4

13. (i) If x= (3/2)2 × (2/3)-4, find the value of x-2.

Solution:

x= (3/2)2 × (2/3)-4

= (3/2)2 × (3/2)4 (we know that 1/an = a-n)

= (3/2)2+4 (we know that an × am = an+m)

= (3/2)6

x-2 = ((3/2)6)-2

= (3/2)-12

= (2/3)12

(ii) If x = (4/5)-2 ÷ (1/4)2, find the value of x-1.

Solution:

x = (4/5)-2 ÷ (1/4)2

= (5/4)2 ÷ (1/4)2 (we know that 1/an = a-n)

= (5/4)2 × (4/1)2 (we know that 1/a ÷ 1/b = 1/a × b/1)

= 25/16 × 16

= 25

x-1 = 1/25

14. Find the value of x for which 52x ÷ 5-3 = 55

Solution:

52x ÷ 5-3 = 55

52x+3 = 55 (we know that an ÷ am = an-m)

When the bases are same we can directly equate the coefficients

2x+3 = 5

2x = 5-3

2x = 2

x = 1

RD Sharma Solutions for Class 8 Maths Exercise 2.2 Chapter 2 Powers

Class 8 Maths Chapter 2 Powers Exercise 2.2 is based on the problems, which include all the six laws of integral exponents and the use of the decimal number system in the laws. To understand the concepts better, download the free RD Sharma Solutions Chapter 2 in PDF format with answers to all the questions. Try solving these solutions to attain good marks in the annual exam.

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