Class 8 RD Sharma Solutions Chapter 2 Powers Exercise 2.2 | Set 2 (original) (raw)
Last Updated : 20 Sep, 2024
Chapter 2 of RD Sharma's Class 8 mathematics textbook covers Powers. Exercise 2.2 Set 2 typically deals with the laws of exponents and their applications. This set helps students understand how to manipulate expressions with exponents, including multiplication and division of powers with the same base.
**Question 11. By what number should (5/3) -2 be multiplied so that the product may be (7/3) -1 ?
**Solution:
Let the number be x
(5/3)-2 × x = (7/3)-1
1/(5/3)2 × x = 1/(7/3) (1/an = a-n)
x = (3/7) / (3/5)2
= (3/7) / (9/25)
= (3/7) × (25/9)
3 is the common factor
= (1/7) × (25/3)
= 25/21
**Question 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 (an × am = an + m)
(1/4)-12 = (1/4)-4x
When the bases are same, exponents are equated
-12 = -4x
x = -12/-4
Transposing -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, exponents are equated
-27 = -2x + 1
Transposing 1
-2x = -27 - 1
-2x = -28
Transposing -2
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 (an × am = an + m)
(3/2)2 = (3/2)2x + 1
When the bases are same, exponents are equated
2 = 2x + 1
Transposing 1
2x = 2 - 1
2x = 1
Transposing 2
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 (an × am = an + m)
(2/5)12 = (2/5)2+3x
When the bases are same, exponents are equated
12 = 2 + 3x
Transposing 2
3x = 12 - 2
3x = 10
Transposing 3
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 (an ÷ am = an - m)
When the bases are same, exponents are equated
-x + 4 = 5
Transposing 4
-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 (an × am = an + m)
(8/3)2x+6 = (8/3) x+2
When the bases are same, exponents are equated
2x + 6 = x + 2
Transposing 6 and x
2x - x = -6 + 2
x = -4
**Question 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 (1/an = a-n)
= (3/2)2 + 4 (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 (1/an = a-n)
= (5/4)2 × (4/1)2
= 25/16 × 16
16 is the common factor
= 25
x-1 = 1/25
**Question 14. Find the value of x for which 5 2x ÷ 5 -3 = 5 5
**Solution:
52x ÷ 5-3 = 55
52x + 3 = 55 (an ÷ am = an - m)
When the bases are same, exponents are equated
2x + 3 = 5
Transposing 3
2x = 5 - 3
2x = 2
Transposing 2
x = 1
Summary
Exercise 2.2 Set 2 in Chapter 2 of RD Sharma's Class 8 mathematics textbook focuses on reinforcing students' understanding of the laws of exponents. It covers key concepts such as multiplying powers with the same base (by adding exponents), dividing powers with the same base (by subtracting exponents), and raising a power to another power (by multiplying exponents). The exercise helps students develop their algebraic manipulation skills and logical thinking. By solving these problems, students learn to simplify complex expressions involving exponents, which is crucial for more advanced mathematical concepts they will encounter in higher grades. The set also includes word problems that require students to apply these laws in practical scenarios, enhancing their problem-solving abilities.Powers