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