Saturday, April 23, 2011

13. Exponents and Powers

Exercise 13.1

Question 1:
Find the value of:
(i) 26 (ii) 93
(iii) 112 (iv)54

Answer:
(i) 26 = 2 × 2 × 2 × 2 × 2 × 2 = 64
(ii) 93 = 9 × 9 × 9 = 729
(iii) 112 = 11 × 11 = 121
(iv)54 = 5 × 5 × 5 × 5 = 625

Question 2:

Express the following in exponential form:
(i) 6 × 6 × 6 × 6 (ii) t × t
(iii) b × b × b × b (iv) 5 × 5 × 7 ×7 × 7
(v) 2 × 2 × a × a (vi) a × a × a × c × c × c × c × d

Answer: 
(i) 6 × 6 × 6 × 6 = 64
(ii) t × t= t2
(iii) b × b × b × b = b4
(iv) 5 × 5 × 7 × 7 × 7 = 52 × 73
(v) 2 × 2 × a × a = 22 × a2
(vi) a × a × a × c × c × c × c × d = a3 c4 d

Question 3:

Express the following numbers using exponential notation:
(i) 512 (ii) 343
(iii) 729 (iv) 3125

Answer:

(i) 512 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 29
(ii) 343 = 7 × 7 × 7 = 73
(iii) 729 = 3 × 3 × 3 × 3 × 3 × 3 = 36
(iv) 3125 = 5 × 5 × 5 × 5 × 5 = 55\

Question 4:
Identify the greater number, wherever possible, in each of the following?
(i) 43 or 34 (ii) 53 or 35
(iii) 28 or 82 (iv) 1002 or 2100
(v) 210 or 102

Answer: 
(i) 43 = 4 × 4 × 4 = 64
34 = 3 × 3 × 3 × 3 = 81
Therefore, 34 > 43
(ii) 53 = 5 × 5 × 5 =125
35 = 3 × 3 × 3 × 3 × 3 = 243
Therefore, 35 > 53
(iii) 28 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 256
82 = 8 × 8 = 64
Therefore, 28 > 82
(iv)1002 or 2100
210 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 1024
2100 = 1024 × 1024 × 1024 × 1024 × 1024 × 1024 × 1024 × 1024 ×1024 × 1024
1002 = 100 × 100 = 10000
Therefore, 2100 > 1002
(v) 210 and 102
210 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 1024
102 = 10 × 10 = 100
Therefore, 210 > 102

Question 5:
Express each of the following as product of powers of their prime factors:
(i) 648 (ii) 405
(iii) 540 (iv) 3,600
Answer: 
(i) 648 = 2 × 2 × 2 × 3 × 3 × 3 × 3 = 23. 34
(ii) 405 = 3 × 3 × 3 × 3 × 5 = 34 . 5
(iii) 540 = 2 × 2 × 3 × 3 × 3 × 5 = 22. 33. 5
(iv) 3600 = 2 × 2 × 2 × 2 × 3 × 3 × 5 × 5 = 24. 32. 52

Question 6:
Simplify:
(i) 2 × 103 (ii) 72 × 22
(iii) 23 × 5 (iv) 3 × 44
(v) 0 × 102 ­­­­­ (vi) 52 × 33
(vii) 24 × 32 (viii) 32 × 104

Answer: 
(i) 2 × 103 = 2 × 10 × 10 × 10 = 2 × 1000 = 2000
(ii) 72 × 22 = 7 × 7 × 2 × 2 = 49 × 4 = 196
(iii) 23 × 5 = 2 × 2 × 2 × 5 = 8 × 5 = 40
(iv) 3 × 44 = 3 × 4 × 4 × 4 × 4 = 3 × 256 = 768
(v) 0 × 102 = 0 × 10 × 10 = 0
(vi) 52 × 33 = 5 × 5 × 3 × 3 × 3 = 25 × 27 = 675
(vii) 24 × 32 = 2 × 2 × 2 × 2 × 3 × 3 = 16 × 9 = 144
(viii) 32 × 104 = 3 × 3 × 10 × 10 × 10 × 10 = 9 × 10000 = 90000

Question 7:

Simplify:
(i) (− 4)3 (ii) (− 3) × (− 2)3
(iii) (− 3)2 × (− 5)2 (iv)(− 2)3 × (−10)3

Answer: 
(i) (−4)3 = (−4) × (−4) × (−4) = −64
(ii) (−3) × (−2)3 = (−3) × (−2) × (−2) × (−2) = 24
(iii) (−3)2 × (−5)2 = (−3) × (−3) × (−5) × (−5) = 9 × 25 = 225
(iv) (−2)3 × (−10)3 = (−2) × (−2) × (−2) × (−10) × (−10) × (−10)
= (−8) × (−1000) = 8000

Question 8:

Compare the following numbers:
(i) 2.7 × 1012; 1.5 × 108
(ii) 4 × 1014; 3 × 1017

Answer: 
(i) 2.7 × 1012; 1.5 × 108
2.7 × 1012 > 1.5 × 108
(ii) 4 × 1014; 3 × 1017
3 × 1017 > 4 × 1014

 Exercise 13.2

Question 1:
Using laws of exponents, simplify and write the answer in exponential form:
(i) 32 × 34 × 38 (ii) 615 ÷ 610 (iii) a3 × a2
(iv) 7x× 72 (v) (vi) 25 × 55
(vii) a4 × b4 (viii) (34)3
(ix) (x) 8t ÷ 82

Answer:
(i) 32 × 34 × 38 = (3)2 + 4 + 8 (am × an = am+n)
= 314
(ii) 615 ÷ 610 = (6)15 − 10 (am ÷ an = amn)
= 65
(iii) a3 × a2 = a(3 + 2) (am × an = am+n)
= a5
(iv) 7x + 72 = 7x + 2 (am × an = am+n)
(v) (52)3 ÷ 53
= 52 × 3 ÷ 53 (am)n = amn
= 56 ÷ 53
= 5(6 − 3) (am ÷ an = amn)
= 53
(vi) 25 × 55
= (2 × 5)5 [am × bm = (a × b)m]
= 105
(vii) a4 × b4
= (ab)4 [am × bm = (a × b)m]
(viii) (34)3 = 34 × 3 = 312 (am)n = amn
(ix) (220 ÷ 215) × 23
= (220 − 15) × 23 (am ÷ an = amn)
= 25 × 23
= (25 + 3) (am × an = am+n)
= 28
(x) 8t ÷ 82 = 8(t − 2) (am ÷ an = amn)

Question 2:

Simplify and express each of the following in exponential form:
(i) (ii) (iii)
(iv) (v) (vi) 20 + 30 + 40
(vii) 20 × 30 × 40 (viii) (30 + 20) × 50 (ix)
(x) (xi) (xii)
Answer: 
(i)

(ii) [(52)3 × 54] ÷ 57
= [52 × 3 × 54] ÷ 57 (am)n = amn
= [56 × 54] ÷ 57
= [56 + 4] ÷ 57 (am × an = am+n)
= 510 ÷ 57
= 510 − 7 (am ÷ an = amn)
= 53
(iii) 254 ÷ 53 = (5 ×5)4 ÷ 53
= (52)4 ÷ 53
= 52 × 4 ÷ 53 (am)n = amn
= 58 ÷ 53
= 58 − 3 (am ÷ an = amn)
= 55
(iv)

= 1 × 7 × 115 = 7 × 115
(v)

(vi) 20 + 30 + 40 = 1 + 1 + 1 = 3
(vii) 20 × 30 × 40 = 1 × 1 × 1 = 1
(viii) (30 + 20) × 50 = (1 + 1) × 1 = 2
(ix)

(x)

(xi)

(xii) (23 × 2)2 = (am × an = am+n)
= (24)2 = 24 × 2 (am)n = amn
= 28
 
 Question 3:
Say true or false and justify your answer:
(i) 10 × 1011 = 10011 (ii) 23 > 52
(iii) 23 × 32 = 65 (iv) 30 = (1000)0

Answer: 
(i) 10 × 1011 = 10011
L.H.S. = 10 × 1011 = 1011 + 1 (am × an = am+n)
= 1012
R.H.S. = 10011 = (10 ×10)11= (102)11
= 102 × 11 = 1022 (am)n = amn
As L.H.S. ≠ R.H.S.,
Therefore, the given statement is false.
(ii) 23 > 52
L.H.S. = 23 = 2 × 2 × 2 = 8
R.H.S. = 52 = 5 × 5 = 25
As 25 > 8,
Therefore, the given statement is false.
(iii) 23 × 32 = 65
L.H.S. = 23 × 32 = 2 × 2 × 2 × 3 × 3 = 72
R.H.S. = 65 = 7776
As L.H.S. ≠ R.H.S.,
Therefore, the given statement is false.
(iv) 30 = (1000)0
L.H.S. = 30 = 1
R.H.S. = (1000)0 = 1 = L.H.S.
Therefore, the given statement is true.

Question 4:

Express each of the following as a product of prime factors only in exponential form:
(i) 108 × 192 (ii) 270
(iii) 729 × 64 (iv) 768

Answer:

(i) 108 × 192
= (2 × 2 × 3 × 3 × 3) × (2 × 2 × 2 × 2 × 2 × 2 × 3)
= (22 × 33) × (26 × 3)
= 26 + 2 × 33 + 1 (am × an = am+n)
= 28 × 34
(ii) 270 = 2 × 3 × 3 × 3 × 5 = 2 × 33 × 5
(iii) 729 × 64 = (3 × 3 × 3 × 3 × 3 × 3) × (2 × 2 × 2 × 2 × 2 × 2)
= 36 × 26
(iv) 768 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3 = 28 × 3

Question 5:

Simplify:
(i) (ii) (iii)
 Answer:
(i)

(ii)

(iii)


 
Exercise 13.3 

Question 1:
Write the following numbers in the expanded forms:
279404, 3006194, 2806196, 120719, 20068

Answer:

279404 = 2 × 105 + 7 × 104 + 9 × 103 + 4 × 102 + 0 × 101 + 4 × 100
3006194 = 3 × 106 + 0 × 105 + 0 × 104 + 6 × 103 + 1 × 102 + 9 × 101 + 4 × 100
2806196 = 2 × 106 + 8 × 105 + 0 × 104 + 6 × 103 + 1 × 102 + 9 × 101 + 6 × 100
120719 = 1 × 105 + 2 × 104 + 0 × 103 + 7 × 102 + 1 × 101 + 9 × 100
20068 = 2 × 104 + 0 × 103 + 0 × 102 + 6 × 101 + 8 × 100


Question 2:
Find the number from each of the following expanded forms:
(a) 8 × 104 + 6 × 103 + 0 × 102 + 4 × 101 + 5 × 100
(b) 4 × 105 + 5 × 103 + 3 × 102 + 2 × 100
(c) 3 × 104 + 7 × 102 + 5 × 100
(d) 9 × 105 + 2 × 102 + 3 × 101

Answer: 
(a) 8 × 104 + 6 × 103 + 0 × 102 + 4 × 101 + 5 × 100
= 86045
(b) 4 × 105 + 5 × 103 + 3 × 102 + 2 × 100
= 405302
(c) 3 × 104 + 7 × 102 + 5 × 100
= 30705
(d) 9 × 105 + 2 × 102 + 3 × 101
= 900230

Question 3:

Express the following numbers in standard form:
(i) 5, 00, 00, 000 (ii) 70, 00, 000
(iii) 3, 18, 65, 00, 000 (iv) 3, 90, 878
(v) 39087.8 (vi) 3908.78

Answer:

(i) 50000000 = 5 × 107
(ii) 7000000 = 7 × 106
(iii) 3186500000 = 3.1865 × 109
(iv) 390878 = 3.90878 × 105
(v) 39087.8 = 3.90878 × 104
(vi) 3908.78 = 3.90878 × 103

Question 4:
Express the number appearing in the following statements in standard form.
(a) The distance between Earth and Moon is 384, 000, 000 m.
(b) Speed of light in vacuum is 300, 000, 000 m/s.
(c) Diameter of the Earth is 1, 27, 56, 000 m.
(d) Diameter of the Sun is 1, 400, 000, 000 m.
(e) In a galaxy there are on an average 100, 000, 000, 000 stars.
(f) The universe is estimated to be about 12, 000, 000, 000 years old.
(g) The distance of the Sun from the centre of the Milky Way Galaxy is estimated to be 300, 000, 000, 000, 000, 000, 000 m.
(h) 60, 230, 000, 000, 000, 000, 000, 000 molecules are contained in a drop of water weighing 1.8 gm.
(i) The earth has 1, 353, 000, 000 cubic km of sea water.
(j) The population of India was about 1, 027, 000, 000 in March, 2001.

Answer:

(a) 3.84 × 108 m
(b) 3 × 108 m/s
(c) 1.2756 × 107 m
(d) 1.4 × 109 m
(e) 1 × 1011 stars
(f) 1.2 × 1010 years
(g) 3 × 1020 m
(h) 6.023 × 1022
(i) 1.353 × 109 cubic km
(j) 1.027 × 109

 
 
 
  
 

 
 
 
 
 

2 comments:

  1. pls give questions(ncert) based on congruence of triangles also...

    ReplyDelete