Exercise 6.1
Question 1: Find the values of each of the following:
(i) 132
(ii) 73
(iii) 34
Solution 1:
(i) 132
= 13 × 13 =169
(ii) 73
= 7 × 7 × 7 = 343
(iii) 34
= 3 × 3 × 3 × 3 = 81
Question 2: Find the value of each of the following:
(i) (-7)2
(ii) (-3)4
(iii) (-5)5
Solution 2:
(i) (-7)2
= (-7) × (-7)
= 49
(ii) (-3)4
= (-3) × (-3) × (-3) × (-3)
= 81
(iii) (-5)5
= (-5) × (-5) × (-5) × (-5) × (-5)
= -3125
Question 3: Simplify:
(i) 3 × 102
(ii) 22 × 53
(iii) 33 × 52
Solution 3:
(i) 3 × 102
= 3 × 10 × 10
= 3 × 100
= 300
(ii) 22 × 53
= 2 × 2 × 5 × 5 × 5
= 4 × 125
= 500
(iii) 33 × 52
= 3 × 3 × 3 × 5 × 5
= 27 × 25
= 675
Question 4: Simply:
(i) 32 × 104
(ii) 24 × 32
(iii) 52 × 34
Solution 4:
(i) 32 × 104
= 3 × 3 × 10 × 10 × 10 × 10
= 9 × 10000
= 90000
(ii) 24 × 32
= 2 × 2 × 2 × 2 × 3 × 3
= 16 × 9
= 144
(iii) 52 × 34
= 5 × 5 × 3 × 3 × 3 × 3
= 25 × 81
= 2025
Question 5: Simplify:
(i) (-2) × (-3)3
(ii) (-3)2 × (-5)3
(iii) (-2)5 × (-10)2
Solution 5:
(i) (-2) × (-3)3
= (-2) × (-3) × (-3) × (-3)
= (-2) × (-27)
= 54
(ii) (-3)2 × (-5)3
= (-3) × (-3) × (-5) × (-5) × (-5)
= 9 × (-125)
= -1125
(iii) (-2)5 × (-10)2
= (-2) × (-2) × (-2) × (-2) × (-2) × (-10) × (-10)
= (-32) × 100
= -3200
Question 6: Simplify:
(i) (3/4)2
(ii) ((-2)/3)4
(iii) ((-4)/5)5
Solution 6:
(i) (3/4)2
= (3/4) × (3/4)
= (9/16)
(ii) ((-2)/3)4
= ((-2)/3) × ((-2)/3) × ((-2)/3) × ((-2)/3)
= (16/81)
(iii) ((-4)/5)5
= ((-4)/5) × ((-4)/5) × ((-4)/5) × ((-4)/5) × ((-4)/5)
= ((-1024)/3125)
Question 7: Identify the greater number in each of the following:
(i) 25 or 52
(ii) 34 or 43
(iii) 35 or 53
Solution 7:
(i) 25 or 52
25 = 2 × 2 × 2 × 2 × 2 = 32
52 = 5 × 5 = 25
So, 25 > 52
(ii) 34 or 43
34 = 3 × 3 × 3 × 3 = 81
43 = 4 × 4 × 4 = 64
So, 34 > 43
(iii) 35 or 53
35 = 3 × 3 × 3 × 3 × 3 = 243
53 = 5 × 5 × 5 = 125
So, 35 > 53
Solution 10:
(i) 512
By the prime factorization of 512
= 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2
= 29
(ii) 625
By the prime factorization of 625
= 5 × 5 × 5 × 5
= 54
(iii) 729
By the prime factorization of 729
= 3 × 3 × 3 × 3 × 3 × 3
= 36
Question 11: Express each of the following numbers as a product of powers of their prime factors:
(i) 36
(ii) 675
(iii) 392
Solution 11:
(i) 36
By the prime factorization of 36
= 2 × 2 × 3 × 3
= 22 × 32
(ii) 675
By the prime factorization of 675
= 3 × 3 × 3 × 5 × 5
= 33 × 52
(iii) 392
By the prime factorization of 392
= 2 × 2 × 2 × 7 × 7
= 23 × 72
Question 12: Express each of the following numbers as a product of powers of their prime factors:
(i) 450
(ii) 2800
(iii) 24000
Solution 12:
(i) 450
By the prime factorization of 450
= 2 × 3 × 3 × 5 × 5
= 2 × 32 × 52
(ii) 2800
By the prime factorization of 2800
= 2 × 2 × 2 × 2 × 5 × 5 × 7
= 24 × 52 × 7
(iii) 24000
By the prime factorization of 24000
= 2 × 2 × 2 × 2 × 2 × 2 × 3 × 5 × 5 × 5
= 26 × 3 × 53
Exercise 6.2
Question 1: Using laws of exponents, simplify and write the answer in exponential form
(i) 23 × 24 × 25
(ii) 512 ÷ 53
(iii) (72)3
(iv) (32)5 ÷ 34
(v) 37 × 27
(vi) (521 ÷ 513) × 57
Solution 1:
(i) 23 × 24 × 25
According to the law of exponents: am × an × ap = a(m+n+p)
So, the equation is written as
= 2(3 + 4 + 5)
= 212
(ii) 512 ÷ 53
According to the law of exponents: am ÷ an = am-n
So, the equation is written as
= 512 – 3
= 59
(iii) (72)3
According to the law of exponents: (am)n = am x n
So the equation is written as
= 72 x 3
= 76
(iv) (32)5 ÷ 34
According to the law of exponents: (am)n = am x n
So the equation is written as
= 32 x 5 ÷ 34
= 3 10 ÷ 34
According to the law of exponents: am ÷ an = am-n
= 3(10 – 4)
= 36
(v) 37 × 27
According to the law of exponents: am × bm = (a × b)m
So the equation is written as
= (3 × 2)7
= 67
(vi) (521 ÷ 513) × 57
According to the law of exponents: am ÷ an = am-n
= 5(21 -13) × 57
So the equation is written as
= 58 × 57
According to the law of exponents: am × an = a(m +n)
So the equation is written as
= 5(8+7)
= 515
Question 2: Simplify and express each of the following in exponential form:
(i) {(23)4 × 28} ÷ 212
(ii) (82 × 84) ÷ 83
(iii) (57/52)× 53
(iv) (54× x10y5)/ (54 × x7y4)
Solution 2:
(i) {(23)4 × 28} ÷ 212
= {212 × 28} ÷ 212
= 2(12 + 8) ÷ 212
= 220 ÷ 212
= 2 (20 – 12)
= 28
(ii) (82 × 84) ÷ 83
= 8(2 + 4) ÷ 83
= 86 ÷ 83
= 8(6-3)
= 83
= (23)3
= 29
(iii) (57/52) × 53
= 5(7-2) × 53
= 55 × 53
= 5(5 + 3)
= 58
(iv) (54× x10y5)/ (54 × x7y4)
= (54-4× x10-7y5-4)
= 50x3y1 [since 50 = 1]
= 1x3y
Question 3: Simplify and express each of the following in exponential form:
(i) {(32)3 × 26} × 56
(ii)(x/y) 12 × y24 × (23)4
(iii)(5/2)6 × (5/2)2
(iv) (2/3)5× (3/5)5
Solution 3:
(i) {(32)3 × 26} × 56
= {36 × 26} × 56
= 66 × 56
= 306
(ii) (x/y)12 × y24 × (23)4
= (x12/y12) × y24 × 212
= x12 × y24-12 × 212
= x12 × y12 × 212
= (2xy)12
(iii)(5/2)6 × (5/2)2
= (5/2)6+2
= (5/2)8
(iv) (2/3)5× (3/5)5
= (2/5)5
Question 4: Write 9 × 9 × 9 × 9 × 9 in exponential form with base 3.
Solution 4:
= 9 × 9 × 9 × 9 × 9
= (9)5 = (32)5
= 310
Question 5: Simplify and write each of the following in exponential form:
(i) (25)3 ÷ 53
(ii) (81)5 ÷ (32)5
(iii) 98 × (x2)5/ (27)4 × (x3)2
(iv) 32 × 78 × 136/ 212 × 913
Solution 5:
(i) (25)3 ÷ 53
= (52)3 ÷ 53
= 56 ÷ 53
= 56 – 3
= 53
(ii) (81)5 ÷ (32)5
= (81)5 ÷ 310 ->81 = 34
= (34)5 ÷ 310
= 320 ÷ 310
= 320-10
= 310
(iii) 98 × (x2)5/ (27)4 × (x3)2
= (32)8 × (x2)5/ (33)4× (x3)2
= 316 × x10/312 × x6
= 316-12 × x10-6
= 34 × x4
= (3x)4
(iv) (32 × 78 × 136)/ (212 × 913)
= (32 × 7276 × 136)/(212× 133 × 73)
= (212 × 76 × 136)/(212× 133 × 73)
= (76 × 136)/(133 × 73)
= 916/913
= 916-3
= 913
Question 6: Simplify:
(i) (35)11 × (315)4 – (35)18 × (35)5
(ii) (16 × 2n+1 – 4 × 2n)/(16 × 2n+2 – 2 × 2n+2)
(iii) (10 × 5n+1 + 25 × 5n)/(3 × 5n+2 + 10 × 5n+1)
(iv) (16)7 ×(25)5× (81)3/(15)7 ×(24)5× (80)3
Solution 6:
(i) (35)11 × (315)4 – (35)18 × (35)5
= (3)55 × (3)60 – (3)90 × (3)25
= 3 55+60 – 390+25
= 3115 – 3115
= 0
(ii) (16 × 2n+1 – 4 × 2n)/(16 × 2n+2 – 2 × 2n+2)
= (24 × 2(n+1) -22 × 2n)/(24 × 2(n+2) -2n+1 × 22)
= 22 × 2(n+3-2n)/)22× 2(n+4-2n+1)
= 2n × 23 – 2n/ 2n × 24 – 2n × 2
= 2n(23 – 1)/ 2n(24 – 1)
= 8 -1 /16 -2
= 7/14
= (1/2)
(iii) (10 × 5n+1 + 25 × 5n)/(3 × 5n+2 + 10 × 5n+1)
= (10 × 5n+1 + 52 × 5n)/(3 × 5n+2 + (2 × 5) × 5n+1)
= (10 × 5n+1 + 5 × 5n+1)/(3 × 5n+2 + (2 × 5) × 5n+1)
= 5n+1 (10+5)/ 5n+1 (10+15)
= 15/25
= (3/5)
(iv) (16)7 ×(25)5× (81)3/(15)7 ×(24)5× (80)3
= (16)7 × (52)5× (34)3/(3 × 5 )7 ×(3 × 8)5× (16 × 5)3
= (16)7 × (52)5× (34)3/37 × 57 × 35 × 85× 163× 53
= (16)7/ 85 × 16 3
= (16)4/85
= (2 × 8)4/85
= 24/8
= (16/8)
= 2
Question 7: Find the values of n in each of the following:
(i) 52n × 53 = 511
(ii) 9 × 3n = 37
(iii) 8 × 2n+2 = 32
(iv) 72n+1 ÷ 49 = 73
(v) (3/2)4 × (3/2) 5 = (3/2)2n+1
(vi) (2/3)10× {(3/2)2}5 = (2/3)2n – 2
Solution:
(i) 52n × 53 = 511
= 52n+3 = 511
On equating the coefficients, we find:
2n + 3 = 11
⇒2n = 11- 3
⇒2n = 8
⇒ n = (8/2)
⇒ n = 4
(ii) 9 × 3n = 37
= (3)2 × 3n = 37
= (3)2+n = 37
On equating the coefficients, we find:
2 + n = 7
⇒ n = 7 – 2 = 5
(iii) 8 × 2n+2 = 32
= (2)3 × 2n+2 = (2)5 [since 23 = 8 and 25 = 32]
= (2)3+n+2 = (2)5
On equating the coefficients, we find:
3 + n + 2 = 5
⇒ n + 5 = 5
⇒ n = 5 -5
⇒ n = 0
(iv) 72n+1 ÷ 49 = 73
= 72n+1 ÷ 72 = 73 [since 49 = 72]
= 72n+1-2 = 73
= 72n-1=73
On equating the coefficients, we find:
2n – 1 = 3
⇒ 2n = 3 + 1
⇒ 2n = 4
⇒ n =4/2 =2
(v) (3/2)4 × (3/2) 5 = (3/2)2n+1
= (3/2)4+5 = (3/2)2n+1
= (3/2)9 = (3/2)2n+1
On equating the coefficients, we find:
2n + 1 = 9
⇒ 2n = 9 – 1
⇒ 2n = 8
⇒ n = 8/2
=4
(vi) (2/3)10× {(3/2)2}5 = (2/3)2n – 2
= (2/3)10 × (3/2)10 = (2/3)2n – 2
= 2 10 × 310/310 × 210 = (2/3)2n – 2
= 1 = (2/3)2n – 2
= (2/3)0 = (2/3)2n – 2
On equating the coefficients, we find:
0 =2n -2
2n -2 =0
2n =2
n = 1
Question 8: If (9n × 32 × 3n – (27)n)/ (33)5 × 23 = (1/27), find the value of n.
Solution 8:
= (9n × 32 × 3n – (27)n)/ (33)5 × 23 = (1/27)
= (32)n × 33 × 3n – (33)n/ (315 × 23) = (1/27)
= 3(2n+2+n) – (33)n/ (315 × 23) = (1/27)
= 3(3n+2)– (33)n/ (315 × 23) = (1/27)
= 33n × 32 – 33n/ (315 × 23) = (1/27)
= 33n × (32 – 1)/ (315 × 23) = (1/27)
= 33n × (9 – 1)/ (315 × 23) = (1/27)
= 33n × (8)/ (315 × 23) = (1/27)
= 33n × 23/ (315 × 23) = (1/27)
= 33n/315 = (1/27)
= 33n-15 = (1/27)
= 33n-15 = (1/33)
= 33n-15 = 3-3
On equating the coefficients, we find:
3n -15 = -3
⇒ 3n = -3 + 15
⇒ 3n = 12
⇒ n = 12/3 = 4
Exercise 6.3
Question 1: Express the following numbers in the standard form:
(i) 3908.78
(ii) 5,00,00,000
(iii) 3,18,65,00,000
(iv) 846 × 107
(v)723 × 109
Solution 1:
(i) 3908.78
= 3.90878 × 103 [Here the decimal point is moved left up to 3 places]
(ii) 5,00,00,000
= 5,00,00,000.00 = 5 × 107 [Here, the decimal point is moved left up to 7 places]
(iii) 3,18,65,00,000
= 3,18,65,00,000.00
= 3.1865 × 109 [here, the decimal point is moved left up to 9 places]
(iv) 846 × 107
= 8.46 × 102 × 10 [here the decimal point is moved left up to 2 places]
= 8.46 × 109 [since am × an = am+n]
(v) 723 × 109
= 7.23 × 102 × 109 [here the decimal point is moved left up to 2 places]
= 7.23 × 1011 [ since am × an = am+n]
Question 2: Write the following numbers in the usual form:
(i) 4.83 × 107
(ii) 3.21 × 105
(iii) 3.5 × 103
Solution 2:
(i) 4.83 × 107
= 483 × 107-2 [hence the decimal point is moved right up to 2 places]
= 483 × 105
= 4, 83, 00,000
(ii) 3.21 × 105
= 321 × 105-2 [hence the decimal point is moved right up to 2 places]
= 321 × 103
= 3, 21,000
(iii) 3.5 × 103
= 35 × 103-1 [hence the decimal point is moved right up to 1 places]
= 35 × 102
= 3,500
Question :3. Express the numbers appearing in the following statements in the standard form:
(i) The distance between the Earth and the Moon is 384,000,000 meters.
(ii) Diameter of the Earth is 1, 27, 56,000 meters.
(iii) Diameter of the Sun is 1,400,000,000 meters.
(iv) The universe is estimated to be about 12,000,000,000 years old.
Solution:
(i) The distance between the Earth and the Moon is 384,000,000 meters.
Distance between the Earth and the Moon is 3.84 × 108 meters.
[Here the decimal point is moved left up to 8 places.]
(ii) Diameter of the Earth is 1, 27, 56,000 meters.
Diameter of the Earth is 1.2756 × 107 meters.
[Here the decimal point is moved left up to 7 places]
(iii) Diameter of the Sun is 1,400,000,000 meters.
Diameter of the Sun is 1.4 × 109 meters.
[Here the decimal point is moved left up to 9 places]
(iv) The universe is estimated to be about 12,000,000,000 years old.
The universe is estimated to be about 1.2× 1010 years old.
[Here the decimal point is moved left up to 10 places]
Exercise 6.4
Question 1: Write the following numbers in the expanded exponential forms:
(i) 20068
(ii) 420719
(iii) 7805192
(iv) 5004132
(v) 927303
Solution 1:
(i) 20068
= 2 × 104 + 0 × 103 + 0 × 102 + 6 × 101 + 8 × 100
(ii) 420719
= 4 × 105 + 2 × 104 + 0 × 103 + 7 × 102 + 1 × 101 + 9 × 100
(iii) 7805192
= 7 × 106 + 8 × 105 + 0 × 104 + 5 × 103 + 1 × 102 + 9 × 101 + 2 × 100
(iv) 5004132
= 5 × 106 + 0 × 105 + 0 × 104 + 4 × 103 + 1 × 102 + 3 × 101 + 2 × 100
(v) 927303
= 9 × 105 + 2 × 104 + 7 × 103 + 3 × 102 + 0 × 101 + 3 × 100
Question 2: Find the number from each of the following expanded forms:
(i) 7 × 104 + 6 × 103 + 0 × 102 + 4 × 101 + 5 × 100
(ii) 5 × 105 + 4 × 104 + 2 × 103 + 3 × 100
(iii) 9 × 105 + 5 × 102 + 3 × 101
(iv) 3 × 104 + 4 × 102 + 5 × 100
Solution 2:
(i) 7 × 104 + 6 × 103 + 0 × 102 + 4 × 101 + 5 × 100
= 7 × 10000 + 6 × 1000 + 0 × 100 + 4 × 10 + 5 × 1
= 70000 + 6000 + 0 + 40 + 5
= 76045
(ii) 5 × 105 + 4 × 104 + 2 × 103 + 3 × 100
= 5 × 100000 + 4 × 10000 + 2 × 1000 + 3 × 1
= 500000 + 40000 + 2000 + 3
= 542003
(iii) 9 × 105 + 5 × 102 + 3 × 101
= 9 × 100000 + 5 × 100 + 3 × 10
= 900000 + 500 + 30
= 900530
(iv) 3 × 104 + 4 × 102 + 5 × 100
= 3 × 10000 + 4 × 100 + 5 × 1
= 30000 + 400 + 5
= 30405