Class 7: Maths Chapter 5 solutions. Complete Class 7 Maths Chapter 5 Notes.
Selina Class 7 ICSE Solutions Mathematics : Chapter 5- Exponents (Including Laws of Exponents)
Selina 7th Maths Chapter 5, Class 7 Maths Chapter 5 solutions
Exercise 5A page: 73
1. Find the value of:
(i) 62
(ii) 73
(iii) 44
(iv) 55
(v) 83
(vi) 75
Solution:
(i) 62
It can be written as
= 6 × 6
= 36
(ii) 73
It can be written as
= 7 × 7 × 7
= 343
(iii) 44
It can be written as
= 4 × 4 × 4 × 4
= 256
(iv) 55
It can be written as
= 5 × 5 × 5 × 5 × 5
= 3125
(v) 83
It can be written as
= 8 × 8 × 8
= 512
(vi) 75
It can be written as
= 7 × 7 × 7 × 7 × 7
= 16807
2. Evaluate:
(i) 23 × 42
(ii) 23 × 52
(iii) 33 × 52
(iv) 22 × 33
(v) 32 × 53
(vi) 53 × 24
(vii) 32 × 42
(viii) (4 × 3)3
(ix) (5 × 4)2
Solution:
(i) 23 × 42
It can be written as
= 2 × 2 × 2 × 4 × 4
On further calculation
= 8 × 16
= 128
(ii) 23 × 52
It can be written as
= 2 × 2 × 2 × 5 × 5
On further calculation
= 8 × 25
= 200
(iii) 33 × 52
It can be written as
= 3 × 3 × 3 × 5 × 5
On further calculation
= 27 × 25
= 675
(iv) 22 × 33
It can be written as
= 2 × 2 × 3 × 3 × 3
On further calculation
= 4 × 27
= 108
(v) 32 × 53
It can be written as
= 3 × 3 × 5 × 5 × 5
On further calculation
= 9 × 125
= 1125
(vi) 53 × 24
It can be written as
= 5 × 5 × 5 × 2 × 2 × 2 × 2
On further calculation
= 125 × 16
= 2000
(vii) 32 × 42
It can be written as
= 3 × 3 × 4 × 4
On further calculation
= 9 × 16
= 144
(viii) (4 × 3)3
It can be written as
= 4 × 4 × 4 × 3 × 3 × 3
On further calculation
= 64 × 27
= 1728
(ix) (5 × 4)2
It can be written as
= 5 × 5 × 4 × 4
On further calculation
= 25 × 16
= 400
3. Evaluate:
(i) (3/4)4
(ii) (-5/6)5
(iii) (-3/-5)3
Solution:
(i) (3/4)4
It can be written as
= (3/4) × (3/4) × (3/4) × (3/4)
On further calculation
= (3 × 3 × 3 × 3)/ (4 × 4 × 4 × 4)
= 81/256
(ii) (-5/6)5
It can be written as
= (-5/6) × (-5/6) × (-5/6) × (-5/6) × (-5/6)
On further calculation
= [(-5) × (-5) × (-5) × (-5) × (-5)]/ (6 × 6 × 6 × 6 × 6)
= -3125/776
(iii) (-3/-5)3
It can be written as
= (-3/-5) × (-3/-5) × (-3/-5)
On further calculation
= [(-3) × (-3) × (-3)]/ [(-5) × (-5) × (-5)]
= 27/125
4. Evaluate:
(i) (2/3)3 × (3/4)2
(ii) (-3/4)3 × (2/3)4
(iii) (3/5)2 × (-2/3)3
Solution:
(i) (2/3)3 × (3/4)2
It can be written as
= (2/3) × (2/3) × (2/3) × (3/4) × (3/4)
On further calculation
= 8/27 × 9/16
= 1/6
(ii) (-3/4)3 × (2/3)4
It can be written as
= (-3/4) × (-3/4) × (-3/4) × (2/3) × (2/3) × (2/3) × (2/3)
On further calculation
= -27/64 × 16/81
= -1/12
(iii) (3/5)2 × (-2/3)3
It can be written as
= (3/5) × (3/5) × (-2/3) × (-2/3) × (-2/3)
On further calculation
= 9/25 × (-8/27)
= -8/75
5. Which is greater:
(i) 23 or 32
(ii) 25 or 52
(iii) 43 or 34
(iv) 54 or 45
Solution:
(i) 23 or 32
It can be written as
23 = 2 × 2 × 2 = 8
32 = 3 × 3 = 9
Hence, 9 is greater than 8 i.e. 32 > 23.
(ii) 25 or 52
It can be written as
25 = 2 × 2 × 2 × 2 × 2 = 32
52 = 5 × 5 = 25
Hence, 32 is greater than 25 i.e. 25 > 52.
(iii) 43 or 34
It can be written as
43 = 4 × 4 × 4 = 64
34 = 3 × 3 × 3 × 3 = 81
Hence, 81 is greater than 64 i.e. 34 > 43.
(iv) 54 or 45
It can be written as
54 = 5 × 5 × 5 × 5 = 625
45 = 4 × 4 × 4 × 4 × 4 = 1024
Hence, 1024 is greater than 625 i.e. 45 > 54.
6. Express each of the following in exponential form:
(i) 512
(ii) 1250
(iii) 1458
(iv) 3600
(v) 1350
(vi) 1176
Solution:
(i) 512
It can be written as
So we get
512 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 29
(ii) 1250
It can be written as
So we get
1250 = 2 × 5 × 5 × 5 × 5 = 2 × 54
(iii) 1458
It can be written as
So we get
1458 = 2 × 3 × 3 × 3 × 3 × 3 × 3 = 2 × 36
(iv) 3600
It can be written as
So we get
3600 = 2 × 2 × 2 × 2 × 3 × 3 × 5 × 5 = 24 × 32 × 52
(v) 1350
It can be written as
So we get
1350 = 2 × 3 × 3 × 3 × 5 × 5 = 2 × 33 × 52
(vi) 1176
It can be written as
So we get
1176 = 2 × 2 × 2 × 3 × 7 × 7 = 23 × 3 × 72
7. If a = 2 and b = 3, find the value of:
(i) (a + b)2
(ii) (b – a)3
(iii) (a × b)a
(iv) (a × b)b
Solution:
(i) (a + b)2
By substituting the values of a and b
= (2 + 3)2
On further calculation
= 52
= 5 × 5
= 25
(ii) (b – a)3
By substituting the values of a and b
= (3 – 2)3
On further calculation
= 13
= 1 × 1 × 1
= 1
(iii) (a × b)a
By substituting the values of a and b
= (2 × 3)2
On further calculation
= 62
= 6 × 6
= 36
(iv) (a × b)b
By substituting the values of a and b
= (2 × 3)3
On further calculation
= 63
= 6 × 6 × 6
= 216
8. Express:
(i) 1024 as a power of 2.
(ii) 343 as a power of 7.
(iii) 729 as a power of 3.
Solution:
(i) 1024 as a power of 2.
It can be written as
So we get
1024 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 210
(ii) 343 as a power of 7.
It can be written as
So we get
343 = 7 × 7 × 7 = 73
(iii) 729 as a power of 3.
It can be written as
So we get
729 = 3 × 3 × 3 × 3 × 3 × 3 = 36
9. If 27 × 32 = 3x × 2y; find the values of x and y.
Solution:
It is given that
27 × 32 = 3x × 2y
So we get
27 = 3x
Here
27 = 3 × 3 × 3 = 33 = 3x
We get
x = 3x
Similarly
32 = 2y
Here
32 = 2 × 2 × 2 × 2 × 2 = 25 = 2y
We get
y = 5
10. If 64 × 625 = 2a × 5b; find: (i) the values of a and b. (ii) 2b × 5a.
Solution:
(i) the values of a and b
It is given that
64 × 625 = 2a × 5b
We know that
64 = 2a
We can write it as
64 = 2 × 2 × 2 × 2 × 2 × 2
So we get
64 = 26
a = 6
Similarly
625 = 5b
We can write it as
625 = 5 × 5 × 5 × 5
So we get
625 = 54
b = 4
(ii) 2b × 5a
Substituting the values of a and b
= 24 × 56
It can be written as
= 2 × 2 × 2 × 2 × 5 × 5 × 5 × 5 × 5 × 5
So we get
= 16 × 15625
= 250000
Exercise 5B page: 75
1. Fill in the blanks:
(i) In 52 = 25, base = ………. and index = …………
(ii) If index = 3x and base = 2y, the number = ………
Solution:
(i) In 52 = 25, base = 5 and index = 2.
(ii) If index = 3x and base = 2y, the number = 2y3x.
2. Evaluate:
(i) 28 ÷ 23
(ii) 23÷ 28
(iii) (26)0
(iv) (3o)6
(v) 83 x 8-5 x 84
(vi) 54 x 53 ÷ 55
(vii) 54 ÷ 53 x 55
(viii) 44 ÷ 43 x 40
(ix) (35 x 47 x 58)0
Solution:
(i) 28 ÷ 23
It can be written as
= 28/ 23
On further calculation
= 2 8-3
= 25
(ii) 23÷ 28
It can be written as
= 23/ 28
On further calculation
= 2 3-8
So we get
= 2 -5
= 1/25
(iii) (26)0
It can be written as
= 2 6 × 0
On further calculation
= 20
So we get
= 1
(iv) (3o) 6
It can be written as
= 3 0 × 6
On further calculation
= 30
So we get
= 1
(v) 83 x 8-5 x 84
It can be written as
= 8 3 + 4 – 5
On further calculation
= 8 7 – 5
So we get
= 82
(vi) 54 x 53 ÷ 55
It can be written as
= (54 x 53)/ 55
On further calculation
= 5 4 + 3 – 5
So we get
= 5 7 – 5
= 52
(vii) 54 ÷ 53 x 55
It can be written as
= 54/ 53 x 55
On further calculation
= 5 4 – 3 + 5
So we get
= 56
(viii) 44 ÷ 43 x 40
It can be written as
= 44/ (43 x 40)
On further calculation
= 44/ (43 x 1)
So we get
= 44/43
= 44-3
= 41
= 4
(ix) (35 x 47 x 58)0
It can be written as
= 35 × 0 × 47 × 0 × 58 × 0
On further calculation
= 30 × 40 × 50
So we get
= 1 × 1 × 1
= 1
3. Simplify, giving answers with positive index:
(i) 2b6. b3. 5b4
(ii) x2y3. 6x5y. 9x3y4
(iii) (-a)5 (a2)
(iv) (-y)2 (-y)3
(v) (-3)2 (3)3
(vi) (-4x) (-5x2)
(vii) (5a2b) (2ab2) (a3b)
(viii) x2a + 7. x2a – 8
(ix) 3y. 32. 3-4
(x) 24a. 23a. 2-a
(xi) 4x2y2 ÷ 9x3y3
(xii) (102)3 (x8)12
(xiii) (a10)10 (16)10
(xiv) (n2)2 (-n2)3
(xv) – (3ab)2 (-5a2bc4)2
(xvi) (-2)2 × (0)3 × (3)3
(xvii) (2a3)4 (4a2)2
(xviii) (4x2y3)3 ÷ (3x2y3)3
Solution:
(i) 2b6. b3. 5b4
It can be written as
= 2 × 5 × b6 + 3 + 4
On further calculation
= 10 b13
(ii) x2y3. 6x5y. 9x3y4
It can be written as
= 6 × 9 × x2 + 5 + 3 × y3 + 1 + 4
On further calculation
= 54 x10 y8
(iii) (-a)5 (a2)
It can be written as
= (-1 × a)5 × a2
On further calculation
= (-1)5 × a5 + 2
So we get
= – 1 × a7
= – a7
(iv) (-y)2 (-y)3
It can be written as
= (-1 × y)2. (-1 × y)3
On further calculation
= (-1)2. y2. (-1)3 × y3
So we get
= 12 + 3. y2 + 3
= 15 y5
= y5
(v) (-3)2 (3)3
It can be written as
= (-1 × 3)2. (3)3
On further calculation
= (-1)2 × 32. 33
So we get
= (-1)2. 32 + 3
= 1. 35
= 35
(vi) (-4x) (-5x2)
It can be written as
= (-1 × 4 × x). (-1 × 5 × x2)1
On further calculation
= (- 1 × 4 × x). (-1 × 5 × x2)
So we get
= – 1 × – 1 × 4 × 5 × x1 + 2
Here
= – 11 + 1. 41. 51 x3
= 20 x3
(vii) (5a2b) (2ab2) (a3b)
It can be written as
= 5 × 2 × a2 + 1 + 3 × b1 + 2 + 1
On further calculation
= 10 a6b4
(viii) x2a + 7. x2a – 8
It can be written as
= x2a + 7 + 2a – 8
On further calculation
= x4a – 1
(ix) 3y. 32. 3-4
It can be written as
= 3y. 32/34
On further calculation
= 3y. (3 × 3)/ (3 × 3 × 3 × 3)
So we get
= 3y × 1/32
= 3y – 2
(x) 24a. 23a. 2-a
It can be written as
= 2 4a + 3a – a
On further calculation
= 2 7a – a
So we get
= 26a
(xi) 4x2y2 ÷ 9x3y3
It can be written as
= 4x2y2/ 9x3y3
On further calculation
= 4x2 – 3 y2 – 3/ 9
So we get
= 4x-1y-1/ 9
= 4/9xy
(xii) (102)3 (x8)12
It can be written as
= 102 × 3. x8 × 12
On further calculation
= 106 x96
(xiii) (a10)10 (16)10
It can be written as
= a10 × 10. 16 × 10
On further calculation
= a100. 160
So we get
= a100
(xiv) (n2)2 (-n2)3
It can be written as
= n2 × 2. (-n)2 × 3
On further calculation
= n4. (-n)6
So we get
= – n4 – 16 n6
= – n4 + 6
= – n 10
(xv) – (3ab)2 (-5a2bc4)2
It can be written as
= – (32a2b2) × (-1)2 × 52a2 × 2b2c4 × 2
On further calculation
= – (32a2b2) (52a4b2c8)
So we get
= – 32. 52. a2 + 4 b2 + 2 c8
= – 225a6b4c8
(xvi) (-2)2 × (0)3 × (3)3
It can be written as
= 4 × 0 × 27
On further calculation
= 0
(xvii) (2a3)4 (4a2)2
It can be written as
= (2a3)4 (22a2)2
On further calculation
= 24 a3 × 4. 22 × 2. a2 × 2
So we get
= 24 a12. 24 a4
Here
= 24 + 4. a12 + 4
= 28 a16
We get
= 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × a16
= 256 a16
(xviii) (4x2y3)3 ÷ (3x2y3)3
It can be written as
It can be written as
It can be written as
It can be written as
It can be written as
4. Simplify and express the answer in the positive exponent form:
Solution:
= – (3)3 – 1 26 – 4
= – (3)2 22
= – 3222
(iv) – 128/2187
So we get
It can be written as
(vi) (a3b-5)-2
It can be written as
= a 3 x -2 b-5 x -2
So we get
= a-6 b10
= b10/ a6
5. Evaluate:
(i) 6-2 ÷ (4-2 × 3-2)
(iii) 53 × 32 + (17)0 × 73
(iv) 25 × 150 + (-3)3 – (2/7)-2
(v) (22)0 + 2-4 ÷ 2-6 + (1/2)-3
(vi) 5n × 25n-1 ÷ (5n-1 × 25n-1)
Solution:
(i) 6-2 ÷ (4-2 × 3-2)
It can be written as
= (1/6)2 ÷ (1/4)2 × (1/3)2
On further calculation
= 1/36 ÷ 1/16 × 1/9
So we get
= 1/36 ÷ 1/144
= 1/36 × 144/1
= 4
It can be written as
(iii) 53 × 32 + (17)0 × 73
It can be written as
= 5 × 5 × 5 × 3 × 3 + (17)0 × 7 × 7 × 7
On further calculation
= 125 × 9 + 1 × 343
So we get
= 1125 + 343
= 1468
(iv) 25 × 150 + (-3)3 – (2/7)-2
It can be written as
= 2 × 2 × 2 × 2 × 2 × 1 + (-3) × (-3) × (-3) – (7/2) × (7/2)
By further calculation
= 32 × 1 – 27 – 49/4
Here we get LCM = 4
= (32 × 4)/ (1 × 4) – (27 × 4)/ (1 × 4) – 49/ (4 × 1)
So we get
= (128 – 108 – 49)/ 4
By subtraction
= -29/ 4
= – 7 1/4
(v) (22)0 + 2-4 ÷ 2-6 + (1/2)-3
It can be written as
= (4)0 + (1/2)4 ÷ (1/2)6 + (2/1)3
By further calculation
= 1 + (1/2 × 1/2 × 1/2 × 1/2) ÷ (1/2 × 1/2 × 1/2 × 1/2 × 1/2 × 1/2) + (2/1 × 2/1 × 2/1)
So we get
= 1 + (1/2 × 1/2 × 1/2 × 1/2 × 2 × 2 × 2 × 2 × 2 × 2) + 8
On further simplification
= 1 + 4 + 8
= 13
(vi) 5n × 25n-1 ÷ (5n-1 × 25n-1)
It can be written as
= 5n × 25n-1 × 1/ (5n-1 × 25n-1)
By further calculation
= 5n × 1/ 5n-1
So we get
= 5n – n + 1
= 51
6. If m = – 2 and n = 2; find the value of:
(i) m2 + n2 – 2mn
(ii) mn + nm
(iii) 6m-3 + 4n2
(iv) 2n3 – 3m
Solution:
(i) m2 + n2 – 2mn
It is given that
m = – 2 and n = 2
Substituting the values we get
= (-2)2 + 22 – 2 (-2) (2)
By further calculation
= 4 + 4 – (-8)
So we get
= 8 + 8
= 16
= 24
(ii) mn + nm
It is given that m = – 2 and n = 2
Substituting the values we get
= (-2)2 + (2)-2
We can write it as
= 4 + 1/2 × 1/2
We get the LCM = 4
= (4 × 4)/ (1 × 4) + 1/4
So we get
= (16 + 1)/ 4
= 17/ 4
= 4 ¼
(iii) 6m-3 + 4n2
It is given that
m = – 2 and n = 2
Substituting the values
= 6 (-2)-3 + 4 (2)2
It can be written as
= 6 × 1/ -2 × 1/ -2 × 1/ -2 + 4 × 2 × 2
So we get
= – 3/4 + 16
Here the LCM = 4
= (-3 + 16 × 4)/ 4
By calculation
= (-3 + 64)/ 4
= – 61/4
= 15 ¼
(iv) 2n3 – 3m
It is given that
m = – 2 and n = 2
By substituting the values
= 2 (2)3 – 3 (-2)
It can be written as
= 2 × (2 × 2 × 2) – 3 × (-2)
By further calculation
= 16 – 3 × (-2)
So we get
= 16 + 6
= 22
Exercise 5B
1. Fill in the blanks:
(i) In 52 = 25, base = ………. and index = …………
(ii) If index = 3x and base = 2y, the number = ………
Solution:
(i) In 52 = 25, base = 5 and index = 2.
(ii) If index = 3x and base = 2y, the number = 2y3x.
2. Evaluate:
(i) 28 ÷ 23
(ii) 23÷ 28
(iii) (26)0
(iv) (3o)6
(v) 83 x 8-5 x 84
(vi) 54 x 53 ÷ 55
(vii) 54 ÷ 53 x 55
(viii) 44 ÷ 43 x 40
(ix) (35 x 47 x 58)0
Solution:
(i) 28 ÷ 23
It can be written as
= 28/ 23
On further calculation
= 2 8-3
= 25
(ii) 23÷ 28
It can be written as
= 23/ 28
On further calculation
= 2 3-8
So we get
= 2 -5
= 1/25
(iii) (26)0
It can be written as
= 2 6 × 0
On further calculation
= 20
So we get
= 1
(iv) (3o) 6
It can be written as
= 3 0 × 6
On further calculation
= 30
So we get
= 1
(v) 83 x 8-5 x 84
It can be written as
= 8 3 + 4 – 5
On further calculation
= 8 7 – 5
So we get
= 82
(vi) 54 x 53 ÷ 55
It can be written as
= (54 x 53)/ 55
On further calculation
= 5 4 + 3 – 5
So we get
= 5 7 – 5
= 52
(vii) 54 ÷ 53 x 55
It can be written as
= 54/ 53 x 55
On further calculation
= 5 4 – 3 + 5
So we get
= 56
(viii) 44 ÷ 43 x 40
It can be written as
= 44/ (43 x 40)
On further calculation
= 44/ (43 x 1)
So we get
= 44/43
= 44-3
= 41
= 4
(ix) (35 x 47 x 58)0
It can be written as
= 35 × 0 × 47 × 0 × 58 × 0
On further calculation
= 30 × 40 × 50
So we get
= 1 × 1 × 1
= 1
3. Simplify, giving answers with positive index:
(i) 2b6. b3. 5b4
(ii) x2y3. 6x5y. 9x3y4
(iii) (-a)5 (a2)
(iv) (-y)2 (-y)3
(v) (-3)2 (3)3
(vi) (-4x) (-5x2)
(vii) (5a2b) (2ab2) (a3b)
(viii) x2a + 7. x2a – 8
(ix) 3y. 32. 3-4
(x) 24a. 23a. 2-a
(xi) 4x2y2 ÷ 9x3y3
(xii) (102)3 (x8)12
(xiii) (a10)10 (16)10
(xiv) (n2)2 (-n2)3
(xv) – (3ab)2 (-5a2bc4)2
(xvi) (-2)2 × (0)3 × (3)3
(xvii) (2a3)4 (4a2)2
(xviii) (4x2y3)3 ÷ (3x2y3)3
Solution:
(i) 2b6. b3. 5b4
It can be written as
= 2 × 5 × b6 + 3 + 4
On further calculation
= 10 b13
(ii) x2y3. 6x5y. 9x3y4
It can be written as
= 6 × 9 × x2 + 5 + 3 × y3 + 1 + 4
On further calculation
= 54 x10 y8
(iii) (-a)5 (a2)
It can be written as
= (-1 × a)5 × a2
On further calculation
= (-1)5 × a5 + 2
So we get
= – 1 × a7
= – a7
(iv) (-y)2 (-y)3
It can be written as
= (-1 × y)2. (-1 × y)3
On further calculation
= (-1)2. y2. (-1)3 × y3
So we get
= 12 + 3. y2 + 3
= 15 y5
= y5
(v) (-3)2 (3)3
It can be written as
= (-1 × 3)2. (3)3
On further calculation
= (-1)2 × 32. 33
So we get
= (-1)2. 32 + 3
= 1. 35
= 35
(vi) (-4x) (-5x2)
It can be written as
= (-1 × 4 × x). (-1 × 5 × x2)1
On further calculation
= (- 1 × 4 × x). (-1 × 5 × x2)
So we get
= – 1 × – 1 × 4 × 5 × x1 + 2
Here
= – 11 + 1. 41. 51 x3
= 20 x3
(vii) (5a2b) (2ab2) (a3b)
It can be written as
= 5 × 2 × a2 + 1 + 3 × b1 + 2 + 1
On further calculation
= 10 a6b4
(viii) x2a + 7. x2a – 8
It can be written as
= x2a + 7 + 2a – 8
On further calculation
= x4a – 1
(ix) 3y. 32. 3-4
It can be written as
= 3y. 32/34
On further calculation
= 3y. (3 × 3)/ (3 × 3 × 3 × 3)
So we get
= 3y × 1/32
= 3y – 2
(x) 24a. 23a. 2-a
It can be written as
= 2 4a + 3a – a
On further calculation
= 2 7a – a
So we get
= 26a
(xi) 4x2y2 ÷ 9x3y3
It can be written as
= 4x2y2/ 9x3y3
On further calculation
= 4x2 – 3 y2 – 3/ 9
So we get
= 4x-1y-1/ 9
= 4/9xy
(xii) (102)3 (x8)12
It can be written as
= 102 × 3. x8 × 12
On further calculation
= 106 x96
(xiii) (a10)10 (16)10
It can be written as
= a10 × 10. 16 × 10
On further calculation
= a100. 160
So we get
= a100
(xiv) (n2)2 (-n2)3
It can be written as
= n2 × 2. (-n)2 × 3
On further calculation
= n4. (-n)6
So we get
= – n4 – 16 n6
= – n4 + 6
= – n 10
(xv) – (3ab)2 (-5a2bc4)2
It can be written as
= – (32a2b2) × (-1)2 × 52a2 × 2b2c4 × 2
On further calculation
= – (32a2b2) (52a4b2c8)
So we get
= – 32. 52. a2 + 4 b2 + 2 c8
= – 225a6b4c8
(xvi) (-2)2 × (0)3 × (3)3
It can be written as
= 4 × 0 × 27
On further calculation
= 0
(xvii) (2a3)4 (4a2)2
It can be written as
= (2a3)4 (22a2)2
On further calculation
= 24 a3 × 4. 22 × 2. a2 × 2
So we get
= 24 a12. 24 a4
Here
= 24 + 4. a12 + 4
= 28 a16
We get
= 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × a16
= 256 a16
(xviii) (4x2y3)3 ÷ (3x2y3)3
It can be written as
It can be written as
It can be written as
It can be written as
It can be written as
4. Simplify and express the answer in the positive exponent form:
Solution:
= – (3)3 – 1 26 – 4
= – (3)2 22
= – 3222
(iv) – 128/2187
So we get
It can be written as
(vi) (a3b-5)-2
It can be written as
= a 3 x -2 b-5 x -2
So we get
= a-6 b10
= b10/ a6
5. Evaluate:
(i) 6-2 ÷ (4-2 × 3-2)
(iii) 53 × 32 + (17)0 × 73
(iv) 25 × 150 + (-3)3 – (2/7)-2
(v) (22)0 + 2-4 ÷ 2-6 + (1/2)-3
(vi) 5n × 25n-1 ÷ (5n-1 × 25n-1)
Solution:
(i) 6-2 ÷ (4-2 × 3-2)
It can be written as
= (1/6)2 ÷ (1/4)2 × (1/3)2
On further calculation
= 1/36 ÷ 1/16 × 1/9
So we get
= 1/36 ÷ 1/144
= 1/36 × 144/1
= 4
It can be written as
(iii) 53 × 32 + (17)0 × 73
It can be written as
= 5 × 5 × 5 × 3 × 3 + (17)0 × 7 × 7 × 7
On further calculation
= 125 × 9 + 1 × 343
So we get
= 1125 + 343
= 1468
(iv) 25 × 150 + (-3)3 – (2/7)-2
It can be written as
= 2 × 2 × 2 × 2 × 2 × 1 + (-3) × (-3) × (-3) – (7/2) × (7/2)
By further calculation
= 32 × 1 – 27 – 49/4
Here we get LCM = 4
= (32 × 4)/ (1 × 4) – (27 × 4)/ (1 × 4) – 49/ (4 × 1)
So we get
= (128 – 108 – 49)/ 4
By subtraction
= -29/ 4
= – 7 1/4
(v) (22)0 + 2-4 ÷ 2-6 + (1/2)-3
It can be written as
= (4)0 + (1/2)4 ÷ (1/2)6 + (2/1)3
By further calculation
= 1 + (1/2 × 1/2 × 1/2 × 1/2) ÷ (1/2 × 1/2 × 1/2 × 1/2 × 1/2 × 1/2) + (2/1 × 2/1 × 2/1)
So we get
= 1 + (1/2 × 1/2 × 1/2 × 1/2 × 2 × 2 × 2 × 2 × 2 × 2) + 8
On further simplification
= 1 + 4 + 8
= 13
(vi) 5n × 25n-1 ÷ (5n-1 × 25n-1)
It can be written as
= 5n × 25n-1 × 1/ (5n-1 × 25n-1)
By further calculation
= 5n × 1/ 5n-1
So we get
= 5n – n + 1
= 51
6. If m = – 2 and n = 2; find the value of:
(i) m2 + n2 – 2mn
(ii) mn + nm
(iii) 6m-3 + 4n2
(iv) 2n3 – 3m
Solution:
(i) m2 + n2 – 2mn
It is given that
m = – 2 and n = 2
Substituting the values we get
= (-2)2 + 22 – 2 (-2) (2)
By further calculation
= 4 + 4 – (-8)
So we get
= 8 + 8
= 16
= 24
(ii) mn + nm
It is given that m = – 2 and n = 2
Substituting the values we get
= (-2)2 + (2)-2
We can write it as
= 4 + 1/2 × 1/2
We get the LCM = 4
= (4 × 4)/ (1 × 4) + 1/4
So we get
= (16 + 1)/ 4
= 17/ 4
= 4 ¼
(iii) 6m-3 + 4n2
It is given that
m = – 2 and n = 2
Substituting the values
= 6 (-2)-3 + 4 (2)2
It can be written as
= 6 × 1/ -2 × 1/ -2 × 1/ -2 + 4 × 2 × 2
So we get
= – 3/4 + 16
Here the LCM = 4
= (-3 + 16 × 4)/ 4
By calculation
= (-3 + 64)/ 4
= – 61/4
= 15 ¼
(iv) 2n3 – 3m
It is given that
m = – 2 and n = 2
By substituting the values
= 2 (2)3 – 3 (-2)
It can be written as
= 2 × (2 × 2 × 2) – 3 × (-2)
By further calculation
= 16 – 3 × (-2)
So we get
= 16 + 6
= 22
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Selina Class 7 ICSE Solutions Mathematics : Chapter 5- Exponents (Including Laws of Exponents)
Chapterwise Selina Publishers ICSE Solutions for Class 7 Mathematics :
- Chapter 1- Integers
- Chapter 2- Rational Numbers
- Chapter 3- Fraction (Including Problems)
- Chapter 4- Decimal Fractions (Decimals)
- Chapter 5- Exponents (Including Laws of Exponents)
- Chapter 6- Ratio and Proportion (Including Sharing in a Ratio)
- Chapter 7- Unitary Method (Including Time and Work)
- Chapter 8- Percent and Percentage
- Chapter 9- Profit, Loss and Discount
- Chapter 10- Simple Interest
- Chapter 11- Fundamental Concepts (Including Fundamental Operations)
- Chapter 12- Simple Linear Equations (Including Word Problems)
- Chapter 13- Set Concepts
- Chapter 14- Lines and Angles (Including Construction of Angles)
- Chapter 15- Triangles
- Chapter 16- Pythagoras Theorem
- Chapter 17- Symmetry (Including Reflection and Rotation)
- Chapter 18- Recognition of Solids (Representing 3-D in 2-D)
- Chapter 19- Congruency: Congruent Triangles
- Chapter 20- Mensuration (Perimeter and Area of Plane Figures)
- Chapter 21- Data Handling
- Chapter 22- Probability
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