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NCERT Solutions for Class 7 Maths Chapter 13, Exponents and Powers, , Exercise 13.1, 1. Find the value of:, (i) 26, Solution:The above value can be written as,, =2×2×2×2×2×2, = 64, (ii) 93, Solution:The above value can be written as,, =9×9×9, = 729, (iii) 112, Solution:The above value can be written as,, = 11 × 11, = 121, (iv) 54, Solution:The above value can be written as,, =5×5×5×5, = 625, 2. Express the following in exponential form:, (i) 6 × 6 × 6 × 6, Solution:The given question can be expressed in the exponential form as 64., (ii) t × t, Solution:The given question can be expressed in the exponential form as t2., (iii) b × b × b × b, , Page: 252

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NCERT Solutions for Class 7 Maths Chapter 13, Exponents and Powers, , Solution:The given question can be expressed in the exponential form as b4., (iv) 5 × 5× 7 × 7 × 7, Solution:The given question can be expressed in the exponential form as 52 × 73., (v) 2 × 2 × a × a, Solution:The given question can be expressed in the exponential form as 22 × a2., (vi) a × a × a × c × c × c × c × d, Solution:The given question can be expressed in the exponential form as a3 × c4 × d., 3. Express each of the following numbers using exponential notation:, (i) 512, Solution:The factors of 512 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2, So it can be expressed in the exponential form as 29., (ii) 343, Solution:The factors of 343 = 7 × 7 × 7, So it can be expressed in the exponential form as 73., (iii) 729, Solution:The factors of 729 = 3 × 3 × 3 × 3 × 3 × 3, So it can be expressed in the exponential form as 36., (iv) 3125, Solution:The factors of 3125 = 5 × 5 × 5 × 5 × 5, So it can be expressed in the exponential form as 55., 4. Identify the greater number, wherever possible, in each of the following?

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NCERT Solutions for Class 7 Maths Chapter 13, Exponents and Powers, , (i) 43 or 34, Solution:The expansion of 43 = 4 × 4 × 4 = 64, The expansion of 34 = 3 × 3 × 3 × 3 = 81, Clearly,, 64 < 81, 3, So, 4 < 34, Hence 34 is the greater number., (ii) 53 or 35, Solution:The expansion of 53 = 5 × 5 × 5 = 125, The expansion of 35 = 3 × 3 × 3 × 3 × 3= 243, Clearly,, 125 < 243, 3, So, 5 < 35, Hence 35 is the greater number., (iii) 28 or 82, Solution:The expansion of 28 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 256, The expansion of 82 = 8 × 8= 64, Clearly,, 256 > 64, So, 28 > 82, Hence 28 is the greater number., (iv) 1002 or 2100, Solution:The expansion of 1002 = 100 × 100 = 10000, The expansion of 2100, 210 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 1024, Then,, 2100 = 1024 × 1024 ×1024 × 1024 ×1024 × 1024 × 1024 × 1024 × 1024 × 1024 =, Clearly,, 1002 < 2100, Hence 2100 is the greater number.

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NCERT Solutions for Class 7 Maths Chapter 13, Exponents and Powers, , (v) 210 or 102, Solution:The expansion of 210 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 1024, The expansion of 102 = 10 × 10= 100, Clearly,, 1024 > 100, 10, So, 2 > 102, Hence 28 is the greater number., 5. Express each of the following as product of powers of their prime factors:, (i) 648, Solution:Factors of 648 = 2 × 2 × 2 × 3 × 3 × 3 × 3, = 23 × 34, (ii) 405, Solution:Factors of 405 = 3 × 3 × 3 × 3 × 5, = 35 × 3, , (iii) 540, Solution:Factors of 540 = 2 × 2 × 3 × 3 × 3 × 5, = 22 × 33 × 5, (iv) 3,600, Solution:Factors of 3600 = 2 × 2 × 2 × 2 × 3 × 3 × 5 × 5, = 24 × 32 × 52, 6. Simplify:, (i) 2 × 103, Solution:The above question can be written as,, = 2 × 10 × 10 × 10, = 2 × 1000

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NCERT Solutions for Class 7 Maths Chapter 13, Exponents and Powers, , = 2000, (ii) 72 × 22, Solution:The above question can be written as,, =7×7×2×2, = 49 × 4, = 196, (iii) 23 × 5, Solution:The above question can be written as,, =2×2×2×5, =8×5, = 40, (iv) 3 × 44, Solution:The above question can be written as,, =3×4×4×4×4, = 3 × 256, = 768, (v) 0 × 102, Solution:The above question can be written as,, = 0 × 10 × 10, = 0 × 100, =0, (vi) 52 × 33, Solution:The above question can be written as,, =5×5×3×3×3, = 25 × 27, = 675

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NCERT Solutions for Class 7 Maths Chapter 13, Exponents and Powers, , (vii) 24 × 32, Solution:The above question can be written as,, =2×2×2×2×3×3, = 16 × 9, = 144, (viii) 32 × 104, Solution:The above question can be written as,, = 3 × 3 × 10 × 10 × 10 × 10, = 9 × 10000, = 90000, 7. Simplify:, (i) (– 4)3, Solution:The expansion of -43, =-4×-4×-4, = - 64, (ii) (–3) × (–2)3, Solution:The expansion of (-3) × (-2)3, =-3×-2×-2×-2, =-3×-8, = 24, (iii) (–3)2 × (–5)2, Solution:The expansion of (-3)2 × (-5)2, =-3×-3×-5×-5, = 9 × 25, = 225, (iv) (–2)3 × (–10)3, Solution:-

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NCERT Solutions for Class 7 Maths Chapter 13, Exponents and Powers, , The expansion of (-2)3 × (-10)3, = - 2 × - 2 × - 2 × - 10 × - 10 × - 10, = - 8 × - 1000, = 8000, 8. Compare the following numbers:, (i) 2.7 × 1012 ; 1.5 × 108, Solution:By observing the question, Comparing the exponents of base 10,, Clearly,, 2.7 × 1012 > 1.5 × 108, (ii) 4 × 1014 ; 3 × 1017, Solution:By observing the question, Comparing the exponents of base 10,, Clearly,, 4 × 1014 < 3 × 1017

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NCERT Solutions for Class 7 Maths Chapter 13, Exponents and Powers, , Exercise 13.2, , Page: 260, , 1. Using laws of exponents, simplify and write the answer in exponential form:, (i) 32 × 34 × 38, Solution:By the rule of multiplying the powers with same base = am × an = am + n, Then,, = (3)2 + 4 + 8, = 314, (ii) 615 ÷ 610, Solution:By the rule of dividing the powers with same base = am ÷ an = am - n, Then,, = (6)15 - 10, = 65, (iii) a3 × a2, Solution:By the rule of multiplying the powers with same base = am × an = am + n, Then,, = (a)3 + 2, = a5, (iv) 7x × 72, Solution:By the rule of multiplying the powers with same base = am × an = am + n, Then,, = (7)x + 2, (v) (52)3 ÷ 53, Solution:By the rule of taking power of as power = (am)n = amn, (52)3 can be written as = (5)2 × 3, = 56, Now, 56 ÷ 53, By the rule of dividing the powers with same base = am ÷ an = am - n

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NCERT Solutions for Class 7 Maths Chapter 13, Exponents and Powers, , Then,, = (5)6 - 3, = 53, (vi) 25 × 55, Solution:By the rule of multiplying the powers with same exponents = am × bm = abm, Then,, = (2 × 5)5, = 105, (vii) a4 × b4, Solution:By the rule of multiplying the powers with same exponents = am × bm = abm, Then,, = (a × b)4, = ab4, (viii) (34)3, Solution:By the rule of taking power of as power = (am)n = amn, (34)3 can be written as = (3)4 × 3, = 312, (ix) (220 ÷ 215) × 23, Solution:By the rule of dividing the powers with same base = am ÷ an = am - n, (220 ÷ 215) can be simplified as,, = (2)20 - 15, = 25, Then,, By the rule of multiplying the powers with same base = am × an = am + n, 25 × 23 can be simplified as,, = (2)5 + 3, = 28, (x) 8t ÷ 82

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NCERT Solutions for Class 7 Maths Chapter 13, Exponents and Powers, , Solution:By the rule of dividing the powers with same base = am ÷ an = am - n, Then,, = (8)t - 2, 2. Simplify and express each of the following in exponential form:, (i) (23 × 34 × 4)/ (3 × 32), Solution:Factors of 32 = 2 × 2 × 2 × 2 × 2, = 25, Factors of 4 = 2 × 2, = 22, Then,, = (23 × 34 × 22)/ (3 × 25), = (23 + 2 × 34) / (3 × 25), … [∵am × an = am + n], = (25 × 34) / (3 × 25), = 25 – 5 × 34 – 1, … [∵am ÷ an = am – n], = 20 × 33, = 1 × 33, = 33, (ii) ((52)3 × 54) ÷ 57, Solution:(52)3 can be written as = (5)2 × 3, = 56, Then,, = (56 × 54) ÷ 57, = (56 + 4) ÷ 57, = 510 ÷ 57, = 510 – 7, = 53, (iii) 254 ÷ 53, Solution:(25)4 can be written as = (5 × 5)4, = (52)4, (52)4 can be written as = (5)2 × 4, , … [∵(am)n = amn], , … [∵am × an = am + n], … [∵am ÷ an = am – n], , … [∵(am)n = amn]

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NCERT Solutions for Class 7 Maths Chapter 13, Exponents and Powers, , = 58, Then,, = 58 ÷ 53, = 58 – 3, = 55, , … [∵am ÷ an = am – n], , (iv) (3 × 72 × 118)/ (21 × 113), Solution:Factors of 21 = 7 × 3, Then,, = (3 × 72 × 118)/ (7 × 3 × 113), = 31-1 × 72-1 × 118 – 3, = 30 × 7 × 115, = 1 × 7 × 115, = 7 × 115, (v) 37/ (34 × 33), Solution:= 37/ (34+3), = 37/ 37, = 37 – 7, = 30, =1, (vi) 20 + 30 + 40, Solution:=1+1+1, =3, (vii) 20 × 30 × 40, Solution:=1×1×1, =1, (viii) (30 + 20) × 50, Solution:-, , … [∵am × an = am + n], … [∵am ÷ an = am – n]

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NCERT Solutions for Class 7 Maths Chapter 13, Exponents and Powers, , = (1 + 1) × 1, = (2) × 1, =2, (ix) (28 × a5)/ (43 × a3), Solution:(4)3 can be written as = (2 × 2)3, = (22)3, (52)4 can be written as = (2)2 × 3, = 26, Then,, = (28 × a5)/ (26 × a3), = 28 – 6 × a5 - 3, = 22 × a2, = 2a2, (x) (a5/a3) × a8, Solution:= (a5 - 3) × a8, = a 2 × a8, = a2 + 8, = a10, (xi) (45 × a8b3)/ (45 × a5b2), Solution:= 45 – 5 × (a8 – 5 × b3 - 2), = 40 × (a3b), = 1 × a3b, = a3b, (xii) (23 × 2)2, Solution:= (23 + 1)2, = (24)2, (24)2 can be written as = (2)4 × 2, = 28, , … [∵(am)n = amn], , … [∵am ÷ an = am – n], … [∵(am)n = amn], , … [∵am ÷ an = am – n], … [∵am × an = am + n], , … [∵am ÷ an = am – n], , … [∵am × an = am + n], … [∵(am)n = amn]

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NCERT Solutions for Class 7 Maths Chapter 13, Exponents and Powers, , 3. Say true or false and justify your answer:, (i) 10 × 1011 = 10011, Solution:Let us consider Left Hand Side (LHS) = 10 × 1011, = 101 + 11, = 1012, Now, consider Right Hand Side (RHS) = 10011, = (10 × 10)11, = (101 + 1)11, = (102)11, = (10)2 × 11, = 1022, By comparing LHS and RHS,, LHS ≠ RHS, Hence, the given statement is false., (ii) 23 > 52, Solution:Let us consider LHS = 23, Expansion of 23 = 2 × 2 × 2, =8, Now, consider RHS = 52, Expansion of 52 = 5 × 5, = 25, By comparing LHS and RHS,, LHS < RHS, 23 < 52, Hence, the given statement is false., (iii) 23 × 32 = 65, Solution:Let us consider LHS = 23 × 32, Expansion of 23 × 32= 2 × 2 × 2 × 3 × 3, = 72, Now, consider RHS = 65, Expansion of 65 = 6 × 6 × 6 × 6 × 6, = 7776, , … [∵am × an = am + n], , … [∵(am)n = amn]

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NCERT Solutions for Class 7 Maths Chapter 13, Exponents and Powers, , By comparing LHS and RHS,, LHS < RHS, 23 < 52, Hence, the given statement is false., (iv) 30 = (1000)0, Solution:Let us consider LHS = 30, =1, Now, consider RHS = 10000, =1, By comparing LHS and RHS,, LHS = RHS, 30 = 10000, Hence, the given statement is true., 4. Express each of the following as a product of prime factors only in exponential, form:, (i) 108 × 192, Solution:The factors of 108 = 2 × 2 × 3 × 3 × 3, = 22 × 33, The factors of 192 = 2 × 2 × 2 × 2 × 2 × 2 × 3, = 26 × 3, Then,, = (22 × 33) × (26 × 3), = 22 + 6 × 33 + 3, … [∵am × an = am + n], = 28 × 36, (ii) 270, Solution:The factors of 270 = 2 × 3 × 3 × 3 × 5, = 2 × 33 × 5, (iii) 729 × 64, The factors of 729 = 3 × 3 × 3 × 3 × 3 × 3, = 36

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NCERT Solutions for Class 7 Maths Chapter 13, Exponents and Powers, , The factors of 64 = 2 × 2 × 2 × 2 × 2 × 2, = 26, Then,, = (36 × 26), = 36 × 26, (iv) 768, Solution:The factors of 768 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3, = 28 × 3, 5. Simplify:, (i) ((25)2 × 73)/ (83 × 7), Solution:83 can be written as = (2 × 2 × 2)3, = (23)3, We have,, = ((25)2 × 73)/ ((23)3 × 7), = (25 × 2 × 73)/ ((23 × 3 × 7), = (210 × 73)/ (29 × 7), = (210 – 9 × 73 - 1), = 2 × 72, =2×7×7, = 98, (ii) (25 × 52 × t8)/ (103 × t4), Solution:25 can be written as = 5 × 5, = 52, 103 can be written as = 103, = (5 × 2)3, = 53 × 23, We have,, = (52 × 52 × t8)/ (53 × 23 × t4), = (52 + 2 × t8)/ (53 × 23 × t4), = (54 × t8)/ (53 × 23 × t4), = (54 – 3 × t8 - 4)/ 23, , … [∵(am)n = amn], … [∵am ÷ an = am – n], , … [∵am × an = am + n], … [∵am ÷ an = am – n]

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NCERT Solutions for Class 7 Maths Chapter 13, Exponents and Powers, , = (5 × t4)/ (2 × 2 × 2), = (5t4)/ 8, (iii) (35 × 105 × 25)/ (57 × 65), Solution:105 can be written as = (5 × 2)5, = 55 × 25, 25 can be written as = 5 × 5, = 52, 65 can be written as = (2 × 3)5, = 25 × 35, Then we have,, = (35 × 55 × 25 × 52)/ (57 × 25 × 35), = (35 × 55 + 2 × 25)/ (57 × 25 × 35), = (35 × 57 × 25)/ (57 × 25 × 35), = (35 – 5 × 57 – 7 × 25 - 5), = (30 × 50 × 20), =1×1×1, =1, , … [∵am × an = am + n], , … [∵am ÷ an = am – n]

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NCERT Solutions for Class 7 Maths Chapter 13, Exponents and Powers, , Exercise 13.3, , Page: 263, , 1. Write the following numbers in the expanded forms:, 279404, Solution:The expanded form of the number 279404 is,, = (2 × 100000) + (7 × 10000) + (9 × 1000) + (4 × 100) + (0 × 10) + (4 × 1), Now we can express it using powers of 10 in the exponent form,, = (2 × 105) + (7 × 104) + (9 × 103) + (4 × 102) + (0 × 101) + (4 × 100), 3006194, Solution:The expanded form of the number 3006194 is,, = (3 × 1000000) + (0 × 100000) + (0 × 10000) + (6 × 1000) + (1 × 100) + (9 × 10) + 4, Now we can express it using powers of 10 in the exponent form,, = (3 × 106) + (0 × 105) + (0 × 104) + (6 × 103) + (1 × 102) + (9 × 101) + (4 × 100), 2806196, Solution:The expanded form of the number 2806196 is,, = (2 × 1000000) + (8 × 100000) + (0 × 10000) + (6 × 1000) + (1 × 100) + (9 × 10) + 6, Now we can express it using powers of 10 in the exponent form,, = (2 × 106) + (8 × 105) + (0 × 104) + (6 × 103) + (1 × 102) + (9 × 101) + (6 × 100), 120719, Solution:The expanded form of the number 120719 is,, = (1 × 100000) + (2 × 10000) + (0 × 1000) + (7 × 100) + (1 × 10) + (9 × 1), Now we can express it using powers of 10 in the exponent form,, = (1 × 105) + (2 × 104) + (0 × 103) + (7 × 102) + (1 × 101) + (9 × 100), 20068, Solution:The expanded form of the number 20068 is,, = (2 × 10000) + (0 × 1000) + (0 × 100) + (6 × 10) + (8 × 1), Now we can express it using powers of 10 in the exponent form,, = (2 × 104) + (0 × 103) + (0 × 102) + (6 × 101) + (8 × 100)

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NCERT Solutions for Class 7 Maths Chapter 13, Exponents and Powers, , 2. Find the number from each of the following expanded forms:, (a) (8 ×10)4 + (6 × 10)3 + (0 × 10)2 + (4 × 10)1 + (5 × 10)0, Solution:The expanded form is,, = (8 × 10000) + (6 × 1000) + (0 × 100) + (4 × 10) + (5 × 1), = 80000 + 6000 + 0 + 40 + 5, = 86045, (b) (4 ×10)5 + (5 × 10)3 + (3 × 10)2 + (2 × 10)0, Solution:The expanded form is,, = (4 × 100000) + (0 × 10000) + (5 × 1000) + (3 × 100) + (0 × 10) + (2 × 1), = 400000 + 0 + 5000 + 300 + 0 + 2, = 405302, (c) (3 ×10)4 + (7 × 10)2 + (5 × 10)0, Solution:The expanded form is,, = (3 × 10000) + (0 × 1000) + (7 × 100) + (0 × 10) + (5 × 1), = 30000 + 0 + 700 + 0 + 5, = 30705, (d) (9 ×10)5 + (2 × 10)2 + (3 × 10)1, Solution:The expanded form is,, = (9 × 100000) + (0 × 10000) + (0 × 1000) + (2 × 100) + (3 × 10) + (0 × 1), = 900000 + 0 + 0 + 200 + 30 + 0, = 900230, 3. Express the following numbers in standard form:, (i) 5,00,00,000, Solution:The standard form of the given number is 5 × 107, (ii) 70,00,000, Solution:The standard form of the given number is 7 × 106

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NCERT Solutions for Class 7 Maths Chapter 13, Exponents and Powers, , (iii) 3,18,65,00,000, Solution:The standard form of the given number is 3.1865 × 109, (iv) 3,90,878, Solution:The standard form of the given number is 3.90878 × 105, (v) 39087.8, Solution:The standard form of the given number is 3.90878 × 104, (vi) 3908.78, Solution:The standard form of the given number is 3.90878 × 103, 4. Express the number appearing in the following statements in standard form., (a) The distance between Earth and Moon is 384,000,000 m., Solution:The standard form of the number appearing in the given statement is 3.84 × 108m., (b) Speed of light in vacuum is 300,000,000 m/s., Solution:The standard form of the number appearing in the given statement is 3 × 108m/s., (c) Diameter of the Earth is 1,27,56,000 m., Solution:The standard form of the number appearing in the given statement is 1.2756 × 107m., (d) Diameter of the Sun is 1,400,000,000 m., Solution:The standard form of the number appearing in the given statement is 1.4 × 109m., (e) In a galaxy there are on an average 100,000,000,000 stars., Solution:The standard form of the number appearing in the given statement is 1 × 1011 stars.

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NCERT Solutions for Class 7 Maths Chapter 13, Exponents and Powers, , (f) The universe is estimated to be about 12,000,000,000 years old., Solution:The standard form of the number appearing in the given statement is 1.2 × 1010 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., Solution:The standard form of the number appearing in the given statement is 3 × 1020m., (h) 60,230,000,000,000,000,000,000 molecules are contained in a drop of water, weighing 1.8 gm., Solution:The standard form of the number appearing in the given statement is 6.023 × 1022, molecules., (i) The earth has 1,353,000,000 cubic km of sea water., Solution:The standard form of the number appearing in the given statement is 1.353 × 109 cubic, km., (j) The population of India was about 1,027,000,000 in March, 2001., Solution:The standard form of the number appearing in the given statement is 1.027 × 109.