NCERT Solutions for Class 7 Maths Chapter 13 Exponents and Powers are given here in detail. Maths is a subject which requires understanding and reasoning skills accompanied by logic. Along with this, it also requires students to practice Maths regularly. Students of Class 7 are suggested to solve NCERT Solutions for Class 7 Maths Chapter 13 to strengthen their fundamentals and be able to solve questions that are usually asked in the examination. A total of 3 exercises are present in the chapter Chapter 13 â€“ Exponents and Powers of NCERT for Class 7 Maths.Â These exercises covers various concepts mentioned in the chapter, which are given below.

- Exponents
- Laws of Exponents
- Multiplying Powers With The Same Base
- Dividing Powers With The Same Base
- Taking Power of a Power
- Multiplying Powers With The Same Base
- Dividing Powers With The Same Exponents
- Miscellaneous Examples Using The Laws of Exponents
- Decimal Number System
- Expressing Large Numbers in The Standard Form

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Exercise 13.1 Page: 252

**1. Find the value of:**

**(i) 2 ^{6}**

**Solution:-**

The above value can be written as,

= 2 Ã— 2 Ã— 2 Ã— 2 Ã— 2 Ã— 2

= 64

**(ii) 9 ^{3}**

**Solution:-**

The above value can be written as,

= 9 Ã— 9 Ã— 9

= 729

**(iii) 11 ^{2}**

**Solution:-**

The above value can be written as,

= 11 Ã— 11

= 121

**(iv) 5 ^{4}**

**Solution:-**

The above value can be written as,

= 5 Ã— 5 Ã— 5 Ã— 5

= 625

**2. Express the following in exponential form:**

**(i) 6 Ã— 6 Ã— 6 Ã— 6 **

**Solution:-**

The given question can be expressed in the exponential form as 6^{4}.

**(ii) t Ã— t **

**Solution:-**

The given question can be expressed in the exponential form as t^{2}.

**(iii) b Ã— b Ã— b Ã— b**

**Solution:-**

The given question can be expressed in the exponential form as b^{4}.

**(iv) 5 Ã— 5Ã— 7 Ã— 7 Ã— 7**

**Solution:-**

The given question can be expressed in the exponential form as 5^{2} Ã— 7^{3}.

Â

**(v) 2 Ã— 2 Ã— a Ã— a**

**Solution:-**

The given question can be expressed in the exponential form as 2^{2} Ã— a^{2}.

**(vi) a Ã— a Ã— a Ã— c Ã— c Ã— c Ã— c Ã— d**

**Solution:-**

The given question can be expressed in the exponential form as a^{3} Ã— c^{4} Ã— d.

**3. Express each of the following numbers using exponential notation:**

**(i) 512 **

**Solution:-**

The factors of 512 = 2 Ã— 2 Ã— 2 Ã— 2 Ã— 2 Ã— 2 Ã— 2 Ã— 2 Ã— 2

So it can be expressed in the exponential form as 2^{9}.

**(ii) 343 **

**Solution:-**

The factors of 343 = 7 Ã— 7 Ã— 7

So it can be expressed in the exponential form as 7^{3}.

**(iii) 729 **

**Solution:-**

The factors of 729 = 3 Ã— 3 Ã— 3 Ã— 3 Ã— 3 Ã— 3

So it can be expressed in the exponential form as 3^{6}.

**(iv) 3125**

**Solution:-**

The factors of 3125 = 5 Ã— 5 Ã— 5 Ã— 5 Ã— 5

So it can be expressed in the exponential form as 5^{5}.

**4. Identify the greater number, wherever possible, in each of the following?**

**(i) 4 ^{3} or 3^{4}**

**Solution:-**

The expansion of 4^{3} = 4 Ã— 4 Ã— 4 = 64

The expansion of 3^{4} = 3 Ã— 3 Ã— 3 Ã— 3 = 81

Clearly,

64 < 81

So, 4^{3} < 3^{4}

Hence 3^{4} is the greater number.

**(ii) 5 ^{3} or 3^{5}**

**Solution:-**

The expansion of 5^{3} = 5 Ã— 5 Ã— 5 = 125

The expansion of 3^{5} = 3 Ã— 3 Ã— 3 Ã— 3 Ã— 3= 243

Clearly,

125 < 243

So, 5^{3} < 3^{5}

Hence 3^{5} is the greater number.

**(iii) 2 ^{8} or 8^{2}**

**Solution:-**

The expansion of 2^{8} = 2 Ã— 2 Ã— 2 Ã— 2 Ã— 2 Ã— 2 Ã— 2 Ã— 2 = 256

The expansion of 8^{2} = 8 Ã— 8= 64

Clearly,

256 > 64

So, 2^{8} > 8^{2}

Hence 2^{8} is the greater number.

**(iv) 100 ^{2} or 2^{100}**

**Solution:-**

The expansion of 100^{2} = 100 Ã— 100 = 10000

The expansion of 2^{100}

2^{10} = 2 Ã— 2 Ã— 2 Ã— 2 Ã— 2 Ã— 2 Ã— 2 Ã— 2 Ã— 2 Ã— 2 = 1024

Then,

2^{100 } = 1024 Ã— 1024 Ã—1024 Ã— 1024 Ã—1024 Ã— 1024 Ã— 1024 Ã— 1024 Ã— 1024 Ã— 1024 =

Clearly,

100^{2} < 2^{100}

Hence 2^{100} is the greater number.

**(v) 2 ^{10} or 10^{2}**

**Solution:-**

The expansion of 2^{10} = 2 Ã— 2 Ã— 2 Ã— 2 Ã— 2 Ã— 2 Ã— 2 Ã— 2 Ã— 2 Ã— 2 = 1024

The expansion of 10^{2} = 10 Ã— 10= 100

Clearly,

1024 > 100

So, 2^{10} > 10^{2}

Hence 2^{10} 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 Ã— 2 Ã— 2 Ã— 3 Ã— 3 Ã— 3 Ã— 3

= 2^{3 }Ã— 3^{4}

**(ii) 405 **

**Solution:-**

Factors of 405 = 3 Ã— 3 Ã— 3 Ã— 3 Ã— 5

= 3^{4} Ã— 5

Â

**(iii) 540 **

**Solution:-**

Factors of 540 = 2 Ã— 2 Ã— 3 Ã— 3 Ã— 3 Ã— 5

= 2^{2 }Ã— 3^{3} Ã— 5

**(iv) 3,600**

**Solution:-**

Factors of 3600 = 2 Ã— 2 Ã— 2 Ã— 2 Ã— 3 Ã— 3 Ã— 5 Ã— 5

= 2^{4 }Ã— 3^{2} Ã— 5^{2}

**6. Simplify:**

**(i) 2 Ã— 10 ^{3}**

**Solution:-**

The above question can be written as,

= 2 Ã— 10 Ã— 10 Ã— 10

= 2 Ã— 1000

= 2000

**(ii) 7 ^{2} Ã— 2^{2}**

**Solution:-**

The above question can be written as,

= 7 Ã— 7 Ã— 2 Ã— 2

= 49 Ã— 4

= 196

**(iii) 2 ^{3} Ã— 5 **

**Solution:-**

The above question can be written as,

= 2 Ã— 2 Ã— 2 Ã— 5

= 8 Ã— 5

= 40

**(iv) 3 Ã— 4 ^{4}**

**Solution:-**

The above question can be written as,

= 3 Ã— 4 Ã— 4 Ã— 4 Ã— 4

= 3 Ã— 256

= 768

**(v) 0 Ã— 10 ^{2}**

**Solution:-**

The above question can be written as,

= 0 Ã— 10 Ã— 10

= 0 Ã— 100

= 0

**(vi) 5 ^{2} Ã— 3^{3}**

**Solution:-**

The above question can be written as,

= 5 Ã— 5 Ã— 3 Ã— 3 Ã— 3

= 25 Ã— 27

= 675

**(vii) 2 ^{4} Ã— 3^{2}**

**Solution:-**

The above question can be written as,

= 2 Ã— 2 Ã— 2 Ã— 2 Ã— 3 Ã— 3

= 16 Ã— 9

= 144

**(viii) 3 ^{2} Ã— 10^{4}**

**Solution:-**

The above question can be written as,

= 3 Ã— 3 Ã— 10 Ã— 10 Ã— 10 Ã— 10

= 9 Ã— 10000

= 90000

**7. Simplify:**

**(i) (â€“ 4) ^{3}**

**Solution:-**

The expansion of -4^{3}

= â€“ 4 Ã— â€“ 4 Ã— â€“ 4

= â€“ 64

**(ii) (â€“3) Ã— (â€“2) ^{3}**

**Solution:-**

The expansion of (-3) Ã— (-2)^{3}

= â€“ 3 Ã— â€“ 2 Ã— â€“ 2 Ã— â€“ 2

= â€“ 3 Ã— â€“ 8

= 24

**(iii) (â€“3) ^{2} Ã— (â€“5)^{2}**

**Solution:-**

The expansion of (-3)^{2} Ã— (-5)^{2}

= â€“ 3 Ã— â€“ 3 Ã— â€“ 5 Ã— â€“ 5

= 9 Ã— 25

= 225

**(iv) (â€“2) ^{3} Ã— (â€“10)^{3}**

**Solution:-**

The expansion of (-2)^{3} Ã— (-10)^{3}

= â€“ 2 Ã— â€“ 2 Ã— â€“ 2 Ã— â€“ 10 Ã— â€“ 10 Ã— â€“ 10

= â€“ 8 Ã— â€“ 1000

= 8000

**8. Compare the following numbers:**

**(i) 2.7 Ã— 10 ^{12} ; 1.5 Ã— 10^{8}**

**Solution:-**

By observing the question

Comparing the exponents of base 10,

Clearly,

2.7 Ã— 10^{12} > 1.5 Ã— 10^{8}

**(ii) 4 Ã— 10 ^{14} ; 3 Ã— 10^{17}**

**Solution:-**

By observing the question

Comparing the exponents of base 10,

Clearly,

4 Ã— 10^{14} < 3 Ã— 10^{17}

Exercise 13.2 Page: 260

**1. Using laws of exponents, simplify and write the answer in exponential form:**

**(i) 3 ^{2} Ã— 3^{4} Ã— 3^{8}**

**Solution:-**

By the rule of multiplying the powers with same base = a^{m }Ã— a^{n} = a^{m + n}

Then,

= (3)^{2 + 4 + 8}

= 3^{14}

**(ii) 6 ^{15} Ã· 6^{10}**

**Solution:-**

By the rule of dividing the powers with same base = a^{m }Ã· a^{n} = a^{m â€“ n}

Then,

= (6)^{15 â€“ 10}

= 6^{5}

**(iii) a ^{3} Ã— a^{2}**

**Solution:-**

By the rule of multiplying the powers with same base = a^{m }Ã— a^{n} = a^{m + n}

Then,

= (a)^{3 + 2 }

= a^{5}

**(iv) 7 ^{x} Ã— 7^{2}**

**Solution:-**

By the rule of multiplying the powers with same base = a^{m }Ã— a^{n} = a^{m + n}

Then,

= (7)^{x + 2}

Â

**(v) (5 ^{2})^{3} Ã· 5^{3}**

**Solution:-**

By the rule of taking power of as power = (a^{m})^{n } = a^{mn}

(5^{2})^{3} can be written as = (5)^{2 Ã— 3}

= 5^{6}

Now, 5^{6 }Ã· 5^{3}

By the rule of dividing the powers with same base = a^{m }Ã· a^{n} = a^{m â€“ n}

Then,

= (5)^{6 â€“ 3}

= 5^{3}

**(vi) 2 ^{5} Ã— 5^{5}**

**Solution:-**

By the rule of multiplying the powers with same exponents = a^{m }Ã— b^{m} = ab^{m}

Then,

= (2 Ã— 5)^{5}

= 10^{5}

**(vii) a ^{4} Ã— b^{4}**

**Solution:-**

By the rule of multiplying the powers with same exponents = a^{m }Ã— b^{m} = ab^{m}

Then,

= (a Ã— b)^{4}

= ab^{4}

**(viii) (3 ^{4})^{3}**

**Solution:-**

By the rule of taking power of as power = (a^{m})^{n } = a^{mn}

(3^{4})^{3} can be written as = (3)^{4 Ã— 3}

= 3^{12}

**(ix) (2 ^{20} Ã· 2^{15}) Ã— 2^{3}**

**Solution:-**

By the rule of dividing the powers with same base = a^{m }Ã· a^{n} = a^{m â€“ n}

(2^{20} Ã· 2^{15}) can be simplified as,

= (2)^{20 â€“ 15}

= 2^{5}

Then,

By the rule of multiplying the powers with same base = a^{m }Ã— a^{n} = a^{m + n}

2^{5} Ã— 2^{3} can be simplified as,

= (2)^{5 + 3}

= 2^{8}

**(x) 8 ^{t} Ã· 8^{2}**

**Solution:-**

By the rule of dividing the powers with same base = a^{m }Ã· a^{n} = a^{m â€“ n}

Then,

= (8)^{t â€“ 2}

**2. Simplify and express each of the following in exponential form:**

**(i) (2 ^{3} Ã— 3^{4} Ã— 4)/ (3 Ã— 32)**

**Solution:-**

Factors of 32 = 2 Ã— 2 Ã— 2 Ã— 2 Ã— 2

= 2^{5}

Factors of 4 = 2 Ã— 2

= 2^{2}

Then,

= (2^{3} Ã— 3^{4} Ã— 2^{2})/ (3 Ã— 2^{5})

= (2^{3 + 2} Ã— 3^{4}) / (3 Ã— 2^{5}) â€¦ [âˆµa^{m }Ã— a^{n} = a^{m + n}]

= (2^{5} Ã— 3^{4}) / (3 Ã— 2^{5})

= 2^{5 â€“ 5} Ã— 3^{4 â€“ 1} â€¦ [âˆµa^{m }Ã· a^{n} = a^{m â€“ n}]

= 2^{0} Ã— 3^{3}

= 1 Ã— 3^{3}

= 3^{3}

**(ii) ((5 ^{2})^{3} Ã— 5^{4}) Ã· 5^{7}**

**Solution:-**

(5^{2})^{3} can be written as = (5)^{2 Ã— 3} â€¦ [âˆµ(a^{m})^{n } = a^{mn}]

= 5^{6}

Then,

= (5^{6 }Ã— 5^{4}) Ã· 5^{7}

= (5^{6 + 4}) Ã· 5^{7} â€¦ [âˆµa^{m }Ã— a^{n} = a^{m + n}]

= 5^{10} Ã· 5^{7}

= 5^{10 â€“ 7} â€¦ [âˆµa^{m }Ã· a^{n} = a^{m â€“ n}]

= 5^{3}

**(iii) 25 ^{4} Ã· 5^{3}**

**Solution:-**

(25)^{4} can be written as = (5 Ã— 5)^{4}

= (5^{2})^{4}

(5^{2})^{4} can be written as = (5)^{2 Ã— 4} â€¦ [âˆµ(a^{m})^{n } = a^{mn}]

= 5^{8}

Then,

= 5^{8} Ã· 5^{3}

= 5^{8 â€“ 3} â€¦ [âˆµa^{m }Ã· a^{n} = a^{m â€“ n}]

= 5^{5}

**(iv) (3 Ã— 7 ^{2} Ã— 11^{8})/ (21 Ã— 11^{3})**

**Solution:-**

Factors of 21 = 7 Ã— 3

Then,

= (3 Ã— 7^{2} Ã— 11^{8})/ (7 Ã— 3 Ã— 11^{3})

= 3^{1-1} Ã— 7^{2-1} Ã— 11^{8 â€“ 3}

= 3^{0} Ã— 7 Ã— 11^{5}

= 1 Ã— 7 Ã— 11^{5}

= 7 Ã— 11^{5}

**(v) 3 ^{7}/ (3^{4} Ã— 3^{3})**

**Solution:-**

= 3^{7}/ (3^{4+3}) â€¦ [âˆµa^{m }Ã— a^{n} = a^{m + n}]

= 3^{7}/ 3^{7}

= 3^{7 â€“ 7} â€¦ [âˆµa^{m }Ã· a^{n} = a^{m â€“ n}]

= 3^{0}

= 1

**(vi) 2 ^{0} + 3^{0} + 4^{0}**

**Solution:-**

= 1 + 1 + 1

= 3

**(vii) 2 ^{0 }Ã— 3^{0} Ã— 4^{0}**

**Solution:-**

= 1 Ã— 1 Ã— 1

= 1

**(viii) (3 ^{0} + 2^{0}) Ã— 5^{0}**

**Solution:-**

= (1 + 1) Ã— 1

= (2) Ã— 1

= 2

**(ix) (2 ^{8} Ã— a^{5})/ (4^{3} Ã— a^{3})**

**Solution:-**

(4)^{3} can be written as = (2 Ã— 2)^{3}

= (2^{2})^{3}

(5^{2})^{4} can be written as = (2)^{2 Ã— 3} â€¦ [âˆµ(a^{m})^{n } = a^{mn}]

= 2^{6}

Then,

= (2^{8} Ã— a^{5})/ (2^{6} Ã— a^{3})

= 2^{8 â€“ 6} Ã— a^{5 â€“ 3} â€¦ [âˆµa^{m }Ã· a^{n} = a^{m â€“ n}]

= 2^{2 }Ã— a^{2}

= 2a^{2} â€¦ [âˆµ(a^{m})^{n } = a^{mn}]

**(x) (a ^{5}/a^{3}) Ã— a^{8}**

**Solution:-**

= (a^{5 â€“ }3) Ã— a^{8} â€¦ [âˆµa^{m }Ã· a^{n} = a^{m â€“ n}]

= a^{2} Ã— a^{8}

= a^{2 + 8} â€¦ [âˆµa^{m }Ã— a^{n} = a^{m + n}]

= a^{10}

**(xi) (4 ^{5} Ã— a^{8}b^{3})/ (4^{5} Ã— a^{5}b^{2})**

**Solution:-**

= 4^{5 â€“ 5} Ã— (a^{8 â€“ 5} Ã— b^{3 â€“ 2}) â€¦ [âˆµa^{m }Ã· a^{n} = a^{m â€“ n}]

= 4^{0} Ã— (a^{3}b)

= 1 Ã— a^{3}b

= a^{3}b

**(xii) (2 ^{3} Ã— 2)^{2}**

**Solution:-**

= (2^{3 + 1})^{2} â€¦ [âˆµa^{m }Ã— a^{n} = a^{m + n}]

= (2^{4})^{2}

(2^{4})^{2} can be written as = (2)^{4 Ã— 2} â€¦ [âˆµ(a^{m})^{n } = a^{mn}]

= 2^{8}

**3. Say true or false and justify your answer:**

**(i) 10 Ã— 10 ^{11} = 100^{11}**

**Solution:-**

Let us consider Left Hand Side (LHS) = 10 Ã— 10^{11}

= 10^{1 + 11} â€¦ [âˆµa^{m }Ã— a^{n} = a^{m + n}]

= 10^{12}

Now, consider Right Hand Side (RHS) = 100^{11}

= (10 Ã— 10)^{11}

= (10^{1 + 1})^{11}

= (10^{2})^{11}

= (10)^{2 Ã— 11} â€¦ [âˆµ(a^{m})^{n } = a^{mn}]

= 10^{22}

By comparing LHS and RHS,

LHS â‰ RHS

Hence, the given statement is false.

**(ii) 2 ^{3} > 5^{2}**

**Solution:-**

Let us consider LHS = 2^{3}

Expansion of 2^{3} = 2 Ã— 2 Ã— 2

= 8

Now, consider RHS = 5^{2}

Expansion of 5^{2} = 5 Ã— 5

= 25

By comparing LHS and RHS,

LHS < RHS

2^{3} < 5^{2}

Hence, the given statement is false.

**(iii) 2 ^{3} Ã— 3^{2} = 6^{5}**

**Solution:-**

Let us consider LHS = 2^{3} Ã— 3^{2}

Expansion of 2^{3} Ã— 3^{2}= 2 Ã— 2 Ã— 2 Ã— 3 Ã— 3

= 72

Now, consider RHS = 6^{5}

Expansion of 6^{5} = 6 Ã— 6 Ã— 6 Ã— 6 Ã— 6

= 7776

By comparing LHS and RHS,

72Â â‰ 7776

LHSÂ â‰ RHS

Hence, the given statement is false.

**(iv) 3 ^{0} = (1000)^{0}**

**Solution:-**

Let us consider LHS = 3^{0}

= 1

Now, consider RHS = 1000^{0}

= 1

By comparing LHS and RHS,

LHS = RHS

3^{0} = 1000^{0}

Hence, the given statement is true.

**4. Express each of the following as a product of prime factors only in exponential form:**

**(i) 108 Ã— 192 **

**Solution:-**

The factors of 108 = 2 Ã— 2 Ã— 3 Ã— 3 Ã— 3

= 2^{2} Ã— 3^{3}

The factors of 192 = 2 Ã— 2 Ã— 2 Ã— 2 Ã— 2 Ã— 2 Ã— 3

= 2^{6} Ã— 3

Then,

= (2^{2} Ã— 3^{3}) Ã— (2^{6} Ã— 3)

= 2^{2 + 6} Ã— 3^{3 + 1} â€¦ [âˆµa^{m }Ã— a^{n} = a^{m + n}]

= 2^{8 }Ã— 3^{4}

**(ii) 270 **

**Solution:-**

The factors of 270 = 2 Ã— 3 Ã— 3 Ã— 3 Ã— 5

= 2 Ã— 3^{3} Ã— 5

**(iii) 729 Ã— 64**

The factors of 729 = 3 Ã— 3 Ã— 3 Ã— 3 Ã— 3 Ã— 3

= 3^{6}

The factors of 64 = 2 Ã— 2 Ã— 2 Ã— 2 Ã— 2 Ã— 2

= 2^{6}

Then,

= (3^{6} Ã— 2^{6})

= 3^{6} Ã— 2^{6}

**(iv) 768**

**Solution:-**

The factors of 768 = 2 Ã— 2 Ã— 2 Ã— 2 Ã— 2 Ã— 2 Ã— 2 Ã— 2 Ã— 3

= 2^{8} Ã— 3

**5. Simplify:**

**(i) ((2 ^{5})^{2} Ã— 7^{3})/ (8^{3} Ã— 7)**

**Solution:-**

8^{3} can be written as = (2 Ã— 2 Ã— 2)^{3}

= (2^{3})^{3}

We have,

= ((2^{5})^{2} Ã— 7^{3})/ ((2^{3})^{3} Ã— 7)

= (2^{5 Ã— 2} Ã— 7^{3})/ ((2^{3 Ã— 3} Ã— 7) â€¦ [âˆµ(a^{m})^{n } = a^{mn}]

= (2^{10 }Ã— 7^{3})/ (2^{9} Ã— 7)

= (2^{10 â€“ 9} Ã— 7^{3 â€“ 1}) â€¦ [âˆµa^{m }Ã· a^{n} = a^{m â€“ n}]

= 2 Ã— 7^{2}

= 2 Ã— 7 Ã— 7

= 98

**(ii) (25 Ã— 5 ^{2} Ã— t^{8})/ (10^{3} Ã— t^{4})**

**Solution:-**

25 can be written as = 5 Ã— 5

= 5^{2}

10^{3} can be written as = 10^{3}

= (5 Ã— 2)^{3}

= 5^{3} Ã— 2^{3}

We have,

= (5^{2} Ã— 5^{2} Ã— t^{8})/ (5^{3} Ã— 2^{3} Ã— t^{4})

= (5^{2 + 2} Ã— t^{8})/ (5^{3} Ã— 2^{3} Ã— t^{4}) â€¦ [âˆµa^{m }Ã— a^{n} = a^{m + n}]

= (5^{4} Ã— t^{8})/ (5^{3} Ã— 2^{3} Ã— t^{4})

= (5^{4 â€“ 3} Ã— t^{8 â€“ 4})/ 2^{3} â€¦ [âˆµa^{m }Ã· a^{n} = a^{m â€“ n}]

= (5 Ã— t^{4})/ (2 Ã— 2 Ã— 2)

= (5t^{4})/ 8

**(iii) (3 ^{5} Ã— 10^{5} Ã— 25)/ (5^{7} Ã— 6^{5})**

**Solution:-**

10^{5 }can be written as = (5 Ã— 2)^{5}

= 5^{5} Ã— 2^{5}

25 can be written as = 5 Ã— 5

= 5^{2}

6^{5} can be written as = (2 Ã— 3)^{5}

= 2^{5} Ã— 3^{5}

Then we have,

= (3^{5} Ã— 5^{5} Ã— 2^{5} Ã— 5^{2})/ (5^{7} Ã— 2^{5} Ã— 3^{5})

= (3^{5} Ã— 5^{5 + 2} Ã— 2^{5})/ (5^{7} Ã— 2^{5} Ã— 3^{5}) â€¦ [âˆµa^{m }Ã— a^{n} = a^{m + n}]

= (3^{5} Ã— 5^{7} Ã— 2^{5})/ (5^{7} Ã— 2^{5} Ã— 3^{5})

= (3^{5 â€“ 5} Ã— 5^{7 â€“ 7 }Ã— 2^{5 â€“ 5})

= (3^{0} Ã— 5^{0} Ã— 2^{0}) â€¦ [âˆµa^{m }Ã· a^{n} = a^{m â€“ n}]

= 1 Ã— 1 Ã— 1

= 1

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 Ã— 100000) + (7 Ã— 10000) + (9 Ã— 1000) + (4 Ã— 100) + (0 Ã— 10) + (4 Ã— 1)

Now we can express it using powers of 10 in the exponent form,

= (2 Ã— 10^{5}) + (7 Ã— 10^{4}) + (9 Ã— 10^{3}) + (4 Ã— 10^{2}) + (0 Ã— 10^{1}) + (4 Ã— 10^{0})

**3006194**

**Solution:-**

The expanded form of the number 3006194 is,

= (3 Ã— 1000000) + (0 Ã— 100000) + (0 Ã— 10000) + (6 Ã— 1000) + (1 Ã— 100) + (9 Ã— 10) + 4

Now we can express it using powers of 10 in the exponent form,

= (3 Ã— 10^{6}) + (0 Ã— 10^{5}) + (0 Ã— 10^{4}) + (6 Ã— 10^{3}) + (1 Ã— 10^{2}) + (9 Ã— 10^{1}) + (4 Ã— 10^{0})

**2806196 **

**Solution:-**

The expanded form of the number 2806196 is,

= (2 Ã— 1000000) + (8 Ã— 100000) + (0 Ã— 10000) + (6 Ã— 1000) + (1 Ã— 100) + (9 Ã— 10) + 6

Now we can express it using powers of 10 in the exponent form,

= (2 Ã— 10^{6}) + (8 Ã— 10^{5}) + (0 Ã— 10^{4}) + (6 Ã— 10^{3}) + (1 Ã— 10^{2}) + (9 Ã— 10^{1}) + (6 Ã— 10^{0})

**120719**

**Solution:-**

The expanded form of the number 120719 is,

= (1 Ã— 100000) + (2 Ã— 10000) + (0 Ã— 1000) + (7 Ã— 100) + (1 Ã— 10) + (9 Ã— 1)

Now we can express it using powers of 10 in the exponent form,

= (1 Ã— 10^{5}) + (2 Ã— 10^{4}) + (0 Ã— 10^{3}) + (7 Ã— 10^{2}) + (1 Ã— 10^{1}) + (9 Ã— 10^{0})

**20068**

**Solution:-**

The expanded form of the number 20068 is,

= (2 Ã— 10000) + (0 Ã— 1000) + (0 Ã— 100) + (6 Ã— 10) + (8 Ã— 1)

Now we can express it using powers of 10 in the exponent form,

= (2 Ã— 10^{4}) + (0 Ã— 10^{3}) + (0 Ã— 10^{2}) + (6 Ã— 10^{1}) + (8 Ã— 10^{0})

**2. Find the number from each of the following expanded forms:**

**(a) (8 Ã—10) ^{4} + (6 Ã— 10)^{3} + (0 Ã— 10)^{2} + (4 Ã— 10)^{1} + (5 Ã— 10)^{0}**

**Solution:-**

The expanded form is,

= (8 Ã— 10000) + (6 Ã— 1000) + (0 Ã— 100) + (4 Ã— 10) + (5 Ã— 1)

= 80000 + 6000 + 0 + 40 + 5

= 86045

**(b) (4 Ã—10) ^{5} + (5 Ã— 10)^{3} + (3 Ã— 10)^{2} + (2 Ã— 10)^{0}**

**Solution:-**

The expanded form is,

= (4 Ã— 100000) + (0 Ã— 10000) + (5 Ã— 1000) + (3 Ã— 100) + (0 Ã— 10) + (2 Ã— 1)

= 400000 + 0 + 5000 + 300 + 0 + 2

= 405302

**(c) (3 Ã—10) ^{4} + (7 Ã— 10)^{2} + (5 Ã— 10)^{0}**

**Solution:-**

The expanded form is,

= (3 Ã— 10000) + (0 Ã— 1000) + (7 Ã— 100) + (0 Ã— 10) + (5 Ã— 1)

= 30000 + 0 + 700 + 0 + 5

= 30705

**(d) (9 Ã—10) ^{5} + (2 Ã— 10)^{2} + (3 Ã— 10)^{1}**

**Solution:-**

The expanded form is,

= (9 Ã— 100000) + (0 Ã— 10000) + (0 Ã— 1000) + (2 Ã— 100) + (3 Ã— 10) + (0 Ã— 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 Ã— 10^{7}

**(ii) 70,00,000 **

**Solution:-**

The standard form of the given number is 7 Ã— 10^{6}

**(iii) 3,18,65,00,000**

**Solution:-**

The standard form of the given number is 3.1865 Ã— 10^{9}

**(iv) 3,90,878**

**Solution:-**

The standard form of the given number is 3.90878 Ã— 10^{5}

**(v) 39087.8**

**Solution:-**

The standard form of the given number is 3.90878 Ã— 10^{4}

**(vi) 3908.78**

**Solution:-**

The standard form of the given number is 3.90878 Ã— 10^{3}

**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 Ã— 10^{8}m.

**(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 Ã— 10^{8}m/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 Ã— 10^{7}m.

**(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 Ã— 10^{9}m.

**(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 Ã— 10^{11} stars.

**(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 Ã— 10^{10} 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 Ã— 10^{20}m.

**(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 Ã— 10^{22} 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 Ã— 10^{9} 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 Ã— 10^{9}.

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