NCERT Solutions for Class 7 Maths Exercise 13.2 Chapter 13 Exponents and Powers in simple PDF are available here. Laws of exponents in that multiplying and dividing powers with the same base, taking power of a power, multiplying and dividing the powers with the same exponents and miscellaneous examples using the laws of exponents are the topics covered in this exercise of NCERT Maths Solutions for Class 7 Chapter 13. While solving the exercise questions from the NCERT Class 7 book, students often face difficulty and eventually tend to pile up their doubts. These Solutions are prepared by our expert tutors with step by step explanations.

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**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

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