### Access answers to Maths NCERT Solutions for Class 7 Chapter 13 – Exponents and Powers Exercise 13.2

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

**(i) 3 ^{2} × 3^{4} × 3^{8}**

**Solution:-**

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

Then,

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

= 3^{14}

**(ii) 6 ^{15} ÷ 6^{10}**

**Solution:-**

By the rule of dividing the powers with same base = a^{m }÷ a^{n} = a^{m – n}

Then,

= (6)^{15 – 10}

= 6^{5}

**(iii) a ^{3} × a^{2}**

**Solution:-**

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

Then,

= (a)^{3 + 2}

= a^{5}

**(iv) 7 ^{x} × 7^{2}**

**Solution:-**

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

Then,

= (7)^{x + 2}

**(v) (5 ^{2})^{3} ÷ 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 × 3}

= 5^{6}

Now, 5^{6 }÷ 5^{3}

By the rule of dividing the powers with same base = a^{m }÷ a^{n} = a^{m – n}

Then,

= (5)^{6 – 3}

= 5^{3}

**(vi) 2 ^{5} × 5^{5}**

**Solution:-**

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

Then,

= (2 × 5)^{5}

= 10^{5}

**(vii) a ^{4} × b^{4}**

**Solution:-**

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

Then,

= (a × 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 × 3}

= 3^{12}

**(ix) (2 ^{20} ÷ 2^{15}) × 2^{3}**

**Solution:-**

By the rule of dividing the powers with same base = a^{m }÷ a^{n} = a^{m – n}

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

= (2)^{20 – 15}

= 2^{5}

Then,

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

2^{5} × 2^{3} can be simplified as,

= (2)^{5 + 3}

= 2^{8}

**(x) 8 ^{t} ÷ 8^{2}**

**Solution:-**

By the rule of dividing the powers with same base = a^{m }÷ a^{n} = a^{m – n}

Then,

= (8)^{t – 2}

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

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

**Solution:-**

Factors of 32 = 2 × 2 × 2 × 2 × 2

= 2^{5}

Factors of 4 = 2 × 2

= 2^{2}

Then,

= (2^{3} × 3^{4} × 2^{2})/ (3 × 2^{5})

= (2^{3 + 2} × 3^{4}) / (3 × 2^{5}) … [∵a^{m }× a^{n} = a^{m + n}]

= (2^{5} × 3^{4}) / (3 × 2^{5})

= 2^{5 – 5} × 3^{4 – 1} … [∵a^{m }÷ a^{n} = a^{m – n}]

= 2^{0} × 3^{3}

= 1 × 3^{3}

= 3^{3}

**(ii) ((5 ^{2})^{3} × 5^{4}) ÷ 5^{7}**

**Solution:-**

(5^{2})^{3} can be written as = (5)^{2 × 3} … [∵(a^{m})^{n }= a^{mn}]

= 5^{6}

Then,

= (5^{6 }× 5^{4}) ÷ 5^{7}

= (5^{6 + 4}) ÷ 5^{7} … [∵a^{m }× a^{n} = a^{m + n}]

= 5^{10} ÷ 5^{7}

= 5^{10 – 7} … [∵a^{m }÷ a^{n} = a^{m – n}]

= 5^{3}

**(iii) 25 ^{4} ÷ 5^{3}**

**Solution:-**

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

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

(5^{2})^{4} can be written as = (5)^{2 × 4} … [∵(a^{m})^{n }= a^{mn}]

= 5^{8}

Then,

= 5^{8} ÷ 5^{3}

= 5^{8 – 3} … [∵a^{m }÷ a^{n} = a^{m – n}]

= 5^{5}

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

**Solution:-**

Factors of 21 = 7 × 3

Then,

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

= 3^{1-1} × 7^{2-1} × 11^{8 – 3}

= 3^{0} × 7 × 11^{5}

= 1 × 7 × 11^{5}

= 7 × 11^{5}

**(v) 3 ^{7}/ (3^{4} × 3^{3})**

**Solution:-**

= 3^{7}/ (3^{4+3}) … [∵a^{m }× a^{n} = a^{m + n}]

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

= 3^{7 – 7} … [∵a^{m }÷ a^{n} = a^{m – n}]

= 3^{0}

= 1

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

**Solution:-**

= 1 + 1 + 1

= 3

**(vii) 2 ^{0 }× 3^{0} × 4^{0}**

**Solution:-**

= 1 × 1 × 1

= 1

**(viii) (3 ^{0} + 2^{0}) × 5^{0}**

**Solution:-**

= (1 + 1) × 1

= (2) × 1

= 2

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

**Solution:-**

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

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

(5^{2})^{4} can be written as = (2)^{2 × 3} … [∵(a^{m})^{n }= a^{mn}]

= 2^{6}

Then,

= (2^{8} × a^{5})/ (2^{6} × a^{3})

= 2^{8 – 6} × a^{5 – 3} … [∵a^{m }÷ a^{n} = a^{m – n}]

= 2^{2 }× a^{2}

= 2a^{2} … [∵(a^{m})^{n }= a^{mn}]

**(x) (a ^{5}/a^{3}) × a^{8}**

**Solution:-**

= (a^{5 – }3) × a^{8} … [∵a^{m }÷ a^{n} = a^{m – n}]

= a^{2} × a^{8}

= a^{2 + 8} … [∵a^{m }× a^{n} = a^{m + n}]

= a^{10}

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

**Solution:-**

= 4^{5 – 5} × (a^{8 – 5} × b^{3 – 2}) … [∵a^{m }÷ a^{n} = a^{m – n}]

= 4^{0} × (a^{3}b)

= 1 × a^{3}b

= a^{3}b

**(xii) (2 ^{3} × 2)^{2}**

**Solution:-**

= (2^{3 + 1})^{2} … [∵a^{m }× a^{n} = a^{m + n}]

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

(2^{4})^{2} can be written as = (2)^{4 × 2} … [∵(a^{m})^{n }= a^{mn}]

= 2^{8}

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

**(i) 10 × 10 ^{11} = 100^{11}**

**Solution:-**

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

= 10^{1 + 11} … [∵a^{m }× a^{n} = a^{m + n}]

= 10^{12}

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

= (10 × 10)^{11}

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

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

= (10)^{2 × 11} … [∵(a^{m})^{n }= a^{mn}]

= 10^{22}

By comparing LHS and RHS,

LHS ≠ RHS

Hence, the given statement is false.

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

**Solution:-**

Let us consider LHS = 2^{3}

Expansion of 2^{3} = 2 × 2 × 2

= 8

Now, consider RHS = 5^{2}

Expansion of 5^{2} = 5 × 5

= 25

By comparing LHS and RHS,

LHS < RHS

2^{3} < 5^{2}

Hence, the given statement is false.

**(iii) 2 ^{3} × 3^{2} = 6^{5}**

**Solution:-**

Let us consider LHS = 2^{3} × 3^{2}

Expansion of 2^{3} × 3^{2}= 2 × 2 × 2 × 3 × 3

= 72

Now, consider RHS = 6^{5}

Expansion of 6^{5} = 6 × 6 × 6 × 6 × 6

= 7776

By comparing LHS and RHS,

72 ≠ 7776

LHS ≠ 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 × 192**

**Solution:-**

The factors of 108 = 2 × 2 × 3 × 3 × 3

= 2^{2} × 3^{3}

The factors of 192 = 2 × 2 × 2 × 2 × 2 × 2 × 3

= 2^{6} × 3

Then,

= (2^{2} × 3^{3}) × (2^{6} × 3)

= 2^{2 + 6} × 3^{3 + 1} … [∵a^{m }× a^{n} = a^{m + n}]

= 2^{8 }× 3^{4}

**(ii) 270**

**Solution:-**

The factors of 270 = 2 × 3 × 3 × 3 × 5

= 2 × 3^{3} × 5

**(iii) 729 × 64**

The factors of 729 = 3 × 3 × 3 × 3 × 3 × 3

= 3^{6}

The factors of 64 = 2 × 2 × 2 × 2 × 2 × 2

= 2^{6}

Then,

= (3^{6} × 2^{6})

= 3^{6} × 2^{6}

**(iv) 768**

**Solution:-**

The factors of 768 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3

= 2^{8} × 3

**5. Simplify:**

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

**Solution:-**

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

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

We have,

= ((2^{5})^{2} × 7^{3})/ ((2^{3})^{3} × 7)

= (2^{5 × 2} × 7^{3})/ ((2^{3 × 3} × 7) … [∵(a^{m})^{n }= a^{mn}]

= (2^{10 }× 7^{3})/ (2^{9} × 7)

= (2^{10 – 9} × 7^{3 – 1}) … [∵a^{m }÷ a^{n} = a^{m – n}]

= 2 × 7^{2}

= 2 × 7 × 7

= 98

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

**Solution:-**

25 can be written as = 5 × 5

= 5^{2}

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

= (5 × 2)^{3}

= 5^{3} × 2^{3}

We have,

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

= (5^{2 + 2} × t^{8})/ (5^{3} × 2^{3} × t^{4}) … [∵a^{m }× a^{n} = a^{m + n}]

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

= (5^{4 – 3} × t^{8 – 4})/ 2^{3} … [∵a^{m }÷ a^{n} = a^{m – n}]

= (5 × t^{4})/ (2 × 2 × 2)

= (5t^{4})/ 8

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

**Solution:-**

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

= 5^{5} × 2^{5}

25 can be written as = 5 × 5

= 5^{2}

6^{5} can be written as = (2 × 3)^{5}

= 2^{5} × 3^{5}

Then we have,

= (3^{5} × 5^{5} × 2^{5} × 5^{2})/ (5^{7} × 2^{5} × 3^{5})

= (3^{5} × 5^{5 + 2} × 2^{5})/ (5^{7} × 2^{5} × 3^{5}) … [∵a^{m }× a^{n} = a^{m + n}]

= (3^{5} × 5^{7} × 2^{5})/ (5^{7} × 2^{5} × 3^{5})

= (3^{5 – 5} × 5^{7 – 7 }× 2^{5 – 5})

= (3^{0} × 5^{0} × 2^{0}) … [∵a^{m }÷ a^{n} = a^{m – n}]

= 1 × 1 × 1

= 1