# Simplify and express each of the following in exponential form:

(i) (2^{3} × 3^{3} × 4) / (3 × 32) (ii) ((5^{2} )^{3} × 5^{4}) ÷ 5^{7 }(iii) 25^{4} ÷ 5^{3 }

(iv) (3 × 7^{2} ×11^{8}) / (21×11^{3}) (v) 3^{7} / (3^{4} × 3^{3}) (vi) 2^{0} + 3^{0} + 4^{0}

(vii) 2^{0} × 3^{0} × 4^{0 }(viii) (3^{0} + 2^{0} )× 5^{0 }(ix) (2^{8} × a^{5}) / (4^{3} × a^{3})^{ }

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

**Solution:**

To solve this question, we be using the laws of exponents.

(i) (2^{3} × 3^{4} × 4) / (3 × 32)

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

= (2^{3 + 2} × 3^{4}) / (3^{1} × 2^{5}) [a^{m} × a^{n} = a^{m + n}]

= (2^{5} × 3^{4 - 1}) / (2^{5}) [a^{m} ÷ a^{n} = a^{m - n}]

= 3^{3}

(ii) ((5^{2} )^{3} × 5^{4}) ÷ 5^{7}

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

= 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}

= (5^{2})^{4} ÷ 5^{3}

= 5^{8} ÷ 5^{3} [(a^{m})^{n} = a^{mn}]

= 5^{8 - 3} [a^{m} ÷ a^{n} = a^{m - n}]

= 5^{5}

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

= (3 × 7^{2 }× 11^{8}) / (3 × 7 × 11^{3}) [Since, 21 = 3 × 7]

= (7^{2 }× 11^{8}) / (7 × 11^{3})

= 7^{2 - 1} × 11^{8 - 3} [a^{m} ÷ a^{n} = a^{m - n}]

= 7 × 11^{5}

(v) 3^{7} / (3^{4} × 3^{3})

= 3^{7} / 3^{4 + 3} [a^{m} × a^{n} = a^{m + n}]

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

= 3^{7 - 7 }[am ÷ an = am-n]

= 3^{0}

= 1 [a^{o} = 1]

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

= 1 + 1 + 1 [a^{o} = 1]

= 3

(vii) 2^{0} × 3^{0} × 4^{0}

= 1 × 1 × 1 [a^{o} = 1]

= 1

(viii) (3^{0} + 2^{0} ) × 5^{0}

= (1 + 1) × 1 [a^{o} = 1]

= 2 × 1

= 2

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

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

= (2^{8} × a^{5}) / (2^{6} × a^{3}) [(a^{m})^{n} = a^{mn}]

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

= 2^{2} × a^{2}

(x) (a^{5} / a^{3}) × a^{8}

= 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})

= 4^{5 - 5} × a^{8 - 5} × b^{3 - 2} [a^{m} ÷ a^{n} = a^{m - n}]

= 4^{0} × a^{3} × b

= a^{3} × b [a^{o} = 1]

(xii) (2^{3} × 2)^{2 }

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

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

= 2^{4 × 2} [(a^{m})^{n} = a^{mn}]

= 2^{8}

**☛ Check: **NCERT Solutions for Class 7 Maths Chapter 13

**Video Solution:**

## Simplify and express each of the following in exponential form: (i) (2³ × 3³ × 4) / (3×32) (ii) [(5² )³ × 5⁴] ÷ 5⁷^{ }(iii) 25⁴ ÷ 5³^{ }(iv) (3 × 7² ×11⁸) / (21×11³) (v) 3⁷ / (3⁴ × 3³) (vi) 2⁰ + 3⁰ + 4⁰(vii) 2⁰ × 3⁰ × 4⁰^{ }(viii) (3⁰ + 2⁰ )× 5⁰^{ }(ix) (2⁸ × a⁵) / (4³ × a³) (x) (a⁵ / a³) × a⁸^{ }(xi) (4⁵ × a⁸b³) / (4⁵ × a⁵b²) (xii) (2³^{ }× 2)²

Maths NCERT Solutions Class 7 Chapter 13 Exercise 13.2 Question 2

**Summary:**

We have simplified and expressed each of the following in exponential form: (i) (2³ × 3³ × 4) / (3×32) (ii) [(5² )³ × 5⁴] ÷ 5⁷^{ }(iii) 25⁴ ÷ 5³^{ }(iv) (3 × 7² ×11⁸) / (21×11³) (v) 3⁷ / (3⁴ × 3³) (vi) 2⁰ + 3⁰ + 4⁰(vii) 2⁰ × 3⁰ × 4⁰^{ }(viii) (3⁰ + 2⁰ )× 5⁰^{ }(ix) (2⁸ × a⁵) / (4³ × a³) (x) (a⁵ / a³) × a⁸^{ }(xi) (4⁵ × a⁸b³) / (4⁵ × a⁵b²) (xii) (2³^{ }× 2)² as follows: (i) 3^{3} , (ii) 5^{3} , (iii) 5^{5} , (iv) 7 × 11^{5} , (v) 1 , (vi) 3 , (vii) 1 , (viii) 2, (ix) 2^{2} × a^{2}, (x) a^{10} , (xi) a^{3} × b, (xii) 2^{8}.

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