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# Find the value of

(i) (3^{0 }× 4^{−1}) × 2^{2 }(ii) (2^{−1 }× 4^{−1}) ÷ 2^{−2}

(iii) (1/2)^{−2 }+ (1/3)^{−2 }+ (1/4)^{−2}

(iv) (3^{−1 }+4^{−1 }+ 5^{−1})^{0}

(v) {(−2/3)^{−2}}^{2}

**Solution:**

The exponent of a number shows how many times the number is multiplied by itself.

(i) (3^{0 }× 4^{−1}) × 2^{2}

According to the rules of exponents,

a^{0} = 1 and a^{−m} = 1/a^{m}

(3^{0 }× 4^{−1}) × 2^{2 }

= (1 + 1/4) × 2^{2}

= [(4 + 1)/4] × 2^{2 }

= (5/4) × 2^{2 }

= 5/2^{2 }× 2^{2 }[Since 4 = 2 × 2 = 2^{2}]

= 5

(ii) (2^{−1 }× 4^{−1}) ÷ 2^{−2}

According to the rules of exponents,

(a^{m})^{n }= a^{mn}, a^{−m} = 1/a^{m}

(2^{−1 }× 4^{−1}) ÷ 2^{−2 }

= [2^{−1 }× {(2)^{2}}^{−1}] ÷ 2^{−2}[Since 4 = 2 × 2 = 2^{2}]

= (2^{−1 }× 2^{−2}) ÷ 2^{−2} [∵(a^{m})^{n }= a^{mn}]

= 2^{−3 }÷ 2^{−2 }[∵a^{m }× a^{n }= a^{m + n}]

= 2^{−3−(−2)} [∵a^{m }÷ a^{n }= a^{m−n}]

= 2^{−3 + 2} = 2^{−1}

= 1/2 [∵a^{−m }= 1/a^{m}]

(iii) (1/2)^{−2 }+ (1/3)^{−2 }+ (1/4)^{−2}

According to the rules of exponents,

(a/b)^{−m} = (b/a)^{m}

(1/2)^{−2 }+ (1/3)^{−2 }+ (1/4)^{−2 }

= (2/1)^{2 }+ (3/1)^{2 }+ (4/1)^{2 }

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

= 4 + 9 + 16 = 29

(iv) (3^{−1 }+4^{−1 }+ 5^{−1})^{0}

According to the rules of exponents,

a^{0} = 1 and a^{−m} = 1/a^{m}

(3^{−1 }+ 4^{−1 }+ 5^{−1})^{0 }

= (1/3 + 1/4 + 1/5)^{0} [Since a^{−m} = 1/a^{m}]

= 1 [Since a^{0} = 1]

(v) {(−2/3)^{−2}}^{2}

According to the rules of exponents,

a^{−m} = 1/a^{m }and (a/b)^{m }= a^{m}/b^{m}

{(−2/3)^{−2}}^{2 }

= {(3/−2)^{2}}^{2} [Sincea^{−m }= 1/a^{m}]

={3^{2}/(−2)^{2}}^{2} [Since(a/b)^{m }= a^{m}/b^{m}]

=(9/4)^{2 }= 81/16

**☛ Check: **NCERT Solutions for Class 8 Maths Chapter 12

**Video Solution:**

## Find the value of (i) (3⁰^{ }× 4⁻¹) × 2²^{ }(ii) (2⁻¹^{ }× 4⁻¹) ÷ 2⁻²^{ }(iii) (1/2)⁻²^{ }+ (1/3)⁻²^{ }+ (1/4)⁻²^{ }(iv) (3⁻¹^{ }+4⁻¹^{ }+ 5⁻¹)⁰^{ } (v) {(−2/3)⁻²}²

Class 8 Maths NCERT Solutions Chapter 12 Exercise 12.1 Question 3

**Summary:**

The value of the following (i) (3^{0 }× 4^{−1}) × 2^{2 }(ii) (2^{−1 }× 4^{−1}) ÷ 2^{−2 }(iii) (1/2)^{−2 }+ (1/3)^{−2 }+ (1/4)^{−2 }(iv) (3^{−1 }+4^{−1 }+ 5^{−1})^{0 }(v) {(−2/3)^{−2}}^{2} are (i) 5 (ii) 1/2 (iii) 29 (iv) 1 and (v) 81/16 respectively.

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