## MP Board Class 7th Maths Solutions Chapter 13 Exponents and Powers Ex 13.2

Question 1.

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

(i) 3^{2} × 3^{4} × 3^{8}

(ii) 6^{15} ÷ 6^{10}

(iii) a^{3} × a^{2}

(iv) 7^{x} × 7^{2}

(v) (5^{2})^{3} 4 5^{3}

(vi) 2^{5} × 5^{5}

(vii) a^{4} × b^{4}

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

(ix) (2^{20} ÷ 2^{15}) × 2^{3}

(x) (8^{t} ÷ 8^{2})

Solution:

(i) 3^{2} × 3^{4} × 3^{8} = (3)^{2 + 4 + 8} = 3^{14} [∵ a^{m} × a^{n} = a^{m+n]}

(ii) 6^{15} ÷ 6^{10} = (6)^{15 – 10} = 6^{5} [∵ a^{m} ÷ a^{n} = a^{m – n}]

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

(iv) 7^{x} × 7^{2} = 7^{x + 2}[∵ a^{m} × a^{n} = a^{m+n}]

(v) (5^{2})^{3} ÷ 5^{3} = 5^{2 × 3} ÷ 5^{3} [∵ (a^{m})^{n} = a^{mn}]

= 5^{6} ÷ 5^{3} = 5^{(6 – 3)} [∵ a^{m} ÷ a^{n} = a^{m – n}]

= 5^{3}

(vi) 2^{5} × 5^{5} = (2 × 5)^{5 }[∵ a^{m} × b^{m} = (a × b)^{m}]

= 10^{5}

(vii) a^{4} × b^{4} = (ab)^{4}

(viii) (3^{4})^{3} = 3^{4 × 3} = 3^{12 } [∵ (a^{m})^{n} = a^{mn}]

(ix) (2^{20} ÷ 2^{15}) × 2^{3} = (2^{20 – 15}) × 2^{3} [∵ a^{m} ÷ a^{n} = a^{m – n}]

= 2^{5} × 2^{3} = 2^{5 + 3} [∵ a^{m} × a^{n} = a^{m+n}]

(x) 8^{t} ÷ 8^{2} = 8^{(t – 2)} [∵ a^{m} ÷ a^{n} = a^{m – n}]

Question 2.

Simplify and express each of the following in exponential form:

Solution:

(ii) [(5^{2})^{3} × 5^{4}] ÷ 5^{7}

= [5^{2 × 3} × 5^{4}] ÷ 5^{7} [∵ (a^{m})^{n} = a^{mn}]

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

= [5^{6 + 4}] ÷ 5^{7}

= 5^{10} ÷ 5^{7}

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

= 5^{3}

(iii) 25^{4} ÷ 5^{3} = (5 × 5)^{4} ÷ 5^{3}

= (5^{2})^{4} ÷ 5^{3} = 5^{2 × 4} ÷ 5^{3} [∵ (a^{m})^{n} = a^{mn}]

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

= 5^{5}

(vi) 2^{0} + 3^{0} + 4^{0} = 1 + 1 + 1 = 3 [∵ a^{0} = 1]

(vii) 2^{0} × 3^{0} × 4^{0} = 1 × 1 × 1 = 1

(viii) (3^{0} + 2^{0}) × 5^{0} = (1 + 1) × 1 = 2

Question 3.

Say true or false and justify your answer:

(i) 10 × 10^{11} = 100^{11 }

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

(iii) 2^{3} × 3^{2} = 6^{5}

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

Solution:

(i) L.H.S = 10 × 10^{11}

= 10^{1 + 11} = 10^{12}

R.H.S = 100^{11} = (10 × 10)^{11} = (10^{0})^{11} = 10^{2 × 11} = 10^{22}

⇒ L.H.S. ≠ R.H.S.

Hence, the given statement is false.

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

L.H.S. = 2^{3} = 2 × 2 × 2 = 8

R.H.S. = 5^{3} = 5 × 5 = 25

⇒ L.H.S ≠ R.H.S

Hence, the given statement is false.

(iii) 2^{3} × 3^{2} = 6^{5}

L.H.S. = 2^{3} × 3^{2} = 2 × 2 × 2 × 3 × 3 = 72

R.H.S. = 6^{5} = 6 × 6 × 6 × 6 × 6 = 7776

⇒ L.H.S. ≠ R.H.S.

Hence, the given statement is false.

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

L.H.S. = 3^{0} = 1

R.H.S. = (1000)^{0} = 1

⇒ L.H.S. = R.H.S.

Hence, the given statement is true.

Question 4.

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

(i) 108 × 192

(ii) 270

(iii) 729 × 64

(iv) 768

Solution:

(i) 108 × 192

= (2 × 2 × 3 × 3 × 3) × (2 × 2 × 2 × 2 × 2 × 2 × 3)

= (2^{2} × 3^{3}) × (2^{6} × 3)

= 2^{2+6} × 3^{3+1} = 2^{8} × 3^{4}

(ii) 270 = 2 × 3 × 3 × 3 × 5 = 2 × 3^{3} × 5

(iii) 729 × 64 = (3 × 3 × 3 × 3 × 3 × 3) × (2 × 2 × 2 × 2 × 2 × 2) = 3^{6} × 2^{6}

(iv) 768 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3 = 2^{8} × 3

Question 5.

Simplify:

Solution: