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## Karnataka State Syllabus Class 7 Maths 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} ÷ 5^{2}

(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} = 5^{6 – 3} [(a^{m})^{n} = a^{mn}

∵ a^{m} ÷ a^{n} = a^{m – n}]

vi) 2^{5} × 5^{5}

= (2 × 5)^{5} = 10^{5} [∵ a^{m} × b^{m} = (ab)^{m}]

vii) a^{4} × b^{4}

= a^{4} × b^{4} = (ab)^{4} [∵ a^{m} × b^{m} = (ab)^{m}]

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

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

ix) (2^{20} ÷ 2^{15})

(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} = 2^{8} [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 :

i)

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}

= 5^{6 + 4} ÷ 5^{7} = 5^{10 – 7} = 5^{3}

iii) 25^{4} ÷ 5^{3}

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

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

iv)

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

= 7^{1} × 11^{5} = 7 × 11^{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 × 1 = 2

ix)

= (2a)^{2}

x)

= a^{ 5 – 3} × a^{8}

= a^{2} × a^{8} a^{2 + 8}

xi)

xii) (2^{3} × 2)^{2}

= (2^{3} × 2^{1})^{2}

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

= 2^{4 × 2} = 2^{8}

Question 3.

Say true or false and justify your answer:

i) 10 × 10^{11} = 100^{11}

L.H.S = 10 × 10^{11} = 10^{1 + 11} = 10^{12} RHS

100^{11} = (10 × 10)^{11}

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

= 1o^{2} × 11 = 10^{22}

∴ 10^{12} ≠ 10^{22}

So 10 × 10^{11} ≠ 100^{11} ∴ It is false.

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

= 2 × 2 × 2 > 5 × 5

= 8 > 25

∴ 2^{3} > 5^{2} ∴ It is false.

iii) 2^{3} × 3^{2} = 6^{5}

= 2 × 2 × 2 × 3 × 3

= 6 × 6 × 6 × 6 × 6

= 72 = 7776

∴ 2^{3} × 3^{2} ≠ 6^{5} ∴ It is false.

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

1 = 1

∴ 3^{0} = (1000)^{0} ∴ It is true

Question 4.

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

i) 108 x 192

= 2 × 2 × 3 × 3 × 3 2 × 2 × 2 × 2 × 2 × 2 × 3

= 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3 × 3

= 2^{8} × 3^{4}

It is the product of prime factor

ii) 270

270 = 2 × 3 × 3 × 3 × 5

= 2 × 3^{3} × 5

It is required prime factor.

iii) 729 × 64

729 = 3 × 3 × 3 × 3 × 3 × 3

64 = 2 × 2 × 2 × 2 × 2 × 2

∴ 729 × 64 = 3^{6} × 2^{6}

It is the required prime factor.

iv) 768

768 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3

= 2^{8} × 3^{1} or = 2^{8} × 3

It is a required prime factor.

Question 5.

Simplify :

i)

Solution:

ii)

Solution:

iii)

Solution: