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^{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})

(i) We know that, (a^{m} × a^{n} = a^{m + n})

Thus,

3^{2} × 3^{4} × 3^{8}

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

= 3^{14}

(ii) We have,

6^{15} 6^{10}

We know that,

(a^{m} a^{n} = a^{m - n})

Thus,

6^{15} 6^{10}

= (6)^{15 - 10}

= 6^{5}

(iii) We have,

a^{3} × a^{2}

We know that,

(a^{m} × a^{n} = a^{m + n})

Therefore,

a^{3} × a^{2}

= (a)^{3 + 2}

= a^{5}

(iv) We have,

7^{x} × 7^{2}

We know that,

(a^{m} × a^{n} = a^{m + n})

Thus,

7^{x} × 7^{2}

= (7)^{x + 2}

(v) We have,

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

Using identity:

(a^{m})^{n} = a^{m × n}

= 5^{2 × 3} 5^{3}

= 5^{6} 5^{3}

We know that,

(a^{m} a^{n} = a^{m - n})

Thus,

5^{6} 5^{3}

= (5)^{6 - 3}

= 5^{3}

(vi) We have,

2^{5} × 5^{5}

We know that,

[a^{m} × b^{m} = (a × b)^{m}]

Thus,

2^{5} × 5^{5}

= (2 × 5)^{5 + 5}

= 10^{5}

(vii) We have,

a^{4} × b^{4}

We know that,

[a^{m} × b^{m} = (a × b)^{m}]

Thus,

a^{4} × b^{4}

= (a × b)^{4}

(viii) We have,

(3^{4})^{3}

We know that,

(a^{m})^{n} = a^{mn})

Thus,

(3^{4})^{3}

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

= 3^{12}

(ix) We have,

(2^{20} 2^{15}) × 2^{3}

We know that,

(a^{m} a^{n} = a^{m - n})

Thus,

(2^{20 - 15}) × 2^{3}

= (2)^{5} × 2^{3}

We know that,

(a^{m} × a^{n} = a^{m + n})

Thus,

(2)^{5} × 2^{3}

= (2^{5 + 3})

= 2^{8}

(x) We have,

(8^{t} 8^{2})

We know that,

(a^{m} a^{n} = a^{m - n})

Thus,

(8^{t} 8^{2})

= (8^{t – 2})

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