Answer :

**(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})

Rate this question :

Simplify and exprNCERT Mathematics

Using laws of expNCERT Mathematics

Express each of tNCERT Mathematics

Say true or falseNCERT Mathematics

Find m so that <sNCERT - Exemplar Mathematics

If <span lang="ENNCERT - Exemplar Mathematics

State whether theNCERT - Exemplar Mathematics

State whether theNCERT - Exemplar Mathematics

State whether theNCERT - Exemplar Mathematics

State whether theNCERT - Exemplar Mathematics