# Using laws of exp

(i) We know that, (am × an = am + n)

Thus,

32 × 34 × 38

= (3)2 + 4 + 8

= 314

(ii) We have,

615 610

We know that,

(am an = am - n)

Thus,

615 610

= (6)15 - 10

= 65

(iii) We have,

a3 × a2

We know that,

(am × an = am + n)

Therefore,

a3 × a2

= (a)3 + 2

= a5

(iv) We have,

7x × 72

We know that,

(am × an = am + n)

Thus,

7x × 72

= (7)x + 2

(v) We have,

(52)3 53

Using identity:

(am)n = am × n

= 52 × 3 53

= 56 53

We know that,

(am an = am - n)

Thus,

56 53

= (5)6 - 3

= 53

(vi) We have,

25 × 55

We know that,

[am × bm = (a × b)m]

Thus,

25 × 55

= (2 × 5)5 + 5

= 105

(vii) We have,

a4 × b4

We know that,

[am × bm = (a × b)m]

Thus,

a4 × b4

= (a × b)4

(viii) We have,

(34)3

We know that,

(am)n = amn)

Thus,

(34)3

= (34)3

= 312

(ix) We have,

(220 215) × 23

We know that,

(am an = am - n)

Thus,

(220 - 15) × 23

= (2)5 × 23

We know that,

(am × an = am + n)

Thus,

(2)5 × 23

= (25 + 3)

= 28

(x) We have,

(8t 82)

We know that,

(am an = am - n)

Thus,

(8t 82)

= (8t – 2)

Rate this question :

How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.
RELATED QUESTIONS :

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