Q. 14.6( 99 Votes )

Using laws of exp

Answer :

(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