-Shresthi Goyal, Subject Matter Expert at Edumarz
(i) 2⅔. 2⅕
[Since, a^p . a^q = a^(p+q)]
=(2)^(⅔+⅕)
=(2)^(13/15)
(ii) [1/(3³)]⁷
[Since, (a^p)^q = a^(pq)]
[3^(-3)]⁷ = (3)^(-21)
= 1/(3)^21
(iii) (11)½/ (11)¼
= [Since a^p / a^q = a^(p-q)]
(11)(½-¼) = (11)¼
(iv) 7½. 8½
[Since, a^p . b^p = (ab)^p]
= (7×8)½ = (56)½