Simplify:
(i) $ 2^{\frac{2}{3}} \cdot 2^{\frac{1}{5}} $
(ii) $ \left(\frac{1}{3^{3}}\right)^{7} $
(iii) $ \frac{11^{\frac{1}{2}}}{11^{\frac{1}{4}}} $
(iv) $ 7^{\frac{1}{2}} \cdot 8^{\frac{1}{2}} $

To do:

We have to simplify
(i) \( 2^{\frac{2}{3}} \cdot 2^{\frac{1}{5}} \)
(ii) \( \left(\frac{1}{3^{3}}\right)^{7} \)
(iii) \( \frac{11^{\frac{1}{2}}}{11^{\frac{1}{4}}} \)
(iv) \( 7^{\frac{1}{2}} \cdot 8^{\frac{1}{2}} \)
Solution: 

We know that,

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

$a^m \times a^n=a^{m+n}$

Therefore,

(i) $2^{\frac{2}{3}} \cdot 2^{\frac{1}{5}}=(2)^{\frac{2}{3}+\frac{1}{5}}$

$=(2)^{\frac{2\times5+1\times3}{15}}$              (LCM of 3 and 5 is 15)

$=(2)^{\frac{10+3}{15}}$

$=(2)^{\frac{13}{15}}$

Hence $2^{\frac{2}{3}} \cdot 2^{\frac{1}{5}}=(2)^{\frac{13}{15}}$

(ii) $(\frac{1}{3^{3}})^{7}=(3^{-3})^{7}$                   [Since $\frac{1}{a^m}=a^{-m}$]

$=(3)^{-3\times7}$

$=(3)^{-21}$

Hence $(\frac{1}{3^{3}})^{7}=(3)^{-21}$

(iii) $\frac{11^{\frac{1}{2}}}{11^{\frac{1}{4}}}=(11)^{\frac{1}{2}-\frac{1}{4}}$           [Since $\frac{a^m}{a^n}=a^{m-n}$]

$=(11)^{\frac{1\times2-1}{4}}$                (LCM of 2 and 4 is 4)

$=(11)^{\frac{2-1}{4}}$

$=(11)^{\frac{1}{4}}$

Hence $\frac{11^{\frac{1}{2}}}{11^{\frac{1}{4}}}=(11)^{\frac{1}{4}}$

(iv) $7^{\frac{1}{2}} \cdot 8^{\frac{1}{2}}=(7\times8)^{\frac{1}{2}}$          [Since $a^m \times b^m = (a\times b)^m$]

$=(56)^{\frac{1}{2}}$

Hence $7^{\frac{1}{2}} \cdot 8^{\frac{1}{2}}=(56)^{\frac{1}{2}}$.