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

Solution:

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

$\left[a^{m} \cdot a^{n}=a^{m+n}\right]$

$=2^{\frac{10+3}{15}}=2^{\frac{13}{15}}$

(ii)$\left(\frac{1}{3^{3}}\right)^{7}=\frac{1}{3^{3 \times 7}}$

$\left[\left(a^{m}\right)^{n}=a^{m m}\right]$

$=\frac{1}{3^{21}}$

$=3^{-21}$

$\left[\frac{1}{a^{m}}=a^{-m}\right]$

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

$\left[\frac{a^{m}}{a^{n}}=a^{m-n}\right]$

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

(iv) $7^{\frac{1}{2}} \cdot 8^{\frac{1}{2}}=(7 \times 8)^{\frac{1}{2}}$

$\left[a^{m} \cdot b^{m}=(a b)^{m}\right]$

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

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