Solution

(i)

$ \begin{aligned} 64^{\frac{1}{2}} & =(2^{6})^{\frac{1}{2}} & =2^{2}=4 \\ & =2^{6 \times \frac{1}{2}} & {[(a^{m})^{n}=a^{m m !}] _{(16)^{\frac{3}{4}}=(2^{4})^{\frac{3}{4}}}^{(iii)} } \\ & =2^{3}=8 & \end{aligned} $

$ =2^{5 \times \frac{2}{3}} \quad[(a^{m})^{n \prime}=a^{m m n}] $

Solution

(ii)

$ \begin{aligned} 32^{\frac{1}{5}} & =(2^{5})^{\frac{1}{5}} \\ & =(2)^{5 \times \frac{1}{5}} \\ & =2^{1}=2 \end{aligned} $

$ \begin{aligned} & =2^{4 \times \frac{3}{4}} \\ & {[(a^{m})^{n}=a^{m m}]} \\ & {[(a^{m})^{n}=a^{m m}] _{(125)^{\frac{-1}{3}}}^{(iv)}=\frac{1}{(125)^{\frac{1}{3}}}} \\ & =2^{3}=8 \end{aligned} $

$(125)^{\frac{1}{3}}=(5^{3})^{\frac{1}{3}}$

$ \begin{matrix} =5^{3 \times \frac{1}{3}} & {[(a^{m})^{n}=a^{m m}]} & =\frac{1}{5^{3 \times \frac{1}{3}}} \\ =5^{1}=5 & =\frac{1}{5} \end{matrix} $

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

Solution

(i)

$ \begin{matrix} 2^{\frac{2}{3}} \cdot 2^{\frac{1}{5}} & =2^{\frac{2}{3}+\frac{1}{5}} & {[a^{m} \cdot a^{n}=a^{m+n}]} \\ & =2^{\frac{10+3}{15}}=2^{\frac{13}{15}} \end{matrix} $

(ii)

$ \begin{aligned} (\frac{1}{3^{3}})^{7} & =\frac{1}{3^{3 \times 7}} & & {[(a^{m})^{n}=a^{m m}] } \\ & =\frac{1}{3^{21}} & & {[\frac{1}{a^{m}}=a^{-m}] } \end{aligned} $

(iii)

$ \begin{matrix} \frac{11^{\frac{1}{2}}}{11^{\frac{1}{4}}} & =11^{\frac{1}{2}-\frac{1}{4}} & {[\frac{a^{m}}{a^{n}}=a^{m-n}]} \\ & =11^{\frac{2-1}{4}}=11^{\frac{1}{4}} & \end{matrix} $

(iv)

$ \begin{aligned} 7^{\frac{1}{2}} \cdot 8^{\frac{1}{2}} & =(7 \times 8)^{\frac{1}{2}} \quad[a^{m} \cdot b^{m}=(a b)^{m}] \\ & =(56)^{\frac{1}{2}} \end{aligned} $