Find:

(i)$9^{\frac{3}{2}}$

(ii)$32^{\frac{2}{5}}$

(iii)$16^{\frac{3}{4}}$

(iv)$125^{\frac{-1}{3}}$
Solution:

(i) $9^{\frac{3}{2}}=\left(3^{2}\right)^{\frac{3}{2}}$

$=3^{2 \times \frac{3}{2}}$

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

$=3^{3}=27$

(ii)$(32)^{\frac{2}{5}}=\left(2^{5}\right)^{\frac{2}{5}}$

$=2^{5 \times \frac{2}{5}}$

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

$=2^{2}=4$

(iii) $(16)^{\frac{3}{4}}=\left(2^{4}\right)^{\frac{3}{4}}$

$=2^{4 \times \frac{3}{4}}$

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

$=2^{3}=8$

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

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

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

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

$=\frac{1}{5}$
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