Find: <br/> <br/> (i)$9^{\frac{3}{2}}$<br/> <br/> (ii)$32^{\frac{2}{5}}$ <br/> <br/>(iii)$16^{\frac{3}{4}}$<br/> <br/> (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)$(125)^{\frac{-1}{3}}=\frac{1}{(125)^{\frac{1}{3}}}$
$\left[a^{-m}=\frac{1}{a^{m}}\right]$
$=\frac{1}{\left(5^{3}\right)^{\frac{1}{3}}}$
$\left[\left(a^{m}\right)^{n}=a^{m n}\right]$
$=\frac{1}{5}$
(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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