# Ex.1.6 Q2 Number System Solution - NCERT Maths Class 9

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## Question

Find: \begin{align} \end{align}

(i) \begin{align} {9^{\frac{3}{2}}}\end{align}

(ii) \begin{align}{32^{\frac{2}{5}}}\end{align}

(iii) \begin{align} {16^{\frac{3}{4}}}\end{align}

(iv) \begin{align}{125^{\frac{{ - 1}}{3}}}\end{align}

Video Solution
Number Systems
Ex 1.6 | Question 2

## Text Solution

Steps:

(i)  \begin{align}{9^{\frac{3}{2}}} & \end{align}

\begin{align}{9^{\frac{3}{2}}}&={\left(3^{2}\right)^{\frac{3}{2}}} \quad {\text { using }\left({a}^{{p}}\right)^{{q}}={a}^{{pq}}} \\ &=(3)^{2 \times \frac{3}{2}}\\&=3^{3}\\&=27\end{align}

(ii) \begin{align}{32^{\frac{2}{5}}}\end{align}

\begin{align}{32^{\frac{2}{5}}}&={\left(2^{5}\right)^{\frac{2}{5}}} \quad {\text { using }\left(a^{p}\right)^{q}=a^{p q}} \\&=(2)^{5 \times \frac{2}{5}}\\&=2^{2}\\&=4\end{align}

(iii) \begin{align}{16^{\frac{3}{4}}}\,\end{align}

\begin{align}{16^{\frac{3}{4}}}&={\left(2^{4}\right)^{\frac{3}{4}}} \quad {\text {using }\left({a}^{{p}}\right)^{{q}}={a}^{{pq}}} \\&={(2)^{4 \times \frac{3}{4}}}\\& =2^{3}\\&=8\end{align}

(iv) \begin{align}{125^{\frac{{ - 1}}{3}}}\end{align}

\begin{align}{125^{\frac{{ - 1}}{3}}}&={\left(5^{3}\right)^{\frac{-1}{3}}} \\&={(5)^{3 \times \frac{-1}{3}}} \quad {\text { using }\left({a}^{{p}}\right)^{{q}}={a}^{{pq}}} \\&={5^{-1}} \\&={\frac{1}{5}}\end{align}

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