# Ex.1.6 Q3 Number System Solution - NCERT Maths Class 9

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## Question

Simplify:

(i) \begin{align}{2^{\frac{2}{3}}}.\;{2^{\frac{1}{5}}}\end{align}

(ii) \begin{align}{\left( {\frac{1}{{{3^3}}}} \right)^7} \end{align}

(iii) \begin{align}\frac{{{{11}^{\frac{1}{2}}}}}{{{{11}^{\frac{1}{4}}}}} \end{align}

(iv) \begin{align} 7^{\frac{1}{2}}\,.\,8^{\frac{1}{2}}\end{align}

Video Solution
Number Systems
Ex 1.6 | Question 3

## Text Solution

Steps:

(i) \begin{align}{2^{\frac{2}{3}}}\;.\;{2^{\frac{1}{5}}} \end{align}

$$\because {{a}^{p}}.\ {{a}^{q}}={{a}^{p+q}}$$ (For $$a>0,$$  $$p$$  and $$q$$  are rational numbers.)

\begin{align} & ={{2}^{\frac{2}{3}+\frac{1}{5}}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\\ & ={{2}^{\frac{10+3}{15}}} \\ & ={{2}^{\frac{13}{15}}}\end{align}

(ii)

\begin{align}{{\left( \frac{1}{{{3}^{3}}} \right)}^{7}}&=\frac{{{1}^{7}}}{{{\left( {{3}^{3}} \right)}^{7}}} \\ & =\frac{1}{{{3}^{21}}} \\ & ={{3}^{-21}}\end{align}

(iii) \begin{align}\frac{{{{11}^{\frac{1}{2}}}}}{{{{11}^{\frac{1}{4}}}}} \end{align}

\begin{align} \frac{{{11}^{\frac{1}{2}}}}{{{11}^{\frac{1}{4}}}}&={{11}^{\frac{1}{2}\,-\,\frac{1}{4}}} \\ &={{11}^{\frac{2-1}{4}}} \\ &={{11}^{\frac{1}{4}}} \\ &=\sqrt[4]{11} \\ \end{align}

(iv) \begin{align} 7^{\frac{1}{2}}\,.\,8^{\frac{1}{2}}\end{align}

\begin{align} & ={{(7\times 8)}^{\frac{1}{2}}} \\ & ={{\left( 56 \right)}^{\frac{1}{2}}} \\ \end{align}

Video Solution
Number Systems
Ex 1.6 | Question 3

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