# Exercise 1.6 Number Systems NCERT Solutions Class 9

Go back to  'Number Systems'

## Chapter 1 Ex.1.6 Question 1

Find:

(i) \begin{align}{64^{\frac{1}{2}}}\end{align}

(ii) \begin{align}{32^{\frac{1}{5}}}\end{align}

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

### Solution

Steps:

(i) \begin{align}{64^{\frac{1}{2}}}\end{align}

\begin{align} {64}^{\frac{1}{2}}&={{({{8}^{2}})}^{\frac{1}{2}}}\\&={{8}^{2\times \frac{1}{2}}}\quad \text{Using (}{{a}^{p}}{{)}^{q}}= {{a}^{pq}}\!\!~\!\! \\&={{8}^{1}}\\&=8\end{align}

Where \begin{align} a > 0, p\, \text{and} \,q \end{align} are rational numbers.

(ii) \begin{align}{32^{\frac{1}{5}}}\end{align}

\begin{align}{32^{\frac{1}{5}}}&={\left(2^{5}\right)^{\frac{1}{5}}} \quad {\text { Using }\left({a}^{{p}}\right)^{{q}}={a}^{{pq}}} \\ &={(2)^{5 \times \frac{1}{5}}} \\ &={2^{1}} \\ &={2}\end{align}

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

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

## Chapter 1 Ex.1.6 Question 2

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}

### 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}

## Chapter 1 Ex.1.6 Question 3

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}

### 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}

Related Sections
Related Sections
Instant doubt clearing with Cuemath Advanced Math Program