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

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

  
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