Question#1: Use the law of exponents to simplify.
\((i)\)
\[\frac{{{\left( 243 \right)}^{-\frac{2}{3}}}{{\left( 32 \right)}^{\frac{1}{5}}}}{\sqrt{{{\left( 196 \right)}^{-1}}}}\]
Solution:
\[\begin{aligned}&=\frac{{{\left( 3^5 \right)}^{-\frac{2}{3}}}{{\left( 2^5 \right)}^{-\frac{1}{5}}}}{\sqrt{{{\left( 14^2 \right)}^{-1}}}}\\
&=\frac{{{\left( 3 \right)}^{-\frac{5\times 2}{3}}}\times{{\left( 2^\cancel{5} \right)}^{-\frac{1}{\cancel{5}}}}}{\sqrt{{{\left( 14^{-1} \right)}^{2}}}}\\
&=\frac{{{\left( 3 \right)}^{-\frac{10}{3}}}\times{{\left( 2 \right)}^{-1}}}{\sqrt{{{\left( 14^{-1} \right)}^{\cancel{2}}}}}\\
&=\frac{{{\left( 3 \right)}^{-\frac{10}{3}}}\times{{ 2}^{-1}}}{{{ 14^{-1}}}}\\
&=\frac{14}{{{3}^{\frac{10}{3}}}\times 2}\\
&=\frac{\cancelto{7}{14}}{3^{\frac{9}{3}}\times {{3}^{\frac{1}{3}}}\times\cancel{2} }\\
&=\frac{7}{3^{\frac{\cancelto{3}{9}}{\cancel{3}}}\times {{3}^{\frac{1}{3}}}}\\
&=\frac{7}{{{3}^{3}}\times \sqrt[3]{3}}\\
&=\frac{7}{27 \sqrt[3]{3}}\end{aligned}\]
\((ii)\)
\[\left( 2{{x}^{5}}{{y}^{-4}} \right)\left( -8{{x}^{-3}}{{y}^{2}} \right)\]
Solution:
\[\begin{aligned}&=\left( 2{{x}^{5}}{{y}^{-4}} \right)\left( -8{{x}^{-3}}{{y}^{2}} \right) \\
&=2\times (-8){{x}^{5-3}}{{y}^{-4+2}} \\
&=-16{{x}^{2}}{{y}^{-2}} \\
&=\frac{-16{{x}^{2}}}{{{y}^{-2}}}\end{aligned}\]
\((iii)\)
\[{{\left( \frac{{{x}^{-2}}{{y}^{-1}}{{z}^{-4}}}{{{x}^{4}}{{y}^{-3}}{{z}^{0}}} \right)}^{-3}}\]
Solution:
\[\begin{aligned}&={{\left( \frac{{{x}^{-2}}{{y}^{-1}}{{z}^{-4}}}{{{x}^{4}}{{y}^{-3}}{{z}^{0}}} \right)}^{-3}}\\
&={{\left( {{x}^{-2-4}}{{y}^{-1+3}}{{z}^{-4-0}} \right)}^{-3}}\\
&={{\left( {{x}^{-6}}{{y}^{2}}{{z}^{-4}} \right)}^{-3}}\\
&={{x}^{-6\times -3}}{{y}^{2\times -3}}{{z}^{-4\times -3}}\\
&={{x}^{18}}{{y}^{-6}}{{z}^{12}}\\
&=\frac{{{x}^{18}}{{z}^{12}}}{{{y}^{-6}}}\end{aligned}\]
\((iv)\)
\[\frac{{{\left( 81 \right)}^{n}}{{.3}^{5}}-{{\left( 3 \right)}^{4n-1}}\left( 243 \right)}{\left( {{9}^{2n}} \right)\left( {{3}^{3}} \right)}\]
Solution:
\[\begin{align}
& =\frac{{{\left( 81 \right)}^{n}}{{.3}^{5}}-{{\left( 3 \right)}^{4n-1}}\left( 243 \right)}{\left( {{9}^{2n}} \right)\left( {{3}^{3}} \right)} \\
& =\frac{{{\left( {{3}^{4}} \right)}^{n}}{{.3}^{5}}-{{3}^{4n}}{{.3}^{-1}}{{.3}^{5}}}{{{\left( {{3}^{2}} \right)}^{2n}}\left( {{3}^{3}} \right)} \\
& =\frac{{{3}^{4n}}{{.3}^{5}}-{{3}^{4n}}{{.3}^{-1+5}}}{{{3}^{4n}}{{.3}^{3}}} \\
& =\frac{{{3}^{4n}}{{.3}^{4}}\left( 3-1 \right)}{{{3}^{4n}}{{.3}^{3}}} \\
& =\frac{\cancel{{{3}^{4n}}}{.\cancel{{3}^{3}}}.3.2}{\cancel{{{3}^{4n}}}{.\cancel{{3}^{3}}}} \\
& =3.2 \\
& =6 \\
\end{align}\]
Question#2: Show that
\[{{\left( \frac{{{x}^{a}}}{{{x}^{b}}} \right)}^{a+b}}\times {{\left( \frac{{{x}^{b}}}{{{x}^{c}}} \right)}^{b+c}}\times {{\left( \frac{{{x}^{c}}}{{{x}^{a}}} \right)}^{c+a}}=1\]
Solution:
\[\begin{align}\text{L}\text{.H}\text{.S}\text{.}&={{\left( \frac{{{x}^{a}}}{{{x}^{b}}} \right)}^{a+b}}\times {{\left( \frac{{{x}^{b}}}{{{x}^{c}}} \right)}^{b+c}}\times {{\left( \frac{{{x}^{c}}}{{{x}^{a}}} \right)}^{c+a}}\\
& ={{\left( {{x}^{a-b}} \right)}^{a+b}}\times {{\left( {{x}^{b-c}} \right)}^{b+c}}\times {{\left( {{x}^{c-a}} \right)}^{c+a}} \\
& ={{x}^{\left( a-b \right)}}^{\left( a+b \right)}\times {{x}^{\left( b-c \right)}}^{\left( b+c \right)}\times {{x}^{\left( c-a \right)}}^{\left( c+a \right)} \\
& ={{x}^{{{a}^{2}}-{{b}^{2}}}}\times {{x}^{{{b}^{2}}-{{c}^{2}}}}\times {{x}^{{{c}^{2}}-{{a}^{2}}}} \\
& ={{x}^{\cancel{{{a}^{2}}}-\cancel{{{b}^{2}}}+\cancel{{{b}^{2}}}-\cancel{{{c}^{2}}}+\cancel{{{c}^{2}}}-\cancel{{{a}^{2}}}}} \\
& ={{x}^{0}} \\
& =1 \\
&=\text{R.H.S.}\\
\end{align}\]
Question#3: Simplify
\((i)\)
\[\frac{{{2}^{\frac{1}{3}}}\times {{\left( 27 \right)}^{\frac{1}{3}}}\times {{\left( 60 \right)}^{\frac{1}{2}}}}{{{\left( 180 \right)}^{\frac{1}{2}}}\times {{\left( 4 \right)}^{-\frac{1}{3}}}\times {{\left( 9 \right)}^{\frac{1}{4}}}}\]
Solution:
\[\begin{align}
& =\frac{{{2}^{\frac{1}{3}}}\times {{\left( 27 \right)}^{\frac{1}{3}}}\times {{\left( 60 \right)}^{\frac{1}{2}}}}{{{\left( 180 \right)}^{\frac{1}{2}}}\times {{\left( 4 \right)}^{-\frac{1}{3}}}\times {{\left( 9 \right)}^{\frac{1}{4}}}} \\
& =\frac{{{2}^{\frac{1}{3}}}\times {{\left( {{3}^{\cancel{3}}} \right)}^{\frac{1}{\cancel{3}}}}\times {{\left( {{2}^{2}}\times 3\times 5 \right)}^{\frac{1}{2}}}}{{{\left( {{2}^{2}}\times {{3}^{2}}\times 5 \right)}^{\frac{1}{2}}}\times {{\left( {{2}^{2}} \right)}^{-\frac{1}{3}}}\times {{\left( {{3}^{\cancel{2}}} \right)}^{\frac{1}{\cancelto{2}{4}}}}} \\
& =\frac{{{2}^{\frac{1}{3}}}\times 3\times {{\left( {{2}^{\cancel{2}}} \right)}^{\frac{1}{\cancel{2}}}}\times \cancel{{{3}^{\frac{1}{2}}}}\times \cancel{{{5}^{\frac{1}{2}}}}}{{{\left( {{2}^{\cancel{2}}} \right)}^{\frac{1}{\cancel{2}}}}\times {{\left( {{3}^{\cancel{2}}} \right)}^{\frac{1}{\cancel{2}}}}\times \cancel{{{5}^{\frac{1}{2}}}}\times {{2}^{-\frac{2}{3}}}\times \cancel{{{3}^{\frac{1}{2}}}}} \\
& =\frac{{{2}^{\frac{1}{3}}}\times \cancel{3}\times \cancel{2}}{\cancel{2}\times \cancel{3}\times {{2}^{-\frac{2}{3}}}} \\
& =\frac{{{2}^{\frac{1}{3}}}}{{{2}^{-\frac{2}{3}}}} \\
& ={{2}^{\frac{1}{3}+\frac{2}{3}}} \\
& ={{2}^{\frac{1+2}{3}}} \\
& ={{2}^{\frac{\cancel{3}}{\cancel{3}}}} \\
& =2 \\
\end{align}\]
\((ii)\)
\[\sqrt{\frac{{{\left( 216 \right)}^{\frac{2}{3}}}\times {{\left( 25 \right)}^{\frac{1}{2}}}}{{{\left( 0.04 \right)}^{-\frac{1}{2}}}}}\]
Solution:
\[\begin{align}
& =\sqrt{\frac{{{\left( 216 \right)}^{\frac{2}{3}}}\times {{\left( 25 \right)}^{\frac{1}{2}}}}{{{\left( 0.04 \right)}^{-\frac{1}{2}}}}} \\
& =\sqrt{\frac{{{\left( {{6}^{\cancel{3}}} \right)}^{\frac{2}{\cancel{3}}}}\times {{\left( {{5}^{\cancel{2}}} \right)}^{\frac{1}{\cancel{2}}}}}{{{\left( \frac{4}{100} \right)}^{-\frac{1}{2}}}}} \\
& =\sqrt{\frac{{{6}^{2}}\times 5}{{{\left( \frac{\cancelto{25}{100}}{\cancel{4}} \right)}^{\frac{1}{2}}}}} \\
& =\sqrt{\frac{{{6}^{2}}\times 5}{{{\left( 25 \right)}^{\frac{1}{2}}}}} \\
& =\sqrt{\frac{{{6}^{2}}\times 5}{{{\left( {{5}^{\cancel{2}}} \right)}^{\frac{1}{\cancel{2}}}}}} \\
& =\sqrt{\frac{{{6}^{2}}\times \cancel{5}}{\cancel{5}}} \\
& =\sqrt{{{6}^{2}}} \\
& =6 \\
\end{align}\]
\((iii)\)
\[{{5}^{{{2}^{3}}}}\div {{\left( {{5}^{2}} \right)}^{3}}\]
Solution:
\[\begin{align}
& ={{5}^{{{2}^{3}}}}\div {{\left( {{5}^{2}} \right)}^{3}} \\
& ={{5}^{8}}\div {{5}^{6}} \\
& =\frac{{{5}^{8}}}{{{5}^{6}}} \\
& ={{5}^{8-6}} \\
& ={{5}^{2}} \\
& =25 \\
\end{align}\]
\((iv)\)
\[{{\left( {{x}^{3}} \right)}^{2}}\div {{x}^{{{3}^{2}}}},x\ne 0\]
Solution:
\[\begin{align}
& ={{\left( {{x}^{3}} \right)}^{2}}\div {{x}^{{{3}^{2}}}} \\
& ={{x}^{6}}\div {{x}^{9}} \\
& =\frac{{{x}^{6}}}{{{x}^{9}}} \\
& ={{x}^{6-9}} \\
& ={{x}^{-3}} \\
& =\frac{1}{{{x}^{3}}} \\
\end{align}\]
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