# EXPONENTS AND RADICALS : EXERCISE (2.2) SOLUTIONS

1.        Evaluate the following.

(a)     $(125)^{\frac{2}{3}}$
(b)     $(81)^{-\frac{3}{2}}$
(c)     $(-27)^{\frac{2}{3}}$
(d)     $\left(\displaystyle\frac{16}{81}\right)^{-\frac{3}{4}}$
(e)     $\left(\displaystyle\frac{-125}{8} \div \frac{1}{64}\right)^{\frac{1}{3}}$
(f)     $(0.125)^{-\frac{2}{3}}$
(g)     $\left(\displaystyle\frac{64}{27}\right)^{-\frac{2}{3}}$
(h)     $(-4)^{-1}+(-1)^{-4}$

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$\displaystyle \begin{array}{l} \text{(a)}\ \ {{(125)}^{{\frac{2}{3}}}}={{({{5}^{3}})}^{{\frac{2}{3}}}}={{5}^{2}}=25\\\\ \text{(b)}\ \ {{(81)}^{{-\frac{3}{2}}}}={{({{9}^{2}})}^{{-\frac{3}{2}}}}={{9}^{{-3}}}=\displaystyle\frac{1}{{{{9}^{3}}}}=\displaystyle\frac{1}{{729}}\\\\ \text{(c)}\ \ {{(-27)}^{{\frac{2}{3}}}}={{\left( {{{{(-3)}}^{3}}} \right)}^{{\frac{2}{3}}}}={{(-3)}^{2}}=9\\\\ \text{(d)}\ \ {{\left( {\displaystyle\frac{{16}}{{81}}} \right)}^{{-\frac{3}{4}}}}={{\left( {{{{\left( {\frac{2}{3}} \right)}}^{4}}} \right)}^{{-\frac{3}{4}}}}={{\displaystyle\left( {\frac{2}{3}} \right)}^{{-3}}}={{\displaystyle\left( {\frac{3}{2}} \right)}^{3}}=\displaystyle\frac{{27}}{8}\\\\ \text{(e)}\ \ \ {{\left( {\displaystyle\frac{{-125}}{8}\div \frac{1}{{64}}} \right)}^{{\frac{1}{3}}}}\\\ \ \ ={{\left( {\displaystyle\frac{{-125}}{8}\times 64} \right)}^{{\frac{1}{3}}}}\\\ \ \ ={{\left( {-125\times 8} \right)}^{{\frac{1}{3}}}}\\\ \ \ ={{\left( {{{{\left( {-5} \right)}}^{3}}\times {{2}^{2}}} \right)}^{{\frac{1}{3}}}}\\\ \ \ =-5\times 2\\\ \ \ =-10\\ \text{(f)}\ \ {{(0.125)}^{{-\frac{2}{3}}}}={{\left( {{{{0.5}}^{3}}} \right)}^{{-\frac{2}{3}}}}={{\left( {0.5} \right)}^{{-2}}}={{\left( {\displaystyle\frac{1}{2}} \right)}^{{-2}}}=4\\\\ \text{(g)}\ \ {{\left( {\displaystyle\frac{{64}}{{27}}} \right)}^{{-\displaystyle\frac{2}{3}}}}={{\left( {\displaystyle\frac{{{{8}^{3}}}}{{{{3}^{3}}}}} \right)}^{{-\frac{2}{3}}}}={{\left( {\displaystyle \frac{8}{3}} \right)}^{{-2}}}={{\left( {\displaystyle \frac{3}{8}} \right)}^{2}}=\frac{9}{{64}}\\\\ \text{(h)}\ \ {{(-4)}^{{-1}}}+{{(-1)}^{{-4}}}=\displaystyle \frac{1}{{(-4)}}+\displaystyle \frac{1}{{{{{(-1)}}^{4}}}}=-\displaystyle \frac{1}{4}+1=\displaystyle \frac{3}{4} \end{array}$

2.        Simplify the following.

(a)     $\sqrt[3]{4^{2}} \cdot 4^{\frac{2}{3}} \cdot\left(\displaystyle\frac{1}{4}\right)^{-\frac{2}{3}}$
(b)     $\sqrt{\displaystyle\frac{512 \times 27^{-3} \times 81 \times 3^{8}}{3^{4}}}$
(c)     $\left(\left(\displaystyle \frac{3}{4}\right)^{-4}\right)^{-0.5} \cdot \sqrt{\left(\displaystyle \frac{4}{3}\right)^{-1}} \div 16^{-0.5}$
(d)     $(27)^{\frac{1}{4}}+\displaystyle \frac{24}{(8)^{-\frac{2}{3}}}+\frac{\sqrt[5]{2}}{(4)^{-\frac{2}{5}}}$
(e)     $\displaystyle \frac{(243)^{\frac{4}{5}}+(64)^{\frac{2}{3}}-(216)^{\frac{1}{3}}}{(225)^{\frac{1}{2}}-(16)^{\frac{3}{4}}}$

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$\displaystyle \begin{array}{l} \text{(a)}\ \ \ \ \sqrt[3]{{{{4}^{2}}}}\cdot {{4}^{{\frac{2}{3}}}}\cdot {{\left( {\frac{1}{4}} \right)}^{{-\frac{2}{3}}}}\\ \ \ \ \ =\ {{4}^{{\frac{2}{3}}}}\cdot {{4}^{{\frac{2}{3}}}}\cdot {{4}^{{\frac{2}{3}}}}\\ \ \ \ \ =\ {{4}^{{\frac{6}{3}}}}\\ \ \ \ \ =\ {{4}^{2}}\\ \ \ \ \ =\ 16\\\\ \text{(b)}\ \ \ \ \sqrt{{\displaystyle \frac{{512\times {{{27}}^{{-3}}}\times 81\times {{3}^{8}}}}{{{{3}^{4}}}}}}\\ \ \ \ \ =\ \sqrt{{\displaystyle \frac{{{{2}^{9}}\times {{3}^{{-9}}}\times {{3}^{4}}\times {{3}^{8}}}}{{{{3}^{4}}}}}}\\ \ \ \ \ =\ \sqrt{{\displaystyle \frac{{{{2}^{9}}}}{3}}}\\ \ \ \ \ =\ 16\sqrt{{\displaystyle \frac{2}{3}}}\\ \ \ \ \ =\ \displaystyle\frac{{16\sqrt{6}}}{3}\\\\ \text{(c)}\ \ \ \ {{\left( {{{{\left( {\displaystyle \frac{3}{4}} \right)}}^{{-4}}}} \right)}^{{-0.5}}}\cdot \sqrt{{{{{\left( {\frac{4}{3}} \right)}}^{{-1}}}}}\div {{16}^{{-0.5}}}\\ \ \ \ \ =\ \ {{\left( {\displaystyle \frac{3}{4}} \right)}^{2}}\cdot \sqrt{{\frac{3}{4}}}\cdot {{16}^{{0.5}}}\\ \ \ \ \ =\ \ \displaystyle \frac{{{{3}^{2}}}}{{{{4}^{2}}}}\cdot \frac{{{{3}^{{\frac{1}{2}}}}}}{{{{4}^{{\frac{1}{2}}}}}}\cdot 4\\ \ \ \ \ =\ \ \displaystyle \frac{{9\sqrt{3}}}{8}\\\\ \text{(d)}\ \ \ \ {{(27)}^{{\frac{1}{4}}}}+\frac{{24}}{{{{{(8)}}^{{-\frac{2}{3}}}}}}+\frac{{\sqrt[5]{2}}}{{{{{(4)}}^{{-\frac{2}{5}}}}}}\\ \ \ \ \ =\ \ \ {{(27)}^{{\frac{1}{4}}}}+\frac{{{{2}^{3}}\times 3}}{{{{{({{2}^{3}})}}^{{-\frac{2}{3}}}}}}+\frac{{{{2}^{{\frac{1}{5}}}}}}{{{{{({{2}^{2}})}}^{{-\frac{2}{5}}}}}}\\ \ \ \ \ =\ \ \ {{(27)}^{{\frac{1}{4}}}}+\frac{{{{2}^{3}}\times 3}}{{{{2}^{{-2}}}}}+\frac{{{{2}^{{\frac{1}{5}}}}}}{{{{2}^{{-\frac{4}{5}}}}}}\\ \ \ \ \ =\ \ {{(27)}^{{\frac{1}{4}}}}+({{2}^{5}}\times 3)+{{2}^{{\frac{5}{5}}}}\\ \ \ \ \ =\ \ \ \sqrt[4]{{27}}+98\\\\ \text{(e)}\ \ \ \ \displaystyle \frac{{{{{(243)}}^{{\frac{4}{5}}}}+{{{(64)}}^{{\frac{2}{3}}}}-{{{(216)}}^{{\frac{1}{3}}}}}}{{{{{(225)}}^{{\frac{1}{2}}}}-{{{(16)}}^{{\frac{3}{4}}}}}}\\ \ \ \ \ =\ \ \ \displaystyle \frac{{{{{({{3}^{5}})}}^{{\frac{4}{5}}}}+{{{({{4}^{3}})}}^{{\frac{2}{3}}}}-{{{({{6}^{3}})}}^{{\frac{1}{3}}}}}}{{{{{({{{15}}^{2}})}}^{{\frac{1}{2}}}}-{{{({{2}^{4}})}}^{{\frac{3}{4}}}}}}\\ \ \ \ \ =\ \ \ \displaystyle \frac{{{{3}^{4}}+{{4}^{2}}-6}}{{15-{{2}^{3}}}}\\ \ \ \ \ =\ \ \ \displaystyle \frac{{81+16-6}}{{15-8}}\\ \ \ \ \ =\ \ \ \displaystyle \frac{{91}}{7}\\ \ \ \ \ =\ \ \ 13\end{array}$

3.        Simplify the following.

(a)     $\displaystyle \frac{x-5 \sqrt{x}}{x-2 \sqrt{x}-15} \div\left(1+\frac{3}{\sqrt{x}}\right)^{-1}$
(b)     $\sqrt[a]{\displaystyle \frac{\sqrt[b]{x}}{\sqrt[b]{x}}} \cdot \sqrt[b]{\displaystyle \frac{\sqrt[a]{x}}{\sqrt[a]{x}}} \cdot \sqrt[c]{\frac{\sqrt[a]{x}}{\sqrt[b]{x}}}$
(c)     $\left[\displaystyle \frac{x^{m}-y^{m}}{x^{\frac{m}{2}}-y^{\frac{m}{2}}}-\displaystyle \frac{x^{m}-y^{m}}{x^{\frac{m}{2}}+y^{\frac{m}{2}}}\right]^{-2}$
(d)     $\left(\displaystyle \frac{a^{\frac{3}{2}}}{b^{-\frac{1}{2}}}\right)^{4}\left(\displaystyle \frac{a^{-2}}{b^{3}}\right)$

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$\begin{array}{l} \text{(a)}\ \ \ \ \displaystyle \frac{{x-5\sqrt{x}}}{{x-2\sqrt{x}-15}}\div {{\left( {1+\displaystyle\frac{3}{{\sqrt{x}}}} \right)}^{{-1}}}\\ \ \ \ \ =\ \ \displaystyle \frac{{{{{\left( {\sqrt{x}} \right)}}^{2}}-5\sqrt{x}}}{{{{{\left( {\sqrt{x}} \right)}}^{2}}-2\sqrt{x}-15}}\div {{\left( {\displaystyle\frac{{\sqrt{x}+3}}{{\sqrt{x}}}} \right)}^{{-1}}}\\ \ \ \ \ =\ \ \displaystyle \frac{{\sqrt{x}\left( {\sqrt{x}-5} \right)}}{{\left( {\sqrt{x}-5} \right)\left( {\sqrt{x}+3} \right)}}\times \displaystyle \frac{{\sqrt{x}+3}}{{\sqrt{x}}}\\ \ \ \ \ =\ 1\\\\ \text{(b)}\ \ \ \ \sqrt[a]{{\displaystyle \frac{{\sqrt[b]{x}}}{{\sqrt[c]{x}}}}}\cdot \sqrt[b]{{\displaystyle \frac{{\sqrt[c]{x}}}{{\sqrt[a]{x}}}}}\cdot \sqrt[c]{{\displaystyle\frac{{\sqrt[a]{x}}}{{\sqrt[b]{x}}}}}\\ \ \ \ \ =\ \displaystyle \frac{{\sqrt[{ab}]{x}}}{{\sqrt[{ac}]{x}}}\cdot \displaystyle\frac{{\sqrt[{bc}]{x}}}{{\sqrt[{ab}]{x}}}\cdot \displaystyle\frac{{\sqrt[{ac}]{x}}}{{\sqrt[{bc}]{x}}}\\ \ \ \ \ =\ 1\\\\ \text{(c)}\ \ \ \ {{\left[ {\displaystyle\frac{{{{x}^{m}}-{{y}^{m}}}}{{{{x}^{{\frac{m}{2}}}}-{{y}^{{\frac{m}{2}}}}}}-\frac{{{{x}^{m}}-{{y}^{m}}}}{{{{x}^{{\frac{m}{2}}}}+{{y}^{{\frac{m}{2}}}}}}} \right]}^{{-2}}}\\ \ \ \ \ =\ {{\left[ {\displaystyle\frac{{{{{\left( {{{x}^{{\frac{m}{2}}}}} \right)}}^{2}}-{{{\left( {{{y}^{{\frac{m}{2}}}}} \right)}}^{2}}}}{{{{x}^{{\frac{m}{2}}}}-{{y}^{{\frac{m}{2}}}}}}-\displaystyle\frac{{{{{\left( {{{x}^{{\frac{m}{2}}}}} \right)}}^{2}}-{{{\left( {{{y}^{{\frac{m}{2}}}}} \right)}}^{2}}}}{{{{x}^{{\frac{m}{2}}}}+{{y}^{{\frac{m}{2}}}}}}} \right]}^{{-2}}}\\ \ \ \ \ =\ {{\left[ {\displaystyle\frac{{\left( {{{x}^{{\frac{m}{2}}}}-{{y}^{{\frac{m}{2}}}}} \right)\left( {{{x}^{{\frac{m}{2}}}}+{{y}^{{\frac{m}{2}}}}} \right)}}{{{{x}^{{\frac{m}{2}}}}-{{y}^{{\frac{m}{2}}}}}}-\displaystyle\frac{{\left( {{{x}^{{\frac{m}{2}}}}-{{y}^{{\frac{m}{2}}}}} \right)\left( {{{x}^{{\frac{m}{2}}}}+{{y}^{{\frac{m}{2}}}}} \right)}}{{{{x}^{{\frac{m}{2}}}}+{{y}^{{\frac{m}{2}}}}}}} \right]}^{{-2}}}\\ \ \ \ \ =\ {{\left[ {\left( {{{x}^{{\frac{m}{2}}}}+{{y}^{{\frac{m}{2}}}}} \right)-\left( {{{x}^{{\frac{m}{2}}}}-{{y}^{{\frac{m}{2}}}}} \right)} \right]}^{{-2}}}\\ \ \ \ \ =\ {{\left[ {{{x}^{{\frac{m}{2}}}}+{{y}^{{\frac{m}{2}}}}-{{x}^{{\frac{m}{2}}}}+{{y}^{{\frac{m}{2}}}}} \right]}^{{-2}}}\\ \ \ \ \ =\ {{\left[ {2{{y}^{{\frac{m}{2}}}}} \right]}^{{-2}}}\\ \ \ \ \ =\displaystyle\frac{1}{{4{{y}^{m}}}} \end{array}$