# Exponents and Radicals : Exercise (2.3) - Solutions

1.         Write the following in radical form.

$\begin{array}{ll} \text{(a)}\quad (5)^{^{\tfrac{1}{2}}} & \text{(b)}\quad (-9)^{^{\tfrac{1}{3}}} \\\\ \text{(c)}\quad (2)^{^{-\frac{1}{2}}} & \text{(d)}\quad \left(-\displaystyle\frac{3}{4}\right)^{^{\tfrac{2}{5}}}\\\\ \text{(e)}\quad \left(\displaystyle\frac{2}{7}\right)^{^{\tfrac{5}{2}}} \end{array}$

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$\begin{array}{l} \text{(a)}\ \ {{\left( 5 \right)}^{{\tfrac{{1\,}}{{2\,}}}}}=\sqrt{5}\\\\ \text{(b)}\ \ {{\left( {-9} \right)}^{{\tfrac{1}{3}}}}=\sqrt[3]{{-9}}\\\\ \text{(c)}\ \ {{\left( 2 \right)}^{{-\tfrac{{1\,}}{{2\,}}}}}=\displaystyle \frac{1}{{{{{\left( 2 \right)}}^{{\tfrac{{1\,}}{{2\,}}}}}}}=\displaystyle \frac{1}{{\sqrt{2}}}\\\\ \text{(d)}\ \ {{\left( {-\displaystyle \frac{3}{4}} \right)}^{{\tfrac{2}{5}}}}=\sqrt[5]{{{{{\left( {-\displaystyle \frac{3}{4}} \right)}}^{2}}}}=\sqrt[5]{{\displaystyle \frac{9}{{16}}}}\\\\ \text{(e)}\ \ {{\left( {\displaystyle \frac{2}{7}} \right)}^{{\tfrac{5}{2}}}}=\sqrt{{{{{\left( {\displaystyle \frac{2}{7}} \right)}}^{5}}}}=\sqrt{{\displaystyle \frac{{32}}{{16807}}}}\\ \end{array}$

2.         Write the following in fractional exponent form.

$\begin{array}{ll} \text{(a)}\quad \sqrt[6]{c^{^{5}}} & \text{(b)}\quad \sqrt[3]{-2} \\\\ \text{(c)}\quad \sqrt[5]{a^{^{4}} \sqrt[3]{b^{^{5}}}} & \text{(d)}\quad \sqrt[4]{\left(\displaystyle\frac{3}{7}\right)^{^{3}}} \end{array}$

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$\begin{array}{l} (\text{a})\ \ \sqrt[6]{{{{c}^{5}}}}={{c}^{{\frac{5}{6}}}}\\\\ (\text{b})\ \ \sqrt[3]{{-2}}={{(-2)}^{{\frac{1}{3}}}}\\\\ (\text{c})\ \ \sqrt[5]{{{{a}^{4}}\sqrt[3]{{{{b}^{5}}}}}}=\sqrt[5]{{{{a}^{4}}{{b}^{{\frac{5}{3}}}}}}={{\left( {{{a}^{4}}{{b}^{{\frac{5}{3}}}}} \right)}^{{\frac{1}{5}}}}={{a}^{{\frac{4}{5}}}}{{b}^{{\frac{1}{3}}}}\\\\ (\text{d})\ \ \sqrt[4]{{{{{\left( {\displaystyle \frac{3}{7}} \right)}}^{3}}}}={{\left( {\displaystyle \frac{3}{7}} \right)}^{{\frac{3}{4}}}} \end{array}$

3.         Change the expression with the same radical and simplify the radicands.

$\begin{array}{ll} \text{(a)}\quad 6 \sqrt{2} \quad& \text{(b)}\quad 3 a \sqrt[3]{x} \\\\ \text{(c)}\quad 2 \sqrt[5]{2} \quad& \text{(d)}\quad \sqrt[4]{\displaystyle\frac{1}{2}}\\\\ \text{(e)}\quad 3 \sqrt{x^{^{3}}} \end{array}$

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$\begin{array}{l} \text{(a)}\quad 6 \sqrt{2}=\sqrt{6^{2} \cdot 2}=\sqrt{72}\\\\ \text{(b)}\quad 3 a \sqrt[3]{x}=\sqrt[3]{3^{3} \cdot a^{3} x}=\sqrt[3]{27 a^{3} x}\\\\ \text{(c)}\quad 2 \sqrt[5]{2}=\sqrt[5]{2^{5} \cdot 2}=\sqrt[5]{64}\\\\ \text{(d)}\quad 3 \sqrt[4]{\displaystyle\frac{1}{2}}=\sqrt[4]{3^{4} \cdot \displaystyle\frac{1}{2}}=\sqrt[4]{\displaystyle\frac{81}{2}}\\\\ \text{(e)}\quad 3 \sqrt{x^{3}}=\sqrt{3^{2} \cdot x^{3}}=\sqrt{9 x^{3}} \end{array}$

4.         Simplify.

$\begin{array}{ll} \text{(a)}\quad \sqrt{32} & \text{(b)}\quad \sqrt[5]{-32} \\\\ \text{(c)}\quad \sqrt[4]{\displaystyle\frac{81 x^{^{16}}}{16 y^{^{4}}}} & \text{(d)}\quad \sqrt[3]{\displaystyle\frac{81 x^{^{2}}}{4 y}} \\\\ \text{(e)}\ \displaystyle\frac{9^{^{\tfrac{1}{2}}}}{\sqrt[3]{27}}& \text{(f)}\quad \sqrt{\displaystyle\frac{2}{3}} \cdot \sqrt{\displaystyle\frac{75}{98}}\\\\ \text{(g)}\quad \sqrt[3]{\displaystyle\frac{-216}{8 \times 10^{^{3}}}} & \text{(h)}\quad \sqrt[n]{\displaystyle\frac{32}{2^{^{5+n}}}} \end{array}$

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$\begin{array}{l} \text{(a)}\quad \sqrt{32}=\sqrt{16 \cdot 2}=4 \sqrt{2}\\\\ \text{(b)}\quad \sqrt[5]{-32}=\sqrt[5]{(-2)^{5}}=-2\\\\ \text{(c)}\quad \sqrt[4]{\displaystyle\frac{81 x^{16}}{16 y^{4}}}=\sqrt[4]{\displaystyle\frac{3^{4}\left(x^{4}\right)^{4}}{2^{4} \cdot y^{4}}}=\displaystyle\frac{3 x^{4}}{2 y}\\\\ \text{(d)}\quad \sqrt[3]{\displaystyle\frac{81 x^{2}}{4 y}}=\sqrt[3]{\displaystyle\frac{3^{3} \cdot 3 x^{2}}{4 y}}=3 \sqrt[3]{\displaystyle\frac{3 x^{2}}{4 y}}\\\\ \text{(e)}\quad \displaystyle\frac{9^{\displaystyle\frac{1}{2}}}{\sqrt[3]{27}}=\displaystyle\frac{\left(3^{2}\right)^{\frac{1}{2}}}{\sqrt[3]{3^{3}}}=\displaystyle\frac{3}{3}=1\\\\ \text{(f)}\quad \sqrt{\displaystyle\frac{2}{3}} \cdot \sqrt{\displaystyle\frac{75}{98}}=\sqrt{\displaystyle\frac{2}{3} \times \displaystyle\frac{75}{98}}=\sqrt{\displaystyle\frac{25}{49}}=\displaystyle\frac{5}{7}\\\\ \text{(g)}\quad \sqrt[3]{\displaystyle\frac{-216}{8 \times 10^{3}}}=\sqrt[3]{\displaystyle\frac{(-6)^{3}}{2^{3} \times 10^{3}}}=\displaystyle\frac{-6}{2 \times 10}=-\displaystyle\frac{3}{10}\\\\ \text{(h)}\quad \sqrt[4]{\displaystyle\frac{32}{2^{5+n}}}=\sqrt[n]{\displaystyle\frac{2^{5}}{2^{5+n}}}=\sqrt[n]{\displaystyle\frac{1}{2^{n}}}=\displaystyle\frac{1}{2} \end{array}$

5.         Rationalize the denominators.

$\begin{array}{ll} \text{(a)}\quad \displaystyle\frac{4 \sqrt{35}}{3 \sqrt{7}}\quad & \text{(b)}\quad \displaystyle\frac{20}{\sqrt{5}} \\\\ \text{(c)}\quad \displaystyle\frac{18}{\sqrt[3]{2}}\quad& \text{(d)}\quad \displaystyle\frac{\sqrt[3]{32}}{\sqrt[4]{27}} \\\\ \text{(e)}\quad \displaystyle\frac{\sqrt[3]{36 a^{^{2}}}}{\sqrt[3]{9 a}}\quad & \text{(f)}\quad \displaystyle\frac{\sqrt[3]{2}}{\sqrt[6]{12}} \\\\ \text{(g)}\quad \displaystyle\frac{1}{\sqrt[3]{x y^{^{2}}}}\quad & \text{(h)}\quad \sqrt[m]{{\displaystyle\frac{{2{{x}^{2}}{{y}^{{3m}}}}}{{9{{x}^{5}}{{y}^{{4m-1}}}}}}} \end{array}$

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$\begin{array}{l}\text{(a)}\ \ \ \ \displaystyle \frac{{4\sqrt{{35}}}}{{3\sqrt{7}}}\\\\\ \ \ \ \ =\displaystyle \frac{4}{3}\sqrt{{\displaystyle \frac{{35}}{7}}}\\\\\ \ \ \ \ =\displaystyle \frac{4}{3}\sqrt{{\displaystyle \frac{{7\times 5}}{7}}}\\\\\ \ \ \ \ =\displaystyle \frac{{4\sqrt{5}}}{3}\\\\\\\text{(b)}\ \ \ \ \displaystyle \frac{{20}}{{\sqrt{5}}}\\\\\ \ \ \ \ =\displaystyle \frac{{20}}{{\sqrt{5}}}\times \displaystyle \frac{{\sqrt{5}}}{{\sqrt{5}}}\ \\\\\ \ \ \ \ =\displaystyle \frac{{20\sqrt{5}}}{5}\\\\\ \ \ \ \ =4\sqrt{5}\\\\\\\text{(c)}\ \ \ \ \displaystyle \frac{{18}}{{\sqrt[3]{2}}}\\\\\ \ \ \ \ =\displaystyle \frac{{18}}{{\sqrt[3]{2}}}\times \displaystyle \frac{{\sqrt[3]{{{{2}^{2}}}}}}{{\sqrt[3]{{{{2}^{2}}}}}}\\\\\ \ \ \ \ =\displaystyle \frac{{18\sqrt[3]{4}}}{{\sqrt[3]{{{{2}^{3}}}}}}\ \\\\\ \ \ \ \ =\displaystyle \frac{{18\sqrt[3]{4}}}{2}\\\ \ \ \ \ =9\sqrt[3]{4}\\\\\text{(d)}\ \ \ \ \displaystyle \frac{{\sqrt[3]{{32}}}}{{\sqrt[4]{{27}}}}\\\\\ \ \ \ \ =\displaystyle \frac{{\sqrt[3]{{{{2}^{3}}\cdot 4}}}}{{\sqrt[4]{{{{3}^{3}}}}}}\times \displaystyle \frac{{\sqrt[4]{3}}}{{\sqrt[4]{3}}}\\\\\ \ \ \ \ =\displaystyle \frac{{2\sqrt[3]{4}\sqrt[4]{3}}}{3}\\\\\\\text{(e)}\ \ \ \ \displaystyle \frac{{\sqrt[3]{{36{{a}^{2}}}}}}{{\sqrt[3]{{9a}}}}\\\\\ \ \ \ \ =\sqrt[3]{{\displaystyle \frac{{36{{a}^{2}}}}{{9a}}}}\\\\\ \ \ \ \ =\sqrt[3]{{4a}}\\\\\\\text{(f)}\ \ \ \ \ \displaystyle \frac{{\sqrt[3]{2}}}{{\sqrt[6]{{12}}}}\\\\\ \ \ \ \ =\displaystyle \frac{{\sqrt[3]{2}}}{{\sqrt[6]{{{{2}^{2}}}}\cdot \sqrt[6]{3}}}\\\\\ \ \ \ \ =\displaystyle \frac{{\sqrt[3]{2}}}{{\sqrt[3]{2}\cdot \sqrt[6]{3}}}\times \displaystyle \frac{{\sqrt[6]{{{{3}^{5}}}}}}{{\sqrt[6]{{{{3}^{5}}}}}}\\\\\ \ \ \ \ =\displaystyle \frac{{\sqrt[6]{{243}}}}{3}\\\\\\\text{(g)}\ \ \ \ \ \displaystyle \frac{1}{{\sqrt[3]{{x{{y}^{2}}}}}}\\\\\ \ \ \ \ =\displaystyle \frac{1}{{\sqrt[3]{{x{{y}^{2}}}}}}\times \displaystyle \frac{{\sqrt[3]{{{{x}^{2}}y}}}}{{\sqrt[3]{{{{x}^{2}}y}}}}\\\\\ \ \ \ \ =\displaystyle \frac{{\sqrt[3]{{{{x}^{2}}y}}}}{{\sqrt[3]{{{{x}^{3}}{{y}^{3}}}}}}\\\\\ \ \ \ \ =\displaystyle \frac{{\sqrt[3]{{{{x}^{2}}y}}}}{{xy}}\\\\\text{(h)}\ \ \ \ \ \sqrt[m]{{\displaystyle \frac{{2{{x}^{2}}{{y}^{{3m}}}}}{{9{{x}^{5}}{{y}^{{4m-1}}}}}}}\\\\\ \ \ \ \ =\sqrt[m]{{\displaystyle \frac{{2y}}{{{{3}^{2}}{{x}^{3}}{{y}^{m}}}}}}\\\\\ \ \ \ \ =\ \displaystyle \frac{{\sqrt[m]{{2y}}}}{{\sqrt[m]{{{{3}^{2}}{{x}^{3}}{{y}^{m}}}}}}\\\\\ \ \ \ \ =\displaystyle \frac{{\sqrt[m]{{2y}}}}{{y\sqrt[m]{{{{3}^{2}}{{x}^{3}}}}}}\times \displaystyle \frac{{\sqrt[m]{{{{3}^{{m-2}}}{{x}^{{m-3}}}}}}}{{\sqrt[m]{{{{3}^{{m-2}}}{{x}^{{m-3}}}}}}}\\\\\ \ \ \ \ =\displaystyle \frac{{\sqrt[m]{{2\cdot {{3}^{{m-2}}}{{x}^{{m-3}}}y}}}}{{3xy}}\end{array}$

6.         Reduce the order as far as possible.

$\begin{array}{ll} \text{(a)}\quad \sqrt[4]{25} \quad& \text{(b)}\quad \sqrt[6]{4} \\\\ \text{(c)}\quad \sqrt[6]{8} \quad& \text{(d)}\quad \sqrt[9]{8 y^{^{3}}} \\\\ \text{(e)}\quad \sqrt[6]{27^{^{3}}} \quad& \text{(f)}\quad \sqrt[8]{a^{2} b^{^{4}}} \\\\ \text{(g)}\quad \sqrt[12]{64 a^{^{2}} b^{^{6}}} & \text{(h)}\quad (72)^{^{\tfrac{3}{5}}} \\\\ \text{(i)}\quad \sqrt[3]{768} \end{array}$

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$\begin{array}{l} \text{(a)}\quad \sqrt[4]{25}=\sqrt[4]{5^{2}}=\sqrt{5}\\\\ \text{(b)}\quad \sqrt[6]{4}=\sqrt[6]{2^{2}}=\sqrt[3]{2}\\\\ \text{(c)}\quad \sqrt[6]{8}=\sqrt[6]{2^{3}}=\sqrt{2}\\\\ \text{(d)}\quad \sqrt[9]{8 y^{3}}=\sqrt[9]{2^{3} \cdot y^{3}}=\sqrt[3]{2 y}\\\\ \text{(e)}\quad \sqrt[6]{27^{3}}=\sqrt{27}=3 \sqrt{3}\\\\ \text{(f)}\quad \sqrt[8]{a^{2} b^{4}}=\sqrt[4]{a b^{2}}\\\\ \text{(g)}\quad \sqrt[12]{64 a^{2} b^{6}}=\sqrt[12]{8^{2} \cdot a^{2} b^{6}}=\sqrt[6]{8 a b^{3}}\\\\ \text{(h)}\quad (72)^{\frac{3}{5}}=\sqrt[5]{72^{3}}=\sqrt[5]{\left(3^{2} \cdot 2^{3}\right)^{3}}=\sqrt[5]{3^{6} \cdot 2^{9}}=6 \sqrt[5]{3 \cdot 16}=6 \sqrt[5]{48}\\\\ \text{(i)}\quad \sqrt[3]{768}=\sqrt[3]{4^{3} \cdot 12}=4 \sqrt[3]{12} \end{array}$

7.         Find the simplified forms.

$\begin{array}{ll} \text{(a)}\quad \sqrt{\displaystyle\frac{9}{50}} \quad& \text{(b)}\quad \sqrt[3]{\displaystyle\frac{-192}{49}} \\\\ \text{(c)}\quad \sqrt[4]{16}\quad& \text{(d)}\quad 2 \sqrt[3]{56} \end{array}$

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$\begin{array}{l} \text{(a)}\quad\sqrt{\displaystyle \frac{9}{50}}=\sqrt{\displaystyle \frac{3^{2}}{5^{2} \cdot 2}}=\displaystyle \frac{3}{5 \sqrt{2}} \times \displaystyle \frac{\sqrt{2}}{\sqrt{2}}=\displaystyle \frac{3 \sqrt{2}}{10}\\\\ \text{(b)}\quad\sqrt[3]{\displaystyle \frac{-192}{49}}=\sqrt[3]{\displaystyle \frac{(-4)^{3} \cdot 3}{7^{2}} \times \displaystyle \frac{7}{7}}=-\displaystyle \frac{4 \sqrt[3]{21}}{7}\\\\ \text{(c)}\quad\sqrt[4]{16}=\sqrt[4]{2^{4}}=2\\\\ \text{(d)}\quad 2 \sqrt[3]{56}=2 \sqrt[3]{2^{3} \cdot 7}=4 \sqrt[3]{7} \end{array}$