Chapter R - Section R.6 - Rational Exponents - R.6 Exercises - Page 56: 47

$\begin{array}{ c c } a) & \mathrm{Choice}\mathbf{\ E}\\ & \\ b) & \mathrm{Choice}\mathbf{\ G}\\ & \\ c) & \mathrm{Choice}\mathbf{\ F}\\ & \\ d) & \mathrm{Choice}\mathbf{\ F} \end{array}$

$\begin{array}{ c l l } a) & =\left(\left(\dfrac{4}{9}\right)^{\dfrac{1}{2}}\right)^{3} & \begin{array}{l} \mathrm{Apply\ the\ rule}\\ a^{nm} =\left( a^{n}\right)^{m} \end{array}\\ & & \\ & =\left(\sqrt{\dfrac{4}{9}}\right)^{3} & \begin{array}{l} \mathrm{Apply\ the\ rule}\\ a^{1/n} =\sqrt[n]{a} \end{array}\\ & & \\ & =\left(\dfrac{\sqrt{4}}{\sqrt{9}}\right)^{3} & \begin{array}{l} \mathrm{Apply\ the\ rule}\\ \sqrt{\dfrac{a}{b}} =\dfrac{\sqrt{a}}{\sqrt{b}} \end{array}\\ & & \\ & =\left(\dfrac{2}{3}\right)^{3} & \begin{array}{l} \mathrm{Simplify\ by}\\ \mathrm{evaluating\ the\ }\\ \mathrm{root} \end{array}\\ & & \\ & =\dfrac{2^{3}}{3^{3}} & \begin{array}{l} \mathrm{Apply\ the\ rule}\\ \left(\dfrac{a}{b}\right)^{n} =\dfrac{a^{n}}{b^{n}} \end{array}\\ & & \\ & =\dfrac{8}{27} & \begin{array}{l} \mathrm{Simplify} .\ \mathrm{The\ result\ matches}\\ \mathrm{Choice}\mathbf{\ E} .\ \end{array}\\ & & \\ b) & =\dfrac{1}{\left(\dfrac{4}{9}\right)^{\dfrac{3}{2}}} & \\ & =\dfrac{1}{\dfrac{8}{27}} & \begin{array}{l} \mathrm{In\ case} \ a) \ \mathrm{we}\\ \mathrm{found\ that\ }\\ \left(\dfrac{4}{9}\right)^{\dfrac{3}{2}} =\dfrac{8}{27} \end{array}\\ & & \\ & =\dfrac{( 27) 1}{( 27)\dfrac{8}{27}} & \begin{array}{l} \mathrm{Multiply\ the\ numerator}\\ \mathrm{and\ denominator\ of\ the}\\ \mathrm{complex\ fraction} \ \mathrm{by\ } 27 \end{array}\\ & & \\ & =\dfrac{27}{8} & \mathrm{Simplify} .\mathrm{\ The\ result\ matches\ Choice\ }\mathbf{G} .\\ & & \\ c) & =-\left(\left(\dfrac{9}{4}\right)^{\dfrac{1}{2}}\right)^{3} & \begin{array}{l} \mathrm{Apply\ the\ rule}\\ a^{nm} =\left( a^{n}\right)^{m} \end{array}\\ & & \\ & =-\left(\sqrt{\dfrac{9}{4}}\right)^{3} & \begin{array}{l} \mathrm{Apply\ the\ rule}\\ a^{1/n} =\sqrt[n]{a} \end{array}\\ & & \\ & =-\left(\dfrac{\sqrt{9}}{\sqrt{4}}\right)^{3} & \begin{array}{l} \mathrm{Apply\ the\ rule}\\ \sqrt{\dfrac{a}{b}} =\dfrac{\sqrt{a}}{\sqrt{b}} \end{array}\\ & & \\ & =-\left(\dfrac{3}{2}\right)^{3} & \begin{array}{l} \mathrm{Simplify\ by}\\ \mathrm{evaluating\ the\ }\\ \mathrm{root} \end{array}\\ & & \\ & =-\dfrac{3^{3}}{2^{3}} & \begin{array}{l} \mathrm{Apply\ the\ rule}\\ \left(\dfrac{a}{b}\right)^{n} =\dfrac{a^{n}}{b^{n}} \end{array}\\ & & \\ & =-\dfrac{27}{8} & \begin{array}{l} \mathrm{Simplify} .\ \mathrm{The\ result\ matches}\\ \mathrm{Choice}\mathbf{\ F} .\ \end{array}\\ & & \\ d) & =-\dfrac{27}{8} & \begin{array}{l} \mathrm{In\ case} \ b) \ \mathrm{we\ found\ that}\\ \left(\dfrac{4}{9}\right)^{-3/2} =\dfrac{27}{8} .\ \mathrm{Therefore\ result\ is\ the\ opposite\ }\\ \mathrm{number\ which\ matches\ }\mathbf{Choice\ F} .\ \end{array} \end{array}$