1. Express the following in exponential form. 1. $\sqrt[3]{64^2}$ is $64^{\frac{2}{3}}$. 2. $\sqrt[4]{81^3}$ is $81^{\frac{3}{4}}$. 3. $\sqrt[5]{32^3}$ is $32^{\frac{3}{5}}$. 4. $\sqrt{81}$ is $81^{\frac{1}{2}}$. 5. $\sqrt{27^2}$ is $27^{\frac{2}{2}}=27^1$. 2. Transform each expression to radical form. 6. $8^{\frac{2}{3}}$ is $\sqrt[3]{8^2}$. 7. $x^{\frac{1}{2}} y^{\frac{3}{2}}$ is $\sqrt{x} \cdot y^{\frac{3}{2}} = \sqrt{x} \cdot \sqrt{y^3}$. 8. $6 x^{\frac{1}{3}}$ is $6 \sqrt[3]{x}$. 9. $q^{\frac{m}{n}}$ is $\sqrt[n]{q^m}$. 10. $125^{-\frac{1}{3}}$ is $\frac{1}{\sqrt[3]{125}}$. 3. Simplify each radical. 11. $\sqrt[5]{243 m^{10} n^5}$: - $243 = 3^5$, so $\sqrt[5]{3^5} = 3$. - $m^{10}$ inside fifth root is $m^{10/5}=m^2$ outside. - $n^5$ inside fifth root is $n^{5/5}=n$ outside. So the simplification is $3 m^2 n$. 12. $\sqrt[4]{16} \sqrt{36} x^1 y^2$: - $\sqrt[4]{16} = \sqrt[4]{2^4} = 2$. - $\sqrt{36} = 6$. - So combined constants: $2 \times 6 = 12$. - Variables $x^1 y^2$ remain as is. So simplified expression is $12 x y^2$. 4. Rationalize the denominator. 13. $\sqrt[3]{\frac{4}{9x^2}} = \frac{\sqrt[3]{4}}{ \sqrt[3]{9x^2}}$. To rationalize denominator, multiply numerator and denominator by $\sqrt[3]{9^2 x^4} = 9^{2/3} x^{4/3}$ to get cube root power 3 in denominator. Denominator becomes $\sqrt[3]{9^3 x^{6}} = 9 x^{2}$. Numerator becomes $\sqrt[3]{4} \cdot \sqrt[3]{81 x^{4}} = \sqrt[3]{324 x^{4}}$. Final expression: $\frac{\sqrt[3]{324 x^{4}}}{9 x^{2}}$. 14. $\sqrt[3]{\frac{6}{27 x^{2}}} = \frac{\sqrt[3]{6}}{ \sqrt[3]{27 x^{2}}}$. Multiply numerator and denominator by $\sqrt[3]{27^{2} x^{4}} = 27^{2/3} x^{4/3}$. Denominator becomes $\sqrt[3]{27^3 x^{6}} = 27 x^{2}$. Numerator becomes $\sqrt[3]{6} \cdot \sqrt[3]{729 x^{4}} = \sqrt[3]{4374 x^{4}}$. Final expression: $\frac{\sqrt[3]{4374 x^{4}}}{27 x^{2}}$.