# How do you simplify (9^(1/2) * 9^(2/3))^(1/6)?

May 22, 2015

Two important law of exponentials here:

• $\left({a}^{n}\right) \left({a}^{m}\right) = {a}^{n + m}$

• ${\left({a}^{n}\right)}^{m} = {a}^{n \cdot m}$

${\left({9}^{\frac{1}{2}} {9}^{\frac{2}{3}}\right)}^{\frac{1}{6}}$=${\left({9}^{\frac{1}{2} + \frac{2}{3}}\right)}^{\frac{1}{6}}$=${\left({9}^{\frac{7}{6}}\right)}^{\frac{1}{6}}$=${9}^{\left(\frac{7}{6}\right) \left(\frac{1}{6}\right)}$=$\textcolor{g r e e n}{{9}^{\frac{7}{36}}}$

May 23, 2015

${\left({9}^{\frac{1}{2}} \cdot {9}^{\frac{2}{3}}\right)}^{\frac{1}{6}} = {\left({9}^{\left(\frac{1}{2} + \frac{2}{3}\right)}\right)}^{\frac{1}{6}}$

$= {\left({9}^{\frac{7}{6}}\right)}^{\frac{1}{6}} = {9}^{\frac{7}{6} \cdot \frac{1}{6}} = {9}^{\frac{7}{36}} = {9}^{\frac{1}{2} \cdot \frac{7}{18}}$

$= {\left({9}^{\frac{1}{2}}\right)}^{\frac{7}{18}} = {\left(\sqrt{9}\right)}^{\frac{7}{18}} = {3}^{\frac{7}{18}}$

The identities we use here are:

${x}^{a} \cdot {x}^{b} = {x}^{a + b}$

${x}^{a b} = {\left({x}^{a}\right)}^{b}$

${x}^{\frac{1}{n}} = \sqrt[n]{x}$