How do you simplify ((16n ^(2/3))/(81n ^(-2/3))) ^(-3/4)?

Jun 12, 2016

${\left(\frac{16 {n}^{\frac{2}{3}}}{81 {n}^{- \frac{2}{3}}}\right)}^{- \frac{3}{4}} = \frac{27}{8 n}$

Explanation:

${\left(\frac{16 {n}^{\frac{2}{3}}}{81 {n}^{- \frac{2}{3}}}\right)}^{- \frac{3}{4}}$

= ${\left(\frac{{2}^{4} {n}^{\frac{2}{3}}}{{3}^{4} {n}^{- \frac{2}{3}}}\right)}^{- \frac{3}{4}}$

= $\left(\frac{{\left({2}^{4}\right)}^{- \frac{3}{4}} {\left({n}^{\frac{2}{3}}\right)}^{- \frac{3}{4}}}{{\left({3}^{4}\right)}^{- \frac{3}{4}} {\left({n}^{- \frac{2}{3}}\right)}^{- \frac{3}{4}}}\right)$

= $\left(\frac{{2}^{\left(4 \times - \frac{3}{4}\right)} {n}^{\left(\frac{2}{3} \times \left(- \frac{3}{4}\right)\right)}}{{3}^{4 \times \left(- \frac{3}{4}\right)} {n}^{\left(- \frac{2}{3}\right) \times \left(- \frac{3}{4}\right)}}\right)$

= $\frac{\left({2}^{- 3} {n}^{- \frac{1}{2}}\right)}{{3}^{- 3} {n}^{\frac{1}{2}}}$

= $\frac{\left({3}^{3} {n}^{- \frac{1}{2}}\right)}{{2}^{3} {n}^{\frac{1}{2}}}$

= $\frac{27}{8 n}$