How do you simplify ${(\frac{16 n^{\frac{2}{3}}}{81 n^{- \frac{2}{3}}})}^{- \frac{3}{4}}$ $((16n ^(2/3))/(81n ^(-2/3))) ^(-3/4)$ ?

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How do you simplify ${(\frac{16 n^{\frac{2}{3}}}{81 n^{- \frac{2}{3}}})}^{- \frac{3}{4}}$ $((16n ^(2/3))/(81n ^(-2/3))) ^(-3/4)$ ?

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Answer 1 · 2016-06-12T17:18:23

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

Explanation:

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

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

= $⎛ ⎜ ⎜ ⎜ ⎝ \frac{{(2^{4})}^{- \frac{3}{4}} {(n^{\frac{2}{3}})}^{- \frac{3}{4}}}{{(3^{4})}^{- \frac{3}{4}} {(n^{- \frac{2}{3}})}^{- \frac{3}{4}}} ⎞ ⎟ ⎟ ⎟ ⎠$ $(((2^4)^(-3/4)(n^(2/3))^(-3/4))/((3^4)^(-3/4)(n^(-2/3))^(-3/4)))$

= $⎛ ⎜ ⎝ \frac{2^{(4 \times - \frac{3}{4})} n^{(\frac{2}{3} \times (- \frac{3}{4}))}}{3^{4 \times (- \frac{3}{4})} n^{(- \frac{2}{3}) \times (- \frac{3}{4})}} ⎞ ⎟ ⎠$ $((2^((4xx-3/4))n^((2/3xx(-3/4))))/(3^(4xx(-3/4))n^((-2/3)xx(-3/4))))$

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

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

= $\frac{27}{8 n}$ $27/(8n)$

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How do you simplify ${(\frac{16 n^{\frac{2}{3}}}{81 n^{- \frac{2}{3}}})}^{- \frac{3}{4}}$ $((16n ^(2/3))/(81n ^(-2/3))) ^(-3/4)$ ?

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Explanation:

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How do you simplify ${(\frac{16 n^{\frac{2}{3}}}{81 n^{- \frac{2}{3}}})}^{- \frac{3}{4}}$ $((16n ^(2/3))/(81n ^(-2/3))) ^(-3/4)$ ?