How do you write $\frac{1}{3 {\sqrt{2}}^{- n}}$ $1/(3sqrt2^-n)$ as a fractional exponent?

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Answer 1 · 2015-08-02T03:15:05

$\frac{1}{3 {(\sqrt{2})}^{- n}} = \frac{2^{\frac{n}{2}}}{3}$ $color(red)(1/(3(sqrt2)^(-n)) = 2^(n/2)/3)$

$\frac{1}{x^{- n}} = x^{n}$ $1/x^(-n)= x^n$ , so

$\frac{1}{3 {(\sqrt{2})}^{- n}} = \frac{{(\sqrt{2})}^{n}}{3} = \frac{{(2^{\frac{1}{2}})}^{n}}{3}$ $1/(3(sqrt2)^(-n)) = (sqrt2)^n/3 = (2^(1/2))^n/3$

Also, ${(x^{a})}^{b} = x^{a b}$ $(x^a)^b = x^(ab)$ , so

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

∴ $\frac{1}{3 {(\sqrt{2})}^{- n}} = \frac{2^{\frac{n}{2}}}{3}$ $1/(3(sqrt2)^(-n)) = 2^(n/2)/3$

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