How do you simplify ${(\frac{49}{81})}^{- \frac{1}{2}}$ $(49/81) ^ (-1/2)$ ?

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How do you simplify ${(\frac{49}{81})}^{- \frac{1}{2}}$ $(49/81) ^ (-1/2)$ ?

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Answer 1 · 2018-05-03T16:16:23

9/7

Explanation:

$\frac{1}{\sqrt{\frac{49}{81}}}$ $1/sqrt(49/81)$ = $\frac{1}{\frac{7}{9}}$ $1/(7/9)$ = $\frac{9}{7}$ $9/7$

Answer 2 · 2018-05-03T16:17:00

$\frac{9}{7}$ $9/7$

Explanation:

${(\frac{a}{b})}^{\frac{m}{n}} = \sqrt[n]{{(\frac{a}{b})}^{m}}$ $(a/b)^(m/n)=root(n)((a/b)^m)$
If $\frac{m}{n}$ $m/n$ is negative
${(\frac{a}{b})}^{- (\frac{m}{n})} = \frac{1}{\sqrt[n]{{(\frac{a}{b})}^{m}}}$ $(a/b)^-(m/n)=1/root(n)((a/b)^m)$

SO:
${(\frac{49}{81})}^{- \frac{1}{2}} = \frac{1}{\sqrt{{(\frac{49}{81})}^{1}}} = \frac{1}{\frac{7}{9}} = \frac{9}{7}$ $(49/81)^(-1/2)=1/sqrt((49/81)^1)=1/(7/9)=9/7$

Answer 3 · 2018-05-03T16:19:00

Simplified we should get $\pm \frac{9}{7}$ $+-9/7$

Explanation:

To solve this we need to remember that $x^{\frac{1}{2}} = \sqrt{x}$ $x^{1/2}= sqrt(x)$
Aslo $x^{- 1} = \frac{1}{x}$ $x^(-1)=1/x$
And lastly we should recognise that
$49 = 7^{2}$ $49=7^2$ and $81 = 9^{2}$ $81=9^2$

If we do, it's quite straightforward:
${(\frac{49}{81})}^{- (\frac{1}{2})} = {(\frac{9^{2}}{7^{2}})}^{\frac{1}{2}} = {(\frac{9}{7})}^{2 \cdot \frac{1}{2}} = \frac{9}{7}$ $(49/81)^-(1/2)=(9^2/7^2)^(1/2)=(9/7)^(2*1/2)=9/7$

As ${(- \sqrt{x})}^{2} = {(\sqrt{x})}^{2} = x$ $(-sqrt(x))^2=(sqrt(x))^2=x$

the full simplification should then be $\pm \frac{9}{7}$ $+-9/7$

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