# How do you simplify (49/81) ^ (-1/2)?

May 3, 2018

9/7

#### Explanation:

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

May 3, 2018

$\frac{9}{7}$

#### Explanation:

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

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

May 3, 2018

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

#### Explanation:

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

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

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

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