Simplify involving indices

**Meggomumsie** · October 2nd 2009, 07:17 AM

Hi. I don't know the answer to this one, could someone check my reasoning and working please?

Simplify $\frac{2a^-2}{a^\frac{-3}{2}}$

Working on the denominator only
$\frac{1}{-\sqrt{a}*-\sqrt{a}*-\sqrt{a}}$

Denominator is $\frac{1}{a\sqrt{a}}$

So I now have $\frac{2a^-2}{1}*\frac{a\sqrt{a}}{1}$

Then $\frac{1}{2a^2}*a\sqrt{a}$

Which gives me my answer $\frac{a\sqrt{a}}{2a^2}$

**ramiee2010** · October 2nd 2009, 07:44 AM

$\frac{2a^{-2}}{a^{\frac{-3}{2}}}= 2 \frac{a^{-2}}{a^{\frac{-3}{2}}}$

$= 2 a^{{-2}-({\frac{-3}{2}})}= 2 a^{\frac{-4+3}{2}}$

$= 2 a^{\frac{-1}{2}} = \frac {2}{\sqrt a }$

or

$2 a^{-2} \times a^{\frac{3}{2}}$

$2 a^{-2+\frac{3}{2}} = = 2 a^{\frac{-4+3}{2}}$

**ramiee2010** · October 2nd 2009, 07:59 AM

Working on the denominator only
$\frac{1}{-\sqrt{a}*-\sqrt{a}*-\sqrt{a}}$

$a^{\frac{-3}{2}} = \frac{1}{a^{\frac{3}{2}}}=\frac{1}{\sqrt {a} \times \sqrt{ a} \times \sqrt {a}}$

$\frac{2a^-2}{1}*\frac{a\sqrt{a}}{1}$

$\frac{2}{a^2}*a\sqrt{a}$

**Meggomumsie** · October 2nd 2009, 08:07 AM

Thank you. Was I correct in my thinking to here?

even though it is probably not the usual way of dealing with this kind of question?

and is $\frac{2}{a^2}*a\sqrt{a}$ the final and correct answer to this question please?