Raising a number to a negative fraction power I am doing a math problem where I need to raise 9 to the -3/2 power. I am unsure how this is done. I believe it's the equivalent of saying 2√9^-3, but I am unsure if this is true. If you can help me it would be much appreciated. Thank you for your time.
 A: $$9^{-\frac{3}{2}} = \frac{1}{9^\frac{3}{2}} = \frac{1}{\sqrt{9^3}} = \frac{1}{\sqrt{729}} = \frac{1}{27} $$
You can read about negative exponents on wikipedia. You can solve problems like that with WolframAlpha.
A: $$9^{-3/2}=\dfrac{1}{9^{3/2}}=\dfrac{1}{9\sqrt{9}}=\dfrac{1}{27}$$
In the second step, we use the following: $9^{3/2}=9^{1/2\,+\,1}=9^{1/2}\cdot9^1=9\sqrt{9}$.
A: Any number raised to the $-1$ power means taking the reciprocal.
For example: $8^{-1}=1/8$.  Now, $9^{-3/2}=(9^{3/2})^{-1}$
A: $$ 9^{-\tfrac{3}{2}} = \frac {1}{9^\tfrac{3}{2}} = \frac {1}{\sqrt{9^3}} = \frac{1}{\left(\sqrt{9}\right)^3} = \frac{1}{3^3} = \boxed {\dfrac{1}{27}}. $$
A: Just to answer the second part of your question, yes you can write
$\displaystyle 9^{-3/2} = \left( 9^{1/2} \right)^{-3} = \left(\sqrt{9}\right)^{-3}$.  In general, $\frac{1}{n}$ powers are equal to $n$th roots.  Just be careful with $n$ even and the base negative (real-number roots may not exist).
This becomes $3^{-3} = \frac{1}{3^3} = \frac{1}{27}$, so you do get the same answer.
