# 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.

• $9^{-\frac{3}{2}} = \left(\frac{1}{9}\right)^{\frac{3}{2}} = \sqrt[2]{\left(\frac{1}{9}\right)^3}$ Aug 23 '14 at 16:29
• Four responses in < 2 minutes. Aug 23 '14 at 16:33
• @AhaanRungta With strikingly different insights, as usual.
– user147263
Aug 23 '14 at 17:11
• @900sit-upsaday Yep; I feel uncannily proud to be the one with no upvotes, especially since each of the other upvotes were made by me. My community service for the day. Aug 23 '14 at 17:12

$$9^{-\frac{3}{2}} = \frac{1}{9^\frac{3}{2}} = \frac{1}{\sqrt{9^3}} = \frac{1}{\sqrt{729}} = \frac{1}{27}$$

You can read about negative exponents on wikipedia. You can solve problems like that with WolframAlpha.

$$9^{-3/2}=\dfrac{1}{9^{3/2}}=\dfrac{1}{9\sqrt{9}}=\dfrac{1}{27}$$

In the second step, we use the following: $9^{3/2}=9^{1/2\,+\,1}=9^{1/2}\cdot9^1=9\sqrt{9}$.

Any number raised to the $-1$ power means taking the reciprocal.

For example: $8^{-1}=1/8$. Now, $9^{-3/2}=(9^{3/2})^{-1}$

$$9^{-\tfrac{3}{2}} = \frac {1}{9^\tfrac{3}{2}} = \frac {1}{\sqrt{9^3}} = \frac{1}{\left(\sqrt{9}\right)^3} = \frac{1}{3^3} = \boxed {\dfrac{1}{27}}.$$

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.