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.
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$\begingroup$$$9^{-\frac{3}{2}} = \frac{1}{9^\frac{3}{2}} = \frac{1}{\sqrt{9^3}} = \frac{1}{\sqrt{729}} = \frac{1}{27} $$
You can read about negative exponents on wikipedia. You can solve problems like that with WolframAlpha.
$\endgroup$ $\begingroup$$$9^{-3/2}=\dfrac{1}{9^{3/2}}=\dfrac{1}{9\sqrt{9}}=\dfrac{1}{27}$$
In the second step, we use the following: $9^{3/2}=9^{1/2\,+\,1}=9^{1/2}\cdot9^1=9\sqrt{9}$.
$\endgroup$ 0 $\begingroup$Any number raised to the $-1$ power means taking the reciprocal.
For example: $8^{-1}=1/8$. Now, $9^{-3/2}=(9^{3/2})^{-1}$
$\endgroup$ 0 $\begingroup$$$ 9^{-\tfrac{3}{2}} = \frac {1}{9^\tfrac{3}{2}} = \frac {1}{\sqrt{9^3}} = \frac{1}{\left(\sqrt{9}\right)^3} = \frac{1}{3^3} = \boxed {\dfrac{1}{27}}. $$
$\endgroup$ $\begingroup$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.
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