negative exponent problem $$\sqrt{\frac{1}{3^0 + 3^{-1} + 3^{-2} + 3^{-3} + 3^{-4}}}$$
Does this equal =
$$
\begin{align*}
& \sqrt{3^0 + 3^1 + 3^2 + 3^3 + 3^4} \\
=&\sqrt{1 + 3 + 9 + 27 + 81} \\
=&\sqrt{121} \\
=&11. 
\end{align*}
$$
The answer is apparently $\frac{9}{11}$ and I'm not sure what rule of negative exponents I got wrong.
The rule I'm using, incorrectly, is this:
$$\frac{1}{3^{-2}} = 3^2 = 9.$$

Note that $3^{-n} = 1/3^{n}$ for all $n \geq 0$, where $3^{0} = 1$, by convention.  First, simplify the denominator utilizing the last fact:
$$3^{0} + 3^{-1} + 3^{-2} + 3^{-3} + 3^{-4} = 1 + 1/3 + 1/9 + 1/27 + 1/81 \\ = 81/81 + 27/81 + 9/81 + 3/81 + 1/81 \\ = 121/81$$
Thus, we have simplified the original expression to:
$$\sqrt{1/(121/81)} = \sqrt{81/121} = 9/11$$
where the last equality comes from knowing perfect squares for natural numbers less than $20$ (recommended for any student of mathematics).  Observe from the above that you had the correct "fact" all along, you simply need to recognize that each term of the denominator cannot be inverted separately.  Explicitly, 
$$1/(3^{0} + 3^{-1} + 3^{-2} + 3^{-3} + 3^{-4}) \neq 1/3^{0} + 1/3^{-1} + 1/3^{-2} + 1/3^{-3} + 1/3^{-4}$$
It might also be useful for you multiply the numerator and denominator by $\sqrt{3^{4}}$ as was mentioned in some of the other answers, but it is not too difficult to think about it without doing this in my opinion. Let me know if you have any further questions.
