Perhaps I am overthinking this but $(-4)^{(-5/2)}$ is not a valid equation, am I correct? Working through the problem gives me $1/(-4^{5/2})$ which then works out to $1/\sqrt{-4^5}$ which leaves a negative number in the square root, which is not valid.
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To avoid another long discussion, you should re-write the expression as (-4)^(-5/2), Assuming you want to raise negative 4 to a power – DJohnM Sep 01 '15 at 02:33
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It is perfectly valid, the result is simply not a Real Number, it is what is called an Imaginary Number (or a Complex Number in the case in which you had a Real Numbered component as well), which you may not be familiar with.
The result you gave simplifies down to $-\frac{1}{32}i$, where $i$ is called the Imaginary Unit, and is equal to $\sqrt{-1}$.
If you cared to investigate this further try reading these for a decent introduction: Complex Numbers on Wikipedia and Imaginary Numbers. Khan Academy also has a good video on this subject for the introductory level.
Rio Alvarado
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Okay thanks, I should clarify with my instructor as to whether or not he expects real numbers in my calculus 1 course, as my pre-calculus teacher lead me to believe that we will not be using imaginary numbers until calc 3. – Rod Sep 01 '15 at 02:34
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One usually deals briefly with complex numbers sometime in one's Algebra 2 or Pre-Calculus career. Calculus at the high school level never involves Complex Numbers to my knowledge, that is usually saved for what is called Complex Analysis. – Rio Alvarado Sep 01 '15 at 02:39
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The square root of a negative number is a complex number. $(-4)^{-{5/2}} = {1 \over (\sqrt{-4})^5} = {1 \over (2i)^5}=-\frac{1}{32} i$. If this sounds like complete gibberish to you, google "imaginary numbers" and read up on it.
Ud779
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