Can someone explain why this happens? (Dividing variables with exponents) Alright, so let's say I have $$\frac{x^{-6}}{-x^{-4}}$$ The answer is $\dfrac{1}{x^2}$, but why isn't it $\dfrac{1}{-x^2}$?
 A: Let's think about what negative exponents really mean. Consider the following list
$$2^3 = 8$$
$$2^2 = 4$$
$$2^1 = 2$$
Notice that every time we decrease an exponent, we divide by 2. Continuing this list, we have that
$$2^3 = 8$$
$$2^2 = 4$$
$$2^1 = 2$$
$$2^0 = 1$$
$$2^{-1} = 1/2$$
$$2^{-2} = 1/4$$
Rather interesting, we see a pattern: $2^{-n} = \frac{1}{2^n}$. Therefore there is no reason for a negative sign on the quantity itself, since we're successively dividing by positive numbers.
A: It might help you see what's going on if you write them as two story fractions: $$\frac{x^{-6}}{-x^{-4}}=\frac{\left(\dfrac{1}{x^6}\right)}{\left(-\dfrac{1}{x^4}\right)}=\underbrace{\color{red}{\frac{1}{x^6}\times -\frac{x^4}{1}}}_{\large \color{blue}{\text{by the reciprocal rule}}}=-\frac{1}{x^2}=\frac{1}{-x^2}$$ So you are correct, and the "answer" is wrong.
A: If the questions is asking; simplify $\frac {x^{-6}}{-x^{-4}}$, then the answer is $\frac {1}{-x^2}$. But if the answer is $\frac {1}{x^2}$, then I believe the question is $\frac {x^{-6}}{(-x)^{-4}}$.
