Originally Posted by

**allyourbass2212** Ive read over your post but I am still a little confused.

For instance

$\displaystyle 3^{-2}=9^{-1}=\frac{1}{9}\ne-9$

So it appears any number to a negative exponent you can rewrite as $\displaystyle n^{-1}$ rather than $\displaystyle -n$.

For instance

$\displaystyle 4^{-2}=16^{-1} $

$\displaystyle 7^{-2}=49^{-1}$

Is this correct?

If so that would allow us to use $\displaystyle a^{-n} = \frac{1}{a^n}$, in which case the $\displaystyle 16^{-1}$would equal $\displaystyle \frac{1}{16}$ and $\displaystyle 49^{-1}$ would equal $\displaystyle \frac{1}{49}$.

Is my logic correct?