In the equation:

(2n^3)^4 (n^-2)^-1

I work it and get 16n^12-n^2. I am not sure what to do with the "-n^2". Help woulld be greatly appreciated.

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- Nov 7th 2010, 11:13 AMmrkingEquations with variables and exponents
In the equation:

(2n^3)^4 (n^-2)^-1

I work it and get 16n^12-n^2. I am not sure what to do with the "-n^2". Help woulld be greatly appreciated. - Nov 7th 2010, 12:00 PMpickslides
$\displaystyle (2n^3)^4(n^{-2})^{-1}$

$\displaystyle 2^4n^{3\times 4}n^{-2\times -1}$

Can you finish it from here using simple index laws?

__Spoiler__: - Nov 7th 2010, 12:05 PMHallsofIvy
First, although it may seem "picky", you don't have an equation here, you have an expression. (It isn't

**equal**to anything.)

What do you**want**to do with the $\displaystyle n^2$? That is, what are you trying to do with the original expression? Simplify it? $\displaystyle 16n^{12}- n^2$ seems simple enough to me! You could, if you wanted, factor out an "$\displaystyle n^2$". $\displaystyle n^{12}= n^{10}n^2$ so [tex]16n^{12}- n^2= n^2(16n^{10}- 1). If you really wanted to you could factor that further- $\displaystyle 16n^10- 1= (4n^5)^2- 1^2$ is the difference of two squares and can be factored as $\displaystyle (4n^5-1)(4n^5+1)$. whether $\displaystyle n^2(4n^5-1)(4n^5+1)$ is "simpler" than $\displaystyle 16n^{12}- n^2$ is a matter of taste or, perhaps better, of what you wanted to do with it from here. - Nov 7th 2010, 12:11 PMpickslides
- Nov 7th 2010, 12:30 PMHallsofIvy
I'm going to slide right out of that one!

- Nov 7th 2010, 12:52 PMWilmer
Hmmm....I get a simple 16n^14 ; where does the - come from?

- Nov 7th 2010, 12:54 PMpickslides
- Nov 7th 2010, 03:56 PMWilmer
- Nov 7th 2010, 03:58 PMpickslides
Its an expression, thanks smarty pants!