Removing the exponent

This is the question and the answer, but I don't understand it.

With an exponent of ^2 it's a simple matter -3(-2)^2 = -3(4). But the problem below has both a negative exponent and an exponent less than 2. Can someone help me understand first how a negative exponent effects a problem and then how an exponent of less than 2 effects a problem? Thanks.

-3(-2)^-1 = -(3/-2)
Who is Participating?
phoffricConnect With a Mentor Commented:
Now consider a^(-2). That is just 1/( a^2 )
So, if a = -2, then a^(-2) = 1/( a^2 ) = 1/(   (-2)^2   ) = 1/(   -2*-2 ) = +1/4
phoffricConnect With a Mentor Commented:
a^(-1) is defined to be 1/a

-3 * (a)^(-1) = -3/a

Let a = -2
-3 * (-2)^(-1) = -3/(-2) = +3/2
phoffricConnect With a Mentor Commented:
Here is something that may help you understand why the negative exponent was defined that way.

Consider a^5/a^3 = (aaaaa)/aaa = aa = a^2
which leads to a natural formula to remember:  a^n / a^k = a^(n-k)i.e., a^5/a^3 = a^(5-3) = a^2

Now consider a^3/a^5 = aaa/(aaaaa) = 1/(aa) = 1/a^2
But, if you try to use the above formula, you get a^3/a^5 = a^(3-5) = a^(-2)

So, someone said, let's define a negative exponent to be:  a^(-n) = 1/a^n
and since this one formula seemed to fit well into algebra, it stuck.
Cloud Class® Course: Microsoft Office 2010

This course will introduce you to the interfaces and features of Microsoft Office 2010 Word, Excel, PowerPoint, Outlook, and Access. You will learn about the features that are shared between all products in the Office suite, as well as the new features that are product specific.

Here's a little more discussion of negative exponents:
kadinAuthor Commented:
Thanks for your help. I think I understand this a little better.
You're welcome.
BTW - I tend not to try to remember a lot of formulas. Instead I try to understand their derivation, and use that approach when thinking about these types of problems.
Question has a verified solution.

Are you are experiencing a similar issue? Get a personalized answer when you ask a related question.

Have a better answer? Share it in a comment.

All Courses

From novice to tech pro — start learning today.