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### Dividing Positive and Negative Exponents

```
Date: 12/18/97 at 21:21:01
From: Matt
Subject: Dividing positive and negative exponents

I have the problem:

(3^2)^-1 M^-1 N^-2
-------------------
3^-3 M^0 N^3

I am pretty sure that I take care of the 3^2 = 9 first. I think after
that I take 9^-1 = 1/9 But the fractions can't be correct. Am I on the
right track? I am totally lost after that. Do I put the variables with
negative exponents on the bottom?

Matt
```

```
Date: 12/19/97 at 09:43:44
From: Doctor Pipe
Subject: Re: Dividing positive and negative exponents

Matt,

It is correct that 3^2 = 9, but this does not move you directly toward
a solution to this problem.

ability to use, the laws of exponents. Let's see how these laws are
used to simplify the terms in this problem where 3 is the base.

One of these laws says that x^-n = 1/(x^n). If you use this law on the
(3^2)^-1 term in the numerator [x is (3^2) and n is -1] then this term
is "moved" into the denominator and n becomes 1 (I've added some
parentheses so that each term in the numerator and denominator is more
easily seen):

(M^-1) (N^-2)
----------------------------
((3^2)^1) (3^-3) (M^0) (N^3)

and since exponents of 1 are not usually written, the problem becomes:

(M^-1) (N^-2)
------------------------
(3^2) (3^-3) (M^0) (N^3)

Now, another law of exponents comes into play.
This one says that (x^n) (x^m) = x^(n+m).
In our problem right now, we can apply this law to the (3^2) (3^-3)
terms in the denominator: x is 3, n is 2 and m is -3.  This gives:

(M^-1) (N^-2)
----------------------
3^(2+(-3)) (M^0) (N^3)

Which simplifies to:

(M^-1) (N^-2)
--------------------
3^(-1) (M^0) (N^3)

Now, the good thing about a mathematical equation is that you can swap
the left and right sides and the equality is preserved. This means
that the law we used in the first step:

x^-n = 1/(x^n)

can also be written:

1/(x^n) = x^-n

Applying this to the 3^(-1) term in the denominator gives:

(3^1) (M^-1) (N^-2)
-------------------
(M^0) (N^3)

Which simplifies to (exponents of 1 are not usually written):

3 (M^-1) (N^-2)
---------------
(M^0) (N^3)

You can complete this problem by reapplying the above technique to the
"M" terms and then to the "N" terms. [Remember that x^0 (anything to
the zeroth power) is equal to 1!]

The final answer that I get is:

3
-------
M (N^5)

Is that what you get??

-Doctor Pipe,  The Math Forum
Check out our web site!  http://mathforum.org/dr.math/
```
Associated Topics:
High School Exponents

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