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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. 

This problem seems designed to test your understanding of, and your 
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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