Negative Fraction Exponents

Date: 01/27/99 at 09:39:26
From: Khalida
Subject: Pre-calculus Exponents

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

I am having trouble with these kinds of problems. If you could help me 
simplify these exponents and also radical exponents, I would be very 
thankful.

Khalida

Date: 01/27/99 at 12:55:13
From: Doctor Rick
Subject: Re: Pre-calculus Exponents

Hi, Khalida.

I would start by breaking this into a difference of two fractions (by
applying the distributive principle).

    1/3      -2/3
  2n       4n
  ------ - -------
    -2/3     -2/3
  2n       2n 

Now, you can see right off that in each fraction, the numerator and
denominator are both divisible by 2. So divide numerators and 
denominators by 2:

   1/3      -2/3
  n       2n
  ----- - -------
   -2/3     -2/3
  n        n 

What's left is the part you are most concerned about, I'm sure. Here 
are two rules you can use:

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

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

Use the second rule to convert each fraction (division) into a product:

   1/3    2/3    -2/3    2/3
  n    * n    - 2n    * n

Then use the first rule to combine the powers of n:

   (1/3 + 2/3)     (-2/3 + 2/3)
  n            - 2n

Do the additions:

   1     0
  n  - 2n

which we can write as

  n - 2

Wow, that's a lot simpler!

You can simplify expressions with radicals in the same way, if you 
first rewrite each radical as a fractional power, for instance, 
sqrt(x) = x^(1/2). If you need more help with radicals, send an 
example, and we'll work through it.

- Doctor Rick, The Math Forum
  http://mathforum.org/dr.math/