Drexel dragonThe Math ForumDonate to the Math Forum

Ask Dr. Math - Questions and Answers from our Archives
_____________________________________________
Associated Topics || Dr. Math Home || Search Dr. Math
_____________________________________________

Rules of Exponents


Date: 10/02/1999 at 18:24:22
From: CHRIS HELTON
Subject: Rules of indices

The question I am having trouble with is:

Use the rules of indices to simplify the following expression:

     (16x^2)^(1/4)*(4x^3)^(1/2)
     --------------------------
            (27^4)^(1/3)

I don't know where to start breaking the question down to attempt to 
simplify it. This question is part of the foundation topic assignment 
given to me in the class titled Mathematics 1 at the University of 
Strathclyde.


Date: 10/11/1999 at 21:29:59
From: Doctor Sandi
Subject: Re: Rules of indices

Hi Chris, 

First of all I want to go through some of the log laws with you.

Consider 64 = 4^3

The 3 is called the exponent; the 4 is called the base. This can also 
be expressed as log(base 4) 64 = 3.

Something else worth committing to your permanent memory is that any 
number raised to the power of 1/2 is the same as the square root of 
that number. For example, 4^(1/2) = sqrt(4) = 2. This also happens 
with numbers raised to the power of 1/3: the answer will be the same 
as the cube root of the number.

Here are some rules for dealing with indices (write these out properly 
without the ^ so that they don't look so clumsy):

     1.  a^m * a^n = a^(m+n)
     2.  a^m / a^n = a^(m-n)
     3.  (a^m)^n   = a^(mn)
     4.  (ab)^m    = (a^m)(b^m)
     5.  a^0       = 1
     6.  a^(-n)    = 1/(a^n), a is not equal to 0
     7.  a^(1/q)   = q-root(a)
     8.  a^(p/q)   = q-root (a^p) = (q-root(a))^p

Generally when simplifying index expressions:

     1. Express all terms in index form (lowest base) where necessary
     2. Remove the brackets
     3. Add and subtract indices
     4. Express with positive indices

Now to your question: [(16x^2)^(1/4)*(4x^3)^(1/2)]/[(27^4)^(1/3)].

When you have an index raised to the power of another index, you 
multiply the indices. You end up with:

First the numerator (top half of the fraction)

       [(16^(1/4))(x^(2*(1/4))][(4^(1/2)(x^(3*1/2)]
     = [2x^(1/2)][2x^(3/2)]
     = (4x^[(1/2)+(3/2)]
     = (4x^2)

then the denominator:

     (27^4)(^1/3)

Let's deal with the power first. It is a power raised to another power 
so we will multiply them together and end up with:

     4 * (1/3) = 4/3

So we have

     27^(4/3) = 81

So the answer is

     (4x^2)/81

I hope that this has been helpful for you. If you have any other other 
questions that you would like to ask Dr. Math, please feel free to 
write back.

- Doctor Sandi, The Math Forum
  http://mathforum.org/dr.math/   
    
Associated Topics:
High School Exponents

Search the Dr. Math Library:


Find items containing (put spaces between keywords):
 
Click only once for faster results:

[ Choose "whole words" when searching for a word like age.]

all keywords, in any order at least one, that exact phrase
parts of words whole words

Submit your own question to Dr. Math

[Privacy Policy] [Terms of Use]

_____________________________________
Math Forum Home || Math Library || Quick Reference || Math Forum Search
_____________________________________

Ask Dr. MathTM
© 1994-2013 The Math Forum
http://mathforum.org/dr.math/