Exponents in AlgebraDate: 8/8/96 at 9:6:45 From: Ralph Ware Subject: Re: Algebra Terms Dear Dr. Math: I continue to study for my upcoming placement test and I really appreciate all the help and advice that the folks at the Dr. Math forum have given me. Now I am studying the properties of exponents from a book about college algebra and doing the exercises. One of the exercise questions is, "Evaluate: (2^-3)-(2^-2)." Well, the answer is, obviously, -1/8 because (2^-3) is 1/8 and (2^-2) is 1/4 and 1/8-1/4= -1/8. That's all well and good when dealing with a simple base like 2 and a simple exponent like -3, but I don't understand the process for arriving at the answer. For example what if the exercise question was, instead, "Evaluate:(x^-16)-(x^-18)," or the question involved two dissimilar bases (like x and y) or exponents of unknown value represented by letters? Surely there is a procedure for evaluating all such expressions. Could you show me? Thank you, Ralph Ware Date: 8/25/96 at 23:46:47 From: Doctor Mike Subject: Re: Algebra Terms Hello, What we need to get into here is Common Denominators. Probably there is an index entry on this in your book. These problems you have listed seem to be testing your understanding of negative exponents, which seems to be good, but an understanding of this other area is holding you back. When you subtract 1/8 and 1/4 you should think of this as 1/8 and 2/8. Whenever the denominators are the same you just add/subtract the numerators. The 8 in this case is the common denominator since it is common to both fractions (after one of them was re-written). In general subtract a/x - b/x to get (a-b)/x. Often what you must do to get 2 fractions to have the same denominator is to multiply one or both of them by one, not just any old one, but by a special form of one. Here is what I mean for your first case : 1 1 1 1 2 1 2 1 --- - --- = --- - (--- * ---) = --- - --- = - --- 8 4 8 4 2 8 8 8 The particular value of one used here is 2/2. In your other one the 2 denominators are x^16 and x^18. The common denominator here is x^18. So how can you start with 1/(x^16) , multiply it by one, and get something with x^18 as a denominator. Answer is to choose (x^2)/(x^2) as the form of one. Then 1/(x^16) times (x^2)/(x^2) becomes (x^2)/(x^18). Keep in mind that this new fraction is THE SAME AS 1/(x^16) since all we did was to multiply by one. Now your original (x^-16)-(x^-18) has been re-written with a common denominator as (x^2)/(x^18) - 1/(x^18) so the difference is (x^2-1)/(x^18). I cannot give you a total lesson in this area now, but this should get you started. Write back if you encounter another sticking point. I hope this helps. -Doctor Mike, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994- The Math Forum at NCTM. All rights reserved.
http://mathforum.org/dr.math/