Dividing Positive and Negative ExponentsDate: 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/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2015 The Math Forum
http://mathforum.org/dr.math/