Exponents Containing Negative Integers
Date: 12/15/98 at 19:44:53 From: Patrick Harrison Subject: n^-x, -n^x, -n^-x I simply don't understand how to work exponents that contain negative integers. Please explain how I'd work the following types of problems: N^-X -N^ X -N^-X Thank you for your time.
Date: 12/16/98 at 08:55:41 From: Doctor Rick Subject: Re: n^-x, -n^x, -n^-x Hi, Patrick, welcome to Ask Dr. Math! First of all, you must be careful about order of operations when using negative numbers and exponents. "-N^X" means to raise N to the X power first, then take the negative, which is not what I think you mean. I make a habit of using parentheses whenever there MIGHT be confusion about what I mean. I would write: N^(-X); (-N)^X; (-N)^(-X). Any number raised to a negative power is the same as 1 over the number raised to the positive power: -X 1 N = --- N^X It is defined this way so that the addition rule for exponents will still work with negative exponents. For instance: N^X * N^(-X) = N^(X - X) = N^0 = 1 To understand what happens when you raise a negative number to a power, you can use the distributive principle of exponents over multiplication: (AB)^X = (A^X)(B^X) So, (-N)^X = (-1 * N)^X = ((-1)^X)(N^X) What is (-1)^X? We know that (-1)^2 = 1. Therefore -1 to any EVEN power is 1, and -1 to any ODD power is -1. Putting it all together, (-N)^X = N^X if X is even, OR = -(N^X) if X is odd. You can combine the rules for N^(-X) and (-N)^X to get this rule: (-N)^(-X) = 1/(N^X) if X is even, -1/(N^X) if X is odd. I hope this helps. - Doctor Rick, The Math Forum http://mathforum.org/dr.math/
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