Does (a^m)^n Always Equal a^(mn)?Date: 09/30/2009 at 21:55:43 From: Adam Subject: Is the law (a^m)^n = a^(mn) always true? I'm in training to be a math teacher--and I want to know the laws of exponents with all their gory exceptions and complications. One law for exponents states: (a^m)^n = a^(mn). However, is this always true? For instance, is it true that (a^2)^(1/2) = a^1 = a? It doesn't seem that it can be for ((-3)^2))^(1/2) since it can be evaluated one of two ways: ONE: evaluating the "inside" first yields 9^(1/2)= 3 OR TWO: evaluating the "outside" first (by multiplying the exponents and applying the law) yields (-3)^1= -3. So which of these answers is correct--and why? My calculator (a TI-89) indicates the first one (positive 3), which would be consistent with evaluating from left to right, as the order of operations requires when operations are on the same level. In addition, I've always read that SQRT(X^2) is the ABS(X), which makes sense and is consistent with my first answer. If this is right, this would mean the above law is NOT always true. However, I never see this exception mentioned in the laws of exponents that are presented to students. Generally, these laws are claimed to hold for all real number bases and exponents, with the exception of 0^0. In addition, I also note that if you evaluate ((-3)^(1/2))^2 you get the same answer either way you proceed: ONE: evaluating the inside first yields(i*SQRT(3))^2= -3 and TWO: evaluating the outside first yields, once again, (-3)^1= -3. My TI-89 gives me the same answer as well (negative) Yet were the above exponent law to apply, I could apply the commutative law and rearrange the exponents from 1/2 * 2 to 2 * 1/2 and thus regenerate the above problem. It seems to me that the particular "law" of exponents in question is really only valid for a > 0. Am I correct? Is there some gory list of rules of how to deal with these situations? Any help would be much appreciated. Date: 09/30/2009 at 23:01:41 From: Doctor Vogler Subject: Re: Is the law (a^m)^n = a^(mn) always true? Hi Adam, Thanks for writing to Dr Math. That's a very good question. The usual convention is that when x is a positive number, then sqrt(x) is also taken to be the positive square root. Of course, really there are two square roots, and ignoring that fact can cause problems. When taking roots of complex numbers, we have no such convention (e.g. why should the square root of -2i be 1-i instead of i-1?), and we simply deal with the fact that every nonzero complex number has exactly n (different) complex n'th roots. For example, the number 9 has TWO square roots, namely +3 and -3. The "rule" (a^b)^c = a^(b*c), which works so nicely for integer exponents and even for non-integer exponents when the base is positive, becomes problematic with negative or complex bases and non-integer exponents. You see, when n is an integer, then x^n has only one value, but when n is a rational number with (least positive) denominator d, then x^n has d different complex values, and when n is irrational, then x^n has infinitely many different complex values. The rule (a^b)^c = a^(b*c) should be interpreted as: Every value of a^(b*c) is one of the values of (a^b)^c. (It doesn't even work in the other direction, since sometimes some of the values of (a^b)^c aren't any of the values of a^(b*c). For example, -1 is one of the square roots of 1^2.) For related answers from the Dr. Math archive, see: Complex Powers http://mathforum.org/library/drmath/view/60383.html Find the Flaw http://mathforum.org/library/drmath/view/51615.html Was Euler Wrong? 2*Pi=0? http://mathforum.org/library/drmath/view/51629.html Inconsistency in Complex Logarithms http://mathforum.org/library/drmath/view/65171.html Raising Integers to Imaginary Powers or Exponents http://mathforum.org/library/drmath/view/65490.html I found these by searching on these keywords: complex exponent You can find other similar answers by doing the same search at Search Dr. Math http://mathforum.org/library/drmath/mathgrepform.html If you have any questions about this or need more help, please write back and show me what you have been able to do, and I will try to offer further suggestions. - Doctor Vogler, The Math Forum http://mathforum.org/dr.math/ Date: 10/01/2009 at 16:34:39 From: Adam Subject: Thank you (Is the law (a^m)^n = a^(mn) always true?) Thanks so much for your detailed and prompt response. It was exactly what I was looking for. Wow, x^pi has infinitely many values--amazing. |
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