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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 

  Find the Flaw 

  Was Euler Wrong? 2*Pi=0? 

  Inconsistency in Complex Logarithms 

  Raising Integers to Imaginary Powers or Exponents 

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 

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 

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.
Associated Topics:
High School Exponents

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