3

I've been searching about exponential rules on google and the first three results was these ones:

Link 1

Link 2

Link 3

I didn't understand why they all say: $(a^b)^c=a^{bc}$. This is wrong, see for example: $((-2)^{2})^{1/2}=2\neq (-2)^1$.

Why does every site I read say this wrong rule? how can I reformulate this to make it true?

user42912
  • 23,582
  • It's true if everything is positive. – Michael Burr Jun 02 '17 at 12:58
  • The wikipedia entry states that $b$ and $c$ should be integers. This would also fix this issue. – Michael Burr Jun 02 '17 at 13:02
  • 1
    Well, the sites you are reading have certainly made the mistake of not being clear about the conditions assumed for $a,b,c$... but you making an equal mistake by assuming you know what they are allowed to be. This can happen often, and one just has to be prepared to meet unqualified statements with skepticism. The right question is something like "what were the intended constraints?" which is basically what you mean by reformulation, I guess. – rschwieb Jun 02 '17 at 13:02
  • @MichaelBurr good job! Someone has to fix it – user42912 Jun 02 '17 at 13:11

4 Answers4

6

Why does every site I read say this wrong rule?

All these sites you found appear to be aimed at a beginning algebra audience. At this level, typically all bases are positive and all exponents are integers (at least to begin with). And with positive bases and integer exponents, it is indeed the case that $(a^m)^n = a^{mn}$. At this level, in the US at least, one should never expect to see something like $(-2)^{1/2}$ because that ventures into the realm of imaginary numbers, which is not typically covered until 2-3 semesters after beginning algebra.

how can I reformulate this to make it true?

All you really need is $a > 0$, then for any $m, n \in \Bbb R$, we have $(a^m)^n = a^{mn}$. If $a = 0$ then you have to worry about negative exponents because they'll lead to division by zero. If $a < 0$ then it is not true in general that $(a^m)^n = a^{mn}$, as you've already noted with your example.

2

One way would be to simply require that $a>0$. Another way (working in the complex plane) would be to note that given $a\neq0$, both $(a^b)^c$ and $a^{bc}$ could have multiple values, but both must have some values (at least one) in common.

Ethan Bolker
  • 95,224
  • 7
  • 108
  • 199
SomeStrangeUser
  • 3,033
0

Remember that, in general, the exponential function is not injective. That is, you have to be careful about which solutions you're pulling out of the equation. Plus, if $a^x = a^y$ then you can't assume that $x = y$

Note that $((-2)^2)^{\frac12} = \pm 2$, one of those solutions remain when you apply the exponential rule. One way to think of it is that the exponential rule is always true, but you should be sure to check your solutions in the original equation.

Kaynex
  • 2,448
  • I've already seen definitions of exponential such that $(2^2)^{1/2}=2$. Thank you for your answer. – user42912 Jun 02 '17 at 13:51
  • Then you're not using a common definition, and that's where the problem lies. – Kaynex Jun 02 '17 at 13:59
  • Could you give me a reference saying this definition is the common one? Thank you again. – user42912 Jun 02 '17 at 14:01
  • Nope. References aren't how one does math. So, please notice how $(-2)(-2) = 4$. That means that it is perfectly permissible to write $\sqrt{4} = -2$. Of course, that's only half the story here, as another obvious square root of $4$ is $2$. In your case, you had $((-2)^2)^{\frac12}$. Now, a clear simplification here is $4^{\frac12}$ and bingo! As I showed above, both $\pm 2$ are solutions to this. Now, I think we're talking about the same algebra here, so there shouldn't be any confusion. – Kaynex Jun 02 '17 at 16:09
  • Especially if you're going to be playing with exponentials, you shouldn't ignore answers just because you like certain definitions of functions. Looks like the rigor mortis that is so common around here is getting in the way of any real mathematics. – Kaynex Jun 02 '17 at 16:18
0

\begin{align} (a^3)^4 & = \overbrace{(a^3)(a^3)(a^3)(a^3)}^{\large\text{4 factors}} = a^{\overbrace{3+3+3+3}^{\Large\text{4 terms}}} = a^{4\times3} \\[12pt] (a^3)^4 & = \underbrace{(aaa)(aaa)(aaa)(aaa)}_{\large\text{Adding four 3s yields } 4\times 3} = a^{4\times3} \end{align}

Reasonable textbooks will state that these things apply when $a\ge0$, and also apply when the exponents (in this case $3$ and $4$) are integers. Understanding what is displayed above should make it clear why the equalities hold when the exponents are nonnegative integers. Why they also hold when the exponents are negative integers is not hard to figure out.

Taking $n$th roots of negative numbers, however, is problematic; for example $25$ has two square roots: $\pm5.$

Michael Hardy
  • 1