I know that in some cases this rule is not true. For example
$$((-1)^2)^\frac{1}{2}\ne(-1)^{(2\cdot\frac{1}{2})}$$
So when is this rule true ?
$\begingroup$
$\endgroup$
-
$\begingroup$ from wikipedia, it's true while $a>0 \quad and \quad b,c \in \mathbb{R}$ $\endgroup$ – Baconaro Oct 24 '15 at 16:42
-
1$\begingroup$ Related: For which complex $a$, $b$, $c$ does $(a^b)^c=$a^{bc}$ hold? $\endgroup$ – Andrew D. Hwang Oct 24 '15 at 16:51
$\begingroup$
$\endgroup$
It is true when $a>0$ and $b, c \in \mathbb R$. In this situation we have a canonical choice for a value for $a^b$, given by $e^{b \ln a}$, using the principal branch of the natural logarithm on positive real numbers (that is, the one that gives $\ln a \in \mathbb R$).
For other values of $a$ there is no natural choice for $\ln a$, so $a^b$ has multiple equally valid answers. What is true is that each value of $a^{bc}$ is equal to a value for $(a^b)^c$, but not conversely. For instance, in the example you gave, $a=-1, b = 2, c = \frac{1}{2}$, the values of $a^{bc}$ are
$$(-1)^1 = e^{\ln(-1)} = e^{\pi i + 2\pi i n} = -1,$$
and the values of $(a^b)^c$ are
$$( (-1)^2)^{\frac{1}{2}} = 1^{\frac{1}{2}} = e^{\frac{1}{2} \ln(1)} = e^{\frac{1}{2} 2\pi i n} = \pm 1.$$
answered Oct 24 '15 at 17:00
-
$\begingroup$ And if we separately discuss $a=0$ on which $\log a$ is undefined we are done. $\endgroup$ – Alice Ryhl Oct 24 '15 at 18:12
-
$\begingroup$ Actually, I've never seen what a good description of $0^z$ should be in general... Obviously $0^a = 0$ for $a \in \mathbb R^{>0}$, and I know $0^0$ is indeterminate, I believe with possible values in the interval $[0,1]$? And it only makes sense to have $0^{a}$ be undefined/infinite for $a \in \mathbb R^{<0}$. Not sure about more general complex values of $z$. $\endgroup$ – Dustan Levenstein Oct 24 '15 at 18:19
-
$\begingroup$ @DustanLevenstein: It is a misunderstanding to even use the word "indeterminate" unless you're looking at a limit. The expression $0^0$ unambiguously has the value $1$ -- "indeterminate" applies only if you have a limit of the form $f(x)^{g(x)}$ where $f(x)$ and $g(x)$ both go towards $0$. In that case "indeterminate" means neither more nor less than the fact that knowing these limits does not determine what the limit of $f(x)^{g(x)}$ is. $\endgroup$ – hmakholm left over Monica Oct 25 '15 at 12:33
-
$\begingroup$ @HenningMakholm yes, I was referring implicitly to limits; it's just been a long time since I've done Calculus, so I didn't make it explicit when I should have. I don't think it's unfair to say that $0^0$ itself is "multivalued" or "indeterminate". Assigning a value of $1$ to $0^0$ is useful, for many reasons in mathematics, but at the same time I think it's misleading to say that the value of $0^0$ is unambiguous. $\endgroup$ – Dustan Levenstein Oct 25 '15 at 12:57
$\begingroup$
$\endgroup$
The main two cases where this rule holds are
- When $a$ is a positive real (and $b, c$ are real).
- When $b$ and $c$ are both integers.
In the latter case it holds not only for multiplication of numbers, but in any group.
The prototypical case of exponentiation, where base and exponent are both positive integers, is a special case of each of these situations.
answered Oct 24 '15 at 16:44
hmakholm left over Monicahmakholm left over Monica
268k20 gold badges370 silver badges624 bronze badges
$\begingroup$
$\endgroup$
If $b,c$ are integer numbers than the rule $(a^b)^c=a^{bc}$ is true for all $a \in \mathbb{R}$. If at least one of the two exponent is a real non integer number than the rule is true only for non negative basis $a$.
answered Oct 24 '15 at 16:43
