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In my understanding, $64^{\frac23}$ can be defined as $(64^{\frac13})^2=4^2=16.$ How can I define $64^x$ when $x$ is real?

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marked as duplicate by mechanodroid, user99914, Ross Millikan, Isaac Browne, Namaste Jul 3 '18 at 18:39

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    $\begingroup$ $\frac 23 \in \mathbb R$. So you're using a real exponent of base 64 to caluculate $64^{2/3}$, and only then asking how to define $64^x$, where $x \in \mathbb R$? $\endgroup$ – Namaste Jul 3 '18 at 18:43
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The usual definition of $a^x$, when $a\in\mathbb{R}^+$ and $x\in\mathbb R$, is$$a^x=e^{x\log a}.$$Of course, this leads to the question “what is $e^x$?”. You can define it as$$\sum_{n=0}^\infty\frac{x^n}{n!}$$or as$$\lim_{n\to\infty}\left(1+\frac xn\right)^n.$$There are still other possibilities.

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  • $\begingroup$ But don't you first have to define $a^b$ when $b$ is real to define "$\log a$"? $\endgroup$ – user503644 Jul 3 '18 at 9:46
  • $\begingroup$ Now. You can define $\log$ as the inverse of the exponential function $x\mapsto e^x$. Or you can define$$\log a=\int_1^a\frac{\mathrm dt}t.$$ $\endgroup$ – José Carlos Santos Jul 3 '18 at 9:47
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For a fixed $a\in \mathbb R$ with $a>0$ you know what $a^x$ is for $x\in \mathbb Q$. That means that we have a function $a^-\colon \mathbb Q \to \mathbb R$. What we want is to extend it to a function $a^-\colon \mathbb R \to \mathbb R$, and we want this new function to agree with the old function on rational values. In other words, we want to extend along the inclusion $\mathbb Q \to \mathbb R$. Such extension problems are common place and represent a healthy philosophy: start simple and extend to more complicated things gradually.

Now, there is a very general mechanism for such extensions due to Kan (and they are called Kan extensions). The full generality requires category theory, but for this case its much simpler. So, given $x\in \mathbb R$, we attempt to approximate what $a^x$ should be, preferably in terms of things we already know. So, we approximate $x$ by a rational value, say $y$, and we declare $a^y$ to be an approximation of $a^x$. Now, there are two options: do we approximate from above, namely $y>x$, or from below, namely $x<y$. Well, we can consider each option. Let's say we approximate from above. So, for all rational $y$ with $y>x$ we have a bunch of approximations $a^y$ for what we would like to call $a^x$. Since the infimum of the approximations $y$ give us $x$ again, we declare the infimum of the approximations $a^y$ to be the value $a^x$. And this works, in the sense that it's a good way to extend the original function, and many pleasant properties of it follow. What happens if we take approximations from below? Well, in this case, we get the same extension (but generally this will not be the case).

Remark: Many other examples of such Kan extensions in nature can be found in the excellent book "Category Theory in Context" by Emily Riehl,

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  • $\begingroup$ Also, there are other ways apart from above and below to approximate it $\endgroup$ – Holo Jul 3 '18 at 10:05
  • $\begingroup$ @Holo this approach does not require any sequences. Not only is it not a pain to prove the extension is well-defined, but in fact it follows (as stated) from very general principles of category theory. $\endgroup$ – Ittay Weiss Jul 3 '18 at 11:58
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I'll assume $a \geq 1$. One definition that I think is intuitive is $$a^x = \sup_{q \leq x, \,q \in \mathbb Q} a^q.$$

With the benefit of hindsight, people have found other definitions that reduce the amount of work that needs to be done to develop calculus rigorously. For example, Spivak defines $\log(x) = \int_1^x \frac{1}{t} \, dt$, then defines $\exp$ as the inverse function of log, then defines $a^x = \exp (\log(a) x)$.

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  • $\begingroup$ How can you be sure $a^x = \sup_{q \leq x, \,q \in \mathbb Q} a^q$ exists? $\endgroup$ – user503644 Jul 3 '18 at 9:54
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    $\begingroup$ @KentaSuzuki Bounded set in $\Bbb R$, so it has a least upper bound! $\endgroup$ – Saucy O'Path Jul 3 '18 at 10:00
  • $\begingroup$ Why is it bounded? $\endgroup$ – user503644 Jul 3 '18 at 10:00
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    $\begingroup$ @littleO The definition with $\sup$ works only with $a\ge 1$. For $0<a< 1$ you need to go the other way around. $\endgroup$ – Saucy O'Path Jul 3 '18 at 10:01
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    $\begingroup$ You could first prove that if $q$ and $p$ are rational numbers with $q< p$, then $a^q < a^p$. Then, select a rational number $p$ that is greater than $x$. Then $a^q < a^p$ for all rational numbers $q$ such that $q \leq x$. So the supremum is finite. $\endgroup$ – littleO Jul 3 '18 at 10:02