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There are many curves that extend integer exponentiation to larger domains, so why was this one chosen?

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Because $e$ is the most awesome number, clearly. – DavidP Feb 7 '14 at 19:57
Arguably, any choice of exponential and its logarithm could be chosen but $e$ and $\ln$ play well with differentiation (in that they don't pick up factors of $\ln(a)$ if your base was $a$). – Cameron Williams Feb 7 '14 at 19:59
Exponentiation can be defined in different manners, for example by sequences. – Jose Antonio Mar 2 '14 at 5:05
up vote 5 down vote accepted

I believe there is only one function $f(x, y)$, $f: \mathbb{R}^{+} \times \mathbb{R} \rightarrow \mathbb{R}$ satisfying

  1. $f$ is continuous,
  2. $f(x, 1) = x$,
  3. $f(x, y + z) = f(x, y) f(x, z)$,

and that is $f(x, y) = x^y = e^{\ln(x) y}$, where $e^x$ is defined by the exponential function. Note that 2) and 3) fix the values of the function at rational $y$ (e.g. you can see that $f(x, 1/3) = \sqrt[3]{x}$, for example) with odd denominator in lowest terms, and 1) fixes them at the other rational and at irrational $y$. 3) is just the exponent law $x^{n + m} = x^n x^m$ for integer exponents assumed to hold for real ones.

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Very cool. Do you know where I can find a formal proof of this? – enthdegree Feb 7 '14 at 22:50
I don't know of a source offhand, though Wikipedia mentions for the $x = e$ case to look at Rudin's 1976 analysis text, chapter 8, but says it's an exercise there. However, I don't think it should be too hard: just show the difference between any two candidates assumed distinct is zero on a dense set, and then by continuity it must be zero everywhere, giving a contradiction, and a proof of the uniqueness. Then just show that $e^{\ln(x) y}$ satisfies these properties. – mike4ty4 Feb 7 '14 at 23:06

It's economic. You define exponentiation to one base, say $e$, rigorously and reduce all other exponentiation to that case.

Edit: The reason why mathematicians chose just $e$ instead of $42$ for that purpose has reasons which go way beyond the scope of the OP's question.

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Why did this answer received two downvotes yesterday? What's wrong? – Michael Hoppe Mar 8 '14 at 18:18

Because it has some very useful properties. For example

$$ \frac{d}{dx} e^x = e^x $$

but with another base you would need a constant.

An analogy is using radians instead of degrees once you get more advanced at trigonometry. Here again

$$ \frac{d}{dx} \sin x = \cos x $$

when $x$ is in radians. In other representations you need a constant.

Unlike the radians case, $e$ has some other very special properties that are both practical and worthy of study for their own sake.

For example $e^x$ has a very simple infinite series representation: $$ e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \ldots $$

where $n!$ is factorial($n$).

and there is a relationship between $e$ and the trigonmetric functions $\sin$ and $\cos$:

$$ e^{ix} = \cos x + i\sin x $$

where $i$ is the imaginary number $\sqrt{-1}$.

So switching to using $e$ has some practical benefits at first, but as you go further it becomes quite fundamental to many branches of mathematics.

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What you want is that $x^y$ be continuous both in $x$ and $y$, and to coincide with the (real branch of) $x^y$ for rational $y$. For simplicity, apply natural logarithms to the relevant equations.

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$f(x)=e^{\ln(x)\cdot y}$ is the unique form satisfying natural exponentiation:

\begin{equation}\tag{1}f(x,0) = 1\end{equation}

\begin{equation}\tag{2} x\cdot f(x,y) = f(x,y+1)\end{equation}

and differentiability: \begin{equation}\tag{3}\frac{d}{dx}f(x,y)=y\cdot f(x,y-1)\end{equation}

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(Except for details around $x=0$) – enthdegree Feb 7 '14 at 21:22
What do you mean by: "satisfying natural exponentiation"? – goblin Feb 28 '14 at 23:27
@user18921 Exponentiation of natural numbers. The first two properties take care of this: $f(x,0)=1$, so by $(2)$, $f(x,1)=x$ and so on inductively. – enthdegree Mar 2 '14 at 4:54
Ah I see. I would prefer to phrase it: "extending natural exponentiation" and "behaving as expected under differentiation" – goblin Mar 2 '14 at 5:35
Thank u 4 sharing, im glad u undrst& now – enthdegree Mar 7 '14 at 5:37

Hint: Do you know the property $a^{\log_bc}=c^{\log_ba}$. Clearly this applies here. Try to think of the proof for the above expression.

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This property relies on the definition of $e^x$. – Michael Hoppe Feb 7 '14 at 22:28
@MichaelHoppe, that doesn't matter though, since $e^x$ is just shorthand for $\exp(x),$ and (somewhat ironically) the $\exp$ function can be defined without reference to exponentiation. – goblin Feb 28 '14 at 23:26

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