Why is exponentiation defined as $x^y=e^{\ln(x)\cdot y}$? There are many curves that extend integer exponentiation to larger domains, so why was this one chosen?
 A: 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.
A: I believe there is only one function $f(x, y)$, $f: \mathbb{R}^{+} \times \mathbb{R} \rightarrow \mathbb{R}$ satisfying


*

*$f$ is continuous,

*$f(x, 1) = x$,

*$f(x, y + z) = f(x, y) f(x, z)$,


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.
A: 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.
A: 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.
A: $f(x)=e^{\ln(x)\cdot y}$ 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}
A: 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.
