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\section{Power Laws}

\subsection{Definitions}

Definitions for power and log:
\begin{equation}\label{powerdef}
x^a \equiv e^{a \ln x}
\end{equation}
\begin{equation}
\ln x \equiv \ln |x| + i \arg(x) \mbox{ where } -\pi < \arg(x) \le \pi
\end{equation}

\subsection{General rules}

\begin{equation}
e^x e^y = e^{x+y}
\end{equation}
for arbitrary complex \(x\) and \(y\) (with~(\ref{powerdef}) we obtain
the rule \(x^ax^b=x^{a+b}\) since \(x^ax^b\equiv e^{a\ln x}e^{b\ln x} = 
e^{(a+b)\ln x}\equiv x^{a+b}\) for arbitrary complex \(a,b,x\))

\begin{equation}
x^{-a} = \frac{1}{x^a}
\end{equation}
for arbitrary complex \(x\) and \(a\)

\subsection{\((ax)^b=a^b x^b\)}

\subsubsection{\(b\) integer, \(x\) and \(a\) arbitrary complex}

assume \(b>0\)

\begin{eqnarray}
(ax)^b & = & \underbrace{(ax) \cdots (ax)}_{b \times}
\nonumber\\
& = & \underbrace{a \cdots a}_{b \times}
      \underbrace{x \cdots x}_{b \times}
\nonumber\\
& = & a^b x^b \mbox{ q.e.d.}
\end{eqnarray}

if \(b<0\) (so \(b=-|b|\))
\begin{eqnarray}
(ax)^b & = & \frac{1}{(ax)^{|b|}}
\nonumber\\
& = & \frac{1}{a^{|b|} x^{|b|}}
\nonumber\\
& = & a^{-|b|} x^{-|b|}
\nonumber\\
& = & a^b x^b
\end{eqnarray}

\subsubsection{\(a>0\), \(x\) and \(b\) arbitrary complex}

\begin{eqnarray}
(ax)^b & = & e^{b \ln(ax)}
\nonumber\\
& = & e^{b (\ln |ax| + i \arg(ax))}
\end{eqnarray}

if \(a\) is real and positive:
\begin{equation}
\ln |ax| = \ln |a| + \ln |x| = \ln a + \ln |x|
\end{equation}
and 
\begin{equation}
\arg(ax) = \arg(x)
\end{equation}

So
\begin{eqnarray}
e^{b (\ln |ax| + i \arg(ax))} & = &
e^{b (\ln a + \ln |x| + i \arg(x))}
\nonumber\\
& = & e^{b (\ln a + \ln x)}
\nonumber\\
& = & e^{b \ln a} e^{b \ln x}
\nonumber\\
& = & a^b x^b \mbox{ q.e.d.}
\end{eqnarray}

\subsection{\((x^a)^b = x^{ab}\)}

\subsubsection{\(b\) integer, \(x\) and \(a\) arbitrary complex}

assume \(b>0\)

\begin{eqnarray}
(x^a)^b & = & \underbrace{(x^a) \cdots (x^a)}_{b \times}
\nonumber\\
& = & \underbrace{e^{a \ln x} \cdots e^{a \ln x}}_{b \times}
\nonumber\\
& = & e^{\underbrace{\scriptstyle a \ln x + \dots + a \ln x}_{b \times}}
\nonumber\\
& = & e^{a b \ln x}
\nonumber\\
& = & x^{ab} \mbox{ q.e.d.}
\end{eqnarray}

if \(b<0\) (so \(b=-|b|\))
\begin{eqnarray}
(x^a)^b & = & \frac{1}{(x^a)^{|b|}}
\nonumber\\
& = & \frac{1}{x^{a|b|}}
\nonumber\\
& = & x^{-a|b|}
\nonumber\\
& = & x^{ab}
\end{eqnarray}

\subsubsection{\(-1 < a \le 1\), \(x\) and \(b\) arbitrary complex}

We have
\begin{equation}
x^a=e^{a \ln|x| + ia\arg(x)}
\end{equation}
if \(a\) is real
\begin{equation}
|x^a|=e^{a\ln|x|}
\end{equation}
and
\begin{equation}
\arg(x^a)-a\arg(x)=2k\pi
\end{equation}
now if \(-1 < a \le 1\), then \(-\pi < a\arg(x) \le \pi\),
and so \(k=0\), i.e.
\begin{equation}
\arg(x^a)=a\arg(x)
\end{equation}
(Note that for \(a=-1\) this may not be true, as \(-1 \arg(x)\) may be equal to \(-\pi\).)
So
\begin{eqnarray}
\ln(x^a) & = & \ln|x^a| + i\arg(x^a)
\nonumber\\
& = & \ln (e^{a\ln|x|})+ia\arg(x)
\nonumber\\
& = & a \ln |x| + ia\arg(x) \mbox{ (because \(a\ln|x|\) is real)}
\nonumber\\
& = & a\ln x
\end{eqnarray}
Hence
\begin{eqnarray}
(x^a)^b & = & e^{b\ln x^a}
\nonumber\\
& = & e^{ba\ln x}
\nonumber\\
& = & x^{ab} \mbox{ q.e.d.}
\end{eqnarray}

proof contributed by Adam Strzebonski from Wolfram Research
({\tt adams@wolfram.com}) in newsgroup {\tt sci.math.symbolic}.

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