\ANG{
\begin{enumerate}
\item Gib die Definitionen der folgenden Potenzschreibweisen an:\\
\begin{tabular}{lll}
(1) & $a^n$ & $\left( a \in \RR; n \in \NN \setminus \lbrace 0 \rbrace \right)$\\
(2) & $a^{-n}$ & $\left( a \in \RR \setminus \lbrace 0 \rbrace; n \in \NN \setminus \lbrace 0 \rbrace \right)$\\
(3) & $a^0$ & $\left( a \in \RR \setminus \lbrace 0 \rbrace; \right)$\\
(4) & $a^{\frac 1 n}$ & $\left( a \in \RR^+; n \in \NN \setminus \lbrace 0 \rbrace \right)$\\
\end{tabular}
\item Berechne ohne Verwendung eines Taschenrechners.\\
(1) $8^3 \qquad$ (2) $8^{-3} \qquad$ (3) $8^{\frac 1 3} \qquad$ (4) $8^{-\frac 2 3}$
\item Vervollständige für $a,b \in \RR^+$ und $p,q \in \RR$ in der linken Spalte die Potenzgesetze in allgemeiner Form und vereinfache damit die Terme in den anderen Spalten.\\[1.5ex]
\begin{tabular}{|l|l|l|}  \hline
Potenzgesetz  & Beispiel 1 & Beispiel 2 \\ \hline
$a^p\cdot a^q = \phantom{\D\frac X Y+1111}$  & $x^2 \cdot x^{-3} = \phantom{\D\frac X Y+1111}$ & $2^x \cdot 2^{2x+1} = \phantom{\D\frac X Y+111111111}$ \\[1.4ex] \hline
$\D\frac {a^p}{a^q} = $ & $\D\frac {x^{\frac 1 2}}{x^{-2}} = $ & $\D\frac{2^x}{2^{x-1}} = $ \\[1.4ex] \hline
$\left(a^p\right)^q = \phantom{\D\frac X Y+1111}$ & $\left(x^4\right)^{\frac 1 2} = $ & $\left(2^x \right)^2 =$ \\[1.4ex] \hline 
$a^p \cdot b^p = \phantom{\D\frac X Y+1111}$ & $\left(-2x\right)^3 \cdot \left( \frac 1 x \right)^3 = \phantom{\D\frac X Y+111111}$ & $2^x \cdot \left( \frac 1 2 \right)^x = $ \\[1.4ex] \hline
$\frac {a^p}{b^p} = \phantom{\D\frac X Y+1111}$ & $\frac {\left(2x \right)^{0,5}}{\sqrt 2} = $ & $\frac {0,5^x}{1,5^x} = $\\[1.4ex] \hline
\end{tabular}
\item Kreuze jeweils auf der rechten Seite alle Terme an, die äquivalent zum Term auf der linken Seite sind.\\[1.25ex]
\begin{tabular}{|l|l|l|} \hline
(1) & $a^3 \cdot a^2 $ &
\begin{tabular}{p{3cm}p{3cm}p{3cm}}
  $\square \; a^6$  & $\square \; a^5$  & $\square \; a^2 \cdot \left( a \cdot 1 \right)$ \\
  $\square \; \left(a^3\right)^2$  & $\square \; a^2 \cdot \left( a + 1 \right)$ & $\square \; a^3+a^2$
\end{tabular}
\\ \hline
(2) & $2^x \cdot 2^{x+1}$ & 
\begin{tabular}{p{3cm}p{3cm}p{3cm}}
  $\square \; 2^{x^2+x}$  
  & $\square \; 4^{2x+1}$  
  & $\square \; 4^{x^2+x}$ \\
  $\square \; 2^{2x+1}$  
  & $\square \; 2^{x^2+1}$ 
  & $\square \; 4^{x^2+1}$
\end{tabular}
\\ \hline
(3) & $\left( \frac 1 3 \right)^p$ & 
\begin{tabular}{p{3cm}p{3cm}p{3cm}}
  $\square \; \sqrt[3] p$  
  & $\square \; p^{-3}$  
  & $\square \; 3^{-p}$ \\
  $\square \; \frac 1 {3^p}$  
  & $\square \; \left(-3\right)^p$ 
  & $\square \; 3^{\frac 1 p}$ \\
  $\square \; 1^p-3^p$  
  & $\square \; \left(3^p\right)^{-1}$ 
  & $\square \; p^1-p^3$ 
\end{tabular}
\\ \hline
(4) & $a^{p+q}-a^p$ & 
\begin{tabular}{p{3cm}p{3cm}p{3cm}}
  $\square \; \frac {a^{p+q}}{a^p}$  
  & $\square \; a^p \cdot \left( a^q - 1\right)$  
  & $\square \; a^q$ \\
  $\square \; a^p \cdot a^q - a^p$  
  & $\square \; \frac {a^p \cdot a^q}{a^p}$ 
  & $\square \; a^q \cdot \left(a^p - a^{p-q} \right)$
\end{tabular}
\\ \hline
\end{tabular}

\end{enumerate}
}
\LSG{
\begin{enumerate}
\item (1) $a^n = \underbrace{a \cdot a \cdot \cdots \cdot a}_{n mal} \qquad$
      (2) $a^{-n} = \frac 1 {a^n} \qquad $
      (3) $a^0 = 1 \qquad$
      (4) $a^{\frac 1 n} = \sqrt [n] a \qquad$
\item (1) $8^3 = 512 \qquad$
      (2) $8^{-3} = \frac 1 {8^3} = \frac 1 {512}$
      (3) $8^{\frac 1 3} = \sqrt[3] 8 = 2 \qquad $
      (4) $8^{-\frac 2 3} = \frac 1 {8^{\frac 2 3}} = \frac 1 {\sqrt [3] 8^2} = \frac 1 4\qquad$
\item
\begin{tabular}{|p{3cm}|p{5cm}|p{5cm}|}  \hline
Potenzgesetz  & Beispiel 1 & Beispiel 2 \\ \hline
$a^p\cdot a^q = a^{p+q}$  
& $x^2 \cdot x^{-3} = x^{-1}$ 
& $2^x \cdot 2^{2x+1} = 2^{3x+1}$ 
\\[1.4ex] \hline
$\D\frac {a^p}{a^q} = a^{p-q}$ 
& $\D\frac {x^{\frac 1 2}}{x^{-2}} = x^{\frac 5 2}$ 
& $\D\frac{2^x}{2^{x-1}} = 2^1 = 2$ 
\\[1.4ex] \hline
$\left(a^p\right)^q = a^{p \cdot q}$ 
& $\left(x^4\right)^{\frac 1 2} = x^2$ 
& $\left(2^x \right)^2 = 2^{2x} = \left(2^2\right)^x = 4^x$ 
\\[1.4ex] \hline 
$a^p \cdot b^p = \left( a \cdot b \right)^p$ 
& $\left(-2x\right)^3 \cdot \left( \frac 1 x \right)^3 = (-2)^3 = -8$ 
& $2^x \cdot \left( \frac 1 2 \right)^x = \left( 2 \cdot \frac 1 2 \right)^x = 1^x = 1$ \\[1.4ex] \hline
$\frac {a^p}{b^p} = \left(\frac a b\right)^p$ 
& $\frac {\left(2x \right)^{0,5}}{\sqrt 2} = \left(\frac {2x} 2\right)^{0,5} = \sqrt x$ 
& $\frac {0,5^x}{1,5^x} = \left(\frac 1 3\right)^x$\\[1.4ex] 
\hline
\end{tabular}
\item 
\begin{tabular}{|l|l|l|} \hline
(1) & $a^3 \cdot a^2 $ &
\begin{tabular}{p{3cm}p{3cm}p{3cm}}
  $\square \; a^6$  & $\blacksquare \; a^5$  & $\square \; a^2 \cdot \left( a \cdot 1 \right)$ \\
  $\square \; \left(a^3\right)^2$  & $\square \; a^2 \cdot \left( a + 1 \right)$ & $\square \; a^3+a^2$
\end{tabular}
\\ \hline
(2) & $2^x \cdot 2^{x+1}$ & 
\begin{tabular}{p{3cm}p{3cm}p{3cm}}
  $\square \; 2^{x^2+x}$  
  & $\square \; 4^{2x+1}$  
  & $\square \; 4^{x^2+x}$ \\
  $\blacksquare \; 2^{2x+1}$  
  & $\square \; 2^{x^2+1}$ 
  & $\square \; 4^{x^2+1}$
\end{tabular}
\\ \hline
(3) & $\left( \frac 1 3 \right)^p$ & 
\begin{tabular}{p{3cm}p{3cm}p{3cm}}
  $\square \; \sqrt[3] p$  
  & $\square \; p^{-3}$  
  & $\square \; 3^{-p}$ \\
  $\blacksquare \; \frac 1 {3^p}$  
  & $\square \; \left(-3\right)^p$ 
  & $\square \; 3^{\frac 1 p}$ \\
  $\square \; 1^p-3^p$  
  & $\blacksquare \; \left(3^p\right)^{-1}$ 
  & $\square \; p^1-p^3$ 
\end{tabular}
\\ \hline
(4) & $a^{p+q}-a^p$ & 
\begin{tabular}{p{3cm}p{3cm}p{3cm}}
  $\square \; \frac {a^{p+q}}{a^p}$  
  & $\blacksquare \; a^p \cdot \left( a^q - 1\right)$  
  & $\square \; a^q$ \\
  $\blacksquare \; a^p \cdot a^q - a^p$  
  & $\square \; \frac {a^p \cdot a^q}{a^p}$ 
  & $\blacksquare \; a^q \cdot \left(a^p - a^{p-q} \right)$
\end{tabular}
\\ \hline
\end{tabular}

\end{enumerate}
}