Lösung 1.3:4a

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{{Abgesetzte Formel||<math>2^{9}\cdot 2^{-7} = 2^{9-7} = 2^{2} = 4\,</math>.}}
{{Abgesetzte Formel||<math>2^{9}\cdot 2^{-7} = 2^{9-7} = 2^{2} = 4\,</math>.}}
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Alternativ schreibt man alle Terme explizit, und kürzt den Bruch
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Alternativ kann man auch alle Terme explizit aufschreiben und kürzt den Bruch
{{Abgesetzte Formel||<math>\begin{align}
{{Abgesetzte Formel||<math>\begin{align}
2^{9-7} &= 2\cdot 2\cdot {}\rlap{/}2\cdot {}\rlap{/}2\cdot {}\rlap{/}2\cdot {}\rlap{/}2\cdot {}\rlap{/}2\cdot {}\rlap{/}2\cdot {}\rlap{/}2\cdot \frac{1}{{}\rlap{/}2\cdot {}\rlap{/}2\cdot {}\rlap{/}2\cdot {}\rlap{/}2\cdot {}\rlap{/}2\cdot {}\rlap{/}2\cdot {}\rlap{/}2}\\[5pt]
2^{9-7} &= 2\cdot 2\cdot {}\rlap{/}2\cdot {}\rlap{/}2\cdot {}\rlap{/}2\cdot {}\rlap{/}2\cdot {}\rlap{/}2\cdot {}\rlap{/}2\cdot {}\rlap{/}2\cdot \frac{1}{{}\rlap{/}2\cdot {}\rlap{/}2\cdot {}\rlap{/}2\cdot {}\rlap{/}2\cdot {}\rlap{/}2\cdot {}\rlap{/}2\cdot {}\rlap{/}2}\\[5pt]
&= 2\cdot 2 = 4\,\textrm{.}\end{align}</math>}}
&= 2\cdot 2 = 4\,\textrm{.}\end{align}</math>}}

Aktuelle Version

Nachdem die beiden Potenzen dieselbe Basis haben, können wir die Rechenregel für Multiplikation von Potenzen verwenden

\displaystyle 2^{9}\cdot 2^{-7} = 2^{9-7} = 2^{2} = 4\,.

Alternativ kann man auch alle Terme explizit aufschreiben und kürzt den Bruch

\displaystyle \begin{align}

2^{9-7} &= 2\cdot 2\cdot {}\rlap{/}2\cdot {}\rlap{/}2\cdot {}\rlap{/}2\cdot {}\rlap{/}2\cdot {}\rlap{/}2\cdot {}\rlap{/}2\cdot {}\rlap{/}2\cdot \frac{1}{{}\rlap{/}2\cdot {}\rlap{/}2\cdot {}\rlap{/}2\cdot {}\rlap{/}2\cdot {}\rlap{/}2\cdot {}\rlap{/}2\cdot {}\rlap{/}2}\\[5pt] &= 2\cdot 2 = 4\,\textrm{.}\end{align}

Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse1-TU-Berlin/index.php/L%C3%B6sung_1.3:4a“