Lösung 1.3:4a

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Because the base is the same in both factors, the exponents can be combined according to the power rules
Because the base is the same in both factors, the exponents can be combined according to the power rules
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{{Displayed math||<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>.}}
Alternatively, the expressions for the powers can be expanded completely and then cancelled out,
Alternatively, the expressions for the powers can be expanded completely and then cancelled out,
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{{Displayed math||<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>}}

Version vom 08:17, 22. Okt. 2008

Because the base is the same in both factors, the exponents can be combined according to the power rules

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

Alternatively, the expressions for the powers can be expanded completely and then cancelled out,

\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“