Lösung 1.3:3b

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If we divide
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<center> [[Image:1_3_3b.gif]] </center>
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<math>243</math>
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successively by
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<math>3</math>, we see that
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<math>\begin{align}
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& 243=3\centerdot 81=3\centerdot 3\centerdot 27=3\centerdot 3\centerdot 3\centerdot 9 \\
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& \\
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& =3\centerdot 3\centerdot 3\centerdot 3\centerdot 3=3^{5} \\
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\end{align}</math>

Version vom 11:34, 15. Sep. 2008

If we divide \displaystyle 243 successively by \displaystyle 3, we see that


\displaystyle \begin{align} & 243=3\centerdot 81=3\centerdot 3\centerdot 27=3\centerdot 3\centerdot 3\centerdot 9 \\ & \\ & =3\centerdot 3\centerdot 3\centerdot 3\centerdot 3=3^{5} \\ \end{align}

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