Lösung 1.3:2a

Aus Online Mathematik Brückenkurs 1

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{{NAVCONTENT_START}}
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We can write every factor in the expression as a power of
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<center> [[Image:1_3_2a.gif]] </center>
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<math>2</math>,
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{{NAVCONTENT_STOP}}
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<math>\begin{align}
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& 2=2^{1} \\
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& \\
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& 4=2\centerdot 2=2^{2} \\
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& \\
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& 8=2\centerdot 4=2\centerdot 2\centerdot 2=2^{3} \\
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\end{align}</math>
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which gives
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<math>2\centerdot 4\centerdot 8=2^{1}\centerdot 2^{2}\centerdot 2^{3}=2^{1+2+3}=2^{6}</math>

Version vom 11:12, 15. Sep. 2008

We can write every factor in the expression as a power of \displaystyle 2,


\displaystyle \begin{align} & 2=2^{1} \\ & \\ & 4=2\centerdot 2=2^{2} \\ & \\ & 8=2\centerdot 4=2\centerdot 2\centerdot 2=2^{3} \\ \end{align}


which gives


\displaystyle 2\centerdot 4\centerdot 8=2^{1}\centerdot 2^{2}\centerdot 2^{3}=2^{1+2+3}=2^{6}

Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse1-TU-Berlin/index.php/L%C3%B6sung_1.3:2a“