Lösung 1.3:2a

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We can write every factor in the expression as a power of 2,
We can write every factor in the expression as a power of 2,
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{{Displayed math||<math>\begin{align}
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{{Abgesetzte Formel||<math>\begin{align}
2 &= 2^{1}\,, \\
2 &= 2^{1}\,, \\
4 &= 2\cdot 2 = 2^{2}\,,\\
4 &= 2\cdot 2 = 2^{2}\,,\\
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which gives
which gives
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{{Displayed math||<math>2\cdot 4\cdot 8 = 2^{1}\cdot 2^{2}\cdot 2^{3} = 2^{1+2+3} = 2^{6}\,</math>.}}
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{{Abgesetzte Formel||<math>2\cdot 4\cdot 8 = 2^{1}\cdot 2^{2}\cdot 2^{3} = 2^{1+2+3} = 2^{6}\,</math>.}}

Version vom 08:16, 22. Okt. 2008

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

\displaystyle \begin{align}

2 &= 2^{1}\,, \\ 4 &= 2\cdot 2 = 2^{2}\,,\\ 8 &= 2\cdot 4 = 2\cdot 2\cdot 2 = 2^{3}\,, \end{align}

which gives

\displaystyle 2\cdot 4\cdot 8 = 2^{1}\cdot 2^{2}\cdot 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“