Lösung 3.3:3f
Aus Online Mathematik Brückenkurs 1
(Unterschied zwischen Versionen)
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K (Robot: Automated text replacement (-{{Displayed math +{{Abgesetzte Formel)) |
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| Zeile 1: | Zeile 1: | ||
If we write 4 and 16 as | If we write 4 and 16 as | ||
| - | {{ | + | {{Abgesetzte Formel||<math>\begin{align} |
4 &= 2\cdot 2 = 2^2\,,\\[5pt] | 4 &= 2\cdot 2 = 2^2\,,\\[5pt] | ||
16 &= 2\cdot 8 = 2\cdot 2\cdot 4 = 2\cdot 2\cdot 2\cdot 2 = 2^4\,, | 16 &= 2\cdot 8 = 2\cdot 2\cdot 4 = 2\cdot 2\cdot 2\cdot 2 = 2^4\,, | ||
| Zeile 8: | Zeile 8: | ||
we obtain | we obtain | ||
| - | {{ | + | {{Abgesetzte Formel||<math>\begin{align} |
\log_2 4 + \log_2\frac{1}{16} | \log_2 4 + \log_2\frac{1}{16} | ||
&= \log_2 2^2 + \log_2\frac{1}{2^4}\\[5pt] | &= \log_2 2^2 + \log_2\frac{1}{2^4}\\[5pt] | ||
Version vom 08:43, 22. Okt. 2008
If we write 4 and 16 as
| \displaystyle \begin{align}
4 &= 2\cdot 2 = 2^2\,,\\[5pt] 16 &= 2\cdot 8 = 2\cdot 2\cdot 4 = 2\cdot 2\cdot 2\cdot 2 = 2^4\,, \end{align} |
we obtain
| \displaystyle \begin{align}
\log_2 4 + \log_2\frac{1}{16} &= \log_2 2^2 + \log_2\frac{1}{2^4}\\[5pt] &= \log_2 2^2 + \log_2 2^{-4}\\[5pt] &= 2\cdot\log_2 2 + (-4)\cdot\log_2 2\\[5pt] &= 2\cdot 1 + (-4)\cdot 1\\[5pt] &= -2\,\textrm{.} \end{align} |
