Lösung 3.3:2a

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The logarithm <math>\mathop{\text{lg}} 0\textrm{.}1</math> is defined as that number which should stand in the coloured box in order that the equality
The logarithm <math>\mathop{\text{lg}} 0\textrm{.}1</math> is defined as that number which should stand in the coloured box in order that the equality
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{{Displayed math||<math>10^{\bbox[#FFEEAA;,1.5pt]{\phantom{\scriptstyle ??}}} = 0\textrm{.}1</math>}}
+
{{Abgesetzte Formel||<math>10^{\bbox[#FFEEAA;,1.5pt]{\phantom{\scriptstyle ??}}} = 0\textrm{.}1</math>}}
should hold. In this case, we see that
should hold. In this case, we see that
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{{Displayed math||<math>10^{-1} = 0\textrm{.}1</math>}}
+
{{Abgesetzte Formel||<math>10^{-1} = 0\textrm{.}1</math>}}
and therefore <math>\mathop{\text{lg}} 0\textrm{.}1 = -1\,</math>.
and therefore <math>\mathop{\text{lg}} 0\textrm{.}1 = -1\,</math>.

Version vom 08:41, 22. Okt. 2008

The logarithm \displaystyle \mathop{\text{lg}} 0\textrm{.}1 is defined as that number which should stand in the coloured box in order that the equality

\displaystyle 10^{\bbox[#FFEEAA;,1.5pt]{\phantom{\scriptstyle ??}}} = 0\textrm{.}1

should hold. In this case, we see that

\displaystyle 10^{-1} = 0\textrm{.}1

and therefore \displaystyle \mathop{\text{lg}} 0\textrm{.}1 = -1\,.

Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse1-TU-Berlin/index.php/L%C3%B6sung_3.3:2a“