Lösung 3.3:2f

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Instead of always going back to the definition of the logarithm, it is better to learn to work with the log laws,
Instead of always going back to the definition of the logarithm, it is better to learn to work with the log laws,
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:*<math>\ \lg (ab) = \lg a + \lg b</math>
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<math>\lg \left( ab \right)=\lg a+\lg b</math>
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:*<math>\ \lg a^{b} = b\lg a</math>
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and to simplify expressions first. By working in this way, one only needs, in principle, to learn that <math>\lg 10 = 1\,</math>.
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<math>\lg a^{b}=b\lg a</math>
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and to simplify expressions first. By working in this way, one only needs, in principle, to learn that
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<math>\text{lg 1}0\text{ }=\text{1}</math>.
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In our case, we have
In our case, we have
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{{Displayed math||<math>\lg 10^{3} = 3\cdot \lg 10 = 3\cdot 1 = 3\,\textrm{.}</math>}}
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<math>\lg 10^{3}=3\centerdot \lg 10=3\centerdot 1=3</math>.
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Version vom 14:37, 1. Okt. 2008

Instead of always going back to the definition of the logarithm, it is better to learn to work with the log laws,

  • \displaystyle \ \lg (ab) = \lg a + \lg b
  • \displaystyle \ \lg a^{b} = b\lg a

and to simplify expressions first. By working in this way, one only needs, in principle, to learn that \displaystyle \lg 10 = 1\,.

In our case, we have

Vorlage:Displayed math

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