Lösung 3.3:2f

Aus Online Mathematik Brückenkurs 1

(Unterschied zwischen Versionen)
Wechseln zu: Navigation, Suche
K (Lösning 3.3:2f moved to Solution 3.3:2f: Robot: moved page)
Zeile 1: Zeile 1:
-
{{NAVCONTENT_START}}
+
Instead of always going back to the definition of the logarithm, it is better to learn to work with the log laws,
-
<center> [[Image:3_3_2f.gif]] </center>
+
 
-
{{NAVCONTENT_STOP}}
+
 
 +
<math>\lg \left( ab \right)=\lg a+\lg b</math>
 +
 
 +
 
 +
<math>\lg a^{b}=b\lg a</math>
 +
 
 +
and to simplify expressions first. By working in this way, one only needs, in principle, to learn that
 +
<math>\text{lg 1}0\text{ }=\text{1}</math>.
 +
 
 +
In our case, we have
 +
 
 +
 
 +
<math>\lg 10^{3}=3\centerdot \lg 10=3\centerdot 1=3</math>.

Version vom 13:22, 25. Sep. 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 \left( ab \right)=\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 \text{lg 1}0\text{ }=\text{1}.

In our case, we have


\displaystyle \lg 10^{3}=3\centerdot \lg 10=3\centerdot 1=3.

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