Lösung 3.3:3c

Aus Online Mathematik Brückenkurs 1

(Unterschied zwischen Versionen)
Wechseln zu: Navigation, Suche
K (Lösning 3.3:3c moved to Solution 3.3:3c: Robot: moved page)
Zeile 1: Zeile 1:
-
{{NAVCONTENT_START}}
+
First, we rewrite the number
-
<center> [[Image:3_3_3c.gif]] </center>
+
<math>0.\text{125 }</math>
-
{{NAVCONTENT_STOP}}
+
as a fraction which we also simplify:
 +
 
 +
 
 +
<math>0.\text{125 }=\frac{\text{125 }}{1000}=\frac{5\centerdot 25}{10^{3}}=\frac{5\centerdot 5\centerdot 5}{\left( 2\centerdot 5 \right)^{3}}=\frac{1}{2^{3}}=2^{-3}</math>
 +
 
 +
 
 +
Because
 +
<math>0.\text{125 }</math>
 +
was expressed as a power of
 +
<math>\text{2}</math>, the logarithm can be calculated in full:
 +
 
 +
 
 +
<math>\log _{2}0.\text{125 }=\log _{2}2^{-3}=\left( -3 \right)\centerdot \log _{2}2=\left( -3 \right)\centerdot 1=-3</math>

Version vom 14:13, 25. Sep. 2008

First, we rewrite the number \displaystyle 0.\text{125 } as a fraction which we also simplify:


\displaystyle 0.\text{125 }=\frac{\text{125 }}{1000}=\frac{5\centerdot 25}{10^{3}}=\frac{5\centerdot 5\centerdot 5}{\left( 2\centerdot 5 \right)^{3}}=\frac{1}{2^{3}}=2^{-3}


Because \displaystyle 0.\text{125 } was expressed as a power of \displaystyle \text{2}, the logarithm can be calculated in full:


\displaystyle \log _{2}0.\text{125 }=\log _{2}2^{-3}=\left( -3 \right)\centerdot \log _{2}2=\left( -3 \right)\centerdot 1=-3

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