Solution 3.3:3c

From Förberedande kurs i matematik 1

(Difference between revisions)
Jump to: navigation, search
Current revision (06:33, 2 October 2008) (edit) (undo)
m
 
Line 1: Line 1:
-
First, we rewrite the number
+
First, we rewrite the number 0.125 as a fraction which we also simplify
-
<math>0.\text{125 }</math>
+
-
as a fraction which we also simplify:
+
 +
{{Displayed math||<math>0\textrm{.}125 = \frac{125}{1000} = \frac{5\cdot 25}{10^3} = \frac{5\cdot 5\cdot 5}{(2\cdot 5)^3} = \frac{1}{2^3} = 2^{-3}\,\textrm{.}</math>}}
-
<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 0.125 was expressed as a power of 2, the logarithm can be calculated in full
-
 
+
{{Displayed math||<math>\log_2 0\textrm{.}125 = \log_2 2^{-3} = (-3)\cdot\log_2 2 = (-3)\cdot 1 = -3\,\textrm{.}</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>
+

Current revision

First, we rewrite the number 0.125 as a fraction which we also simplify

\displaystyle 0\textrm{.}125 = \frac{125}{1000} = \frac{5\cdot 25}{10^3} = \frac{5\cdot 5\cdot 5}{(2\cdot 5)^3} = \frac{1}{2^3} = 2^{-3}\,\textrm{.}

Because 0.125 was expressed as a power of 2, the logarithm can be calculated in full

\displaystyle \log_2 0\textrm{.}125 = \log_2 2^{-3} = (-3)\cdot\log_2 2 = (-3)\cdot 1 = -3\,\textrm{.}
Retrieved from "http://wiki.sommarmatte.se/wikis/2008/bridgecourse1-ImperialCollege/index.php/Solution_3.3:3c"