Solution 3.3:4b

From Förberedande kurs i matematik 1

(Difference between revisions)
Jump to: navigation, search
Current revision (07:09, 2 October 2008) (edit) (undo)
m
 
Line 1: Line 1:
The two terms can be combined into one logarithmic expression using the log law
The two terms can be combined into one logarithmic expression using the log law
-
<math>\lg a+\lg b=\lg \left( ab \right)</math>.
+
<math>\lg a + \lg b = \lg (ab)</math>,
-
 
+
{{Displayed math||<math>\lg 23 + \lg\frac{1}{23} = \lg \Bigl(23\cdot\frac{1}{23}\Bigr) = \lg 1 = 0\,\textrm{.}</math>}}
-
<math>\lg 23+\lg \frac{1}{23}=\lg 23\centerdot \frac{1}{23}=\lg 1=0</math>
+

Current revision

The two terms can be combined into one logarithmic expression using the log law \displaystyle \lg a + \lg b = \lg (ab),

\displaystyle \lg 23 + \lg\frac{1}{23} = \lg \Bigl(23\cdot\frac{1}{23}\Bigr) = \lg 1 = 0\,\textrm{.}
Retrieved from "http://wiki.sommarmatte.se/wikis/2009/bridgecourse1-ImperialCollege/index.php/Solution_3.3:4b"