Solution 3.3:5b

From Förberedande kurs i matematik 1

(Difference between revisions)
Jump to: navigation, search
m (Lösning 3.3:5b moved to Solution 3.3:5b: Robot: moved page)
Current revision (07:27, 2 October 2008) (edit) (undo)
m
 
(One intermediate revision not shown.)
Line 1: Line 1:
-
{{NAVCONTENT_START}}
+
By using the logarithm laws,
-
<center> [[Image:3_3_5b.gif]] </center>
+
 
-
{{NAVCONTENT_STOP}}
+
{{Displayed math||<math>\begin{align}
 +
\ln a + \ln b &= \ln (a\cdot b)\,,\\[5pt]
 +
\ln a - \ln b &= \ln\frac{a}{b}\,,
 +
\end{align}</math>}}
 +
 
 +
we can collect together the terms into one logarithmic expression
 +
 
 +
{{Displayed math||<math>\begin{align}
 +
\ln 8 - \ln 4 - \ln 2 &= \ln 8 - (\ln 4 + \ln 2)\\[5pt]
 +
&= \ln 8 - \ln(4\cdot 2)\\[5pt]
 +
&= \ln\frac{8}{4\cdot 2}\\[5pt]
 +
&= \ln 1\\[5pt]
 +
&= 0\,,
 +
\end{align}</math>}}
 +
 
 +
where <math>\ln 1 = 0</math>, since <math>e^{0}=1</math> (the equality <math>a^{0}=1</math> holds for all <math>a\ne 0</math>).

Current revision

By using the logarithm laws,

\displaystyle \begin{align}

\ln a + \ln b &= \ln (a\cdot b)\,,\\[5pt] \ln a - \ln b &= \ln\frac{a}{b}\,, \end{align}

we can collect together the terms into one logarithmic expression

\displaystyle \begin{align}

\ln 8 - \ln 4 - \ln 2 &= \ln 8 - (\ln 4 + \ln 2)\\[5pt] &= \ln 8 - \ln(4\cdot 2)\\[5pt] &= \ln\frac{8}{4\cdot 2}\\[5pt] &= \ln 1\\[5pt] &= 0\,, \end{align}

where \displaystyle \ln 1 = 0, since \displaystyle e^{0}=1 (the equality \displaystyle a^{0}=1 holds for all \displaystyle a\ne 0).

Retrieved from "http://wiki.sommarmatte.se/wikis/2009/bridgecourse1-ImperialCollege/index.php/Solution_3.3:5b"