Solution 3.3:5b - Revision history

Tek at 07:27, 2 October 2008

2008-10-02T07:27:48Z

←Older revision		Revision as of 07:27, 2 October 2008
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	By using the logarithm laws,		By using the logarithm laws,

-		+	{{Displayed math\|\|<math>\begin{align}
-	<math>\lg a+\lg b=\~~lg \left~~( a\~~centerdot~~ b \~~right)</math>~~	+	\ln a + \ln b &= \ln (a\cdot b)\,,\\[5pt]
-		+	\ln a - \ln b &= \ln\frac{a}{b}\,,
-		+	\end{align}</math>}}
-	~~<math>~~\~~text{log }~~a-\~~text{ log }~~b=\~~text{log}\left(~~ \frac{a}{b} \~~right)~~</math>	+

	we can collect together the terms into one logarithmic expression		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>}}

-	<math>~~\begin{align}~~	+	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>).
-	~~& \ln 8-\ln 4-\ln 2=\ln 8-\left( \ln 4+\ln 2 \right)=\ln 8-\ln \left( 4\centerdot 2 \right) \\~~	+
-	~~& =\ln \frac{8}{4\centerdot 2}=~~\ln 1=0~~, \\~~	+
-	~~\end{align}~~</math>	+
-		+
-		+
-	~~where ln 1 =0~~, since	+
-	<math>e^{0}=1</math>	+
-	(the equality	+
-	<math>a^{0}=1</math>	+
-	holds for all	+
-	<math>a\ne 0</math>	+
-	).	+

2008-09-26T08:26:17Z

←Older revision		Revision as of 08:26, 26 September 2008
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-	{{~~NAVCONTENT_START~~}}	+	By using the logarithm laws,
-	<~~center~~> ~~[[Image:3_3_5b.gif]]~~ </~~center~~>	+
-	{{~~NAVCONTENT_STOP}~~}	+
		+	<math>\lg a+\lg b=\lg \left( a\centerdot b \right)</math>
		+
		+
		+	<math>\text{log }a-\text{ log }b=\text{log}\left( \frac{a}{b} \right)</math>
		+
		+	we can collect together the terms into one logarithmic expression
		+
		+
		+	<math>\begin{align}
		+	& \ln 8-\ln 4-\ln 2=\ln 8-\left( \ln 4+\ln 2 \right)=\ln 8-\ln \left( 4\centerdot 2 \right) \\
		+	& =\ln \frac{8}{4\centerdot 2}=\ln 1=0, \\
		+	\end{align}</math>
		+
		+
		+	where ln 1 =0, since
		+	<math>e^{0}=1</math>
		+	(the equality
		+	<math>a^{0}=1</math>
		+	holds for all
		+	<math>a\ne 0</math>
		+	).

2008-09-09T09:27:23Z

←Older revision

Revision as of 09:27, 9 September 2008

2008-08-21T07:00:49Z

Robot: Automated text replacement (-[[Bild: +[[Image:)

2008-04-02T09:51:52Z

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