Solution 3.3:3g - Revision history

Tek at 07:00, 2 October 2008

2008-10-02T07:00:44Z

←Older revision		Revision as of 07:00, 2 October 2008
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-	Using the logarithm law,	+	Using the logarithm law, <math>\lg a-\lg b = \lg\frac{a}{b}\,</math>, the expression can be calculated as
-	<math>\lg a-\lg b=\lg ~~\left(~~ \frac{a}{b} \~~right)~~</math>, the expression can be calculated as	+

		+	{{Displayed math\|\|<math>\log_3 12 - \log_3 4 = \log_3\frac{12}{4} = \log _3 3 = 1\,\textrm{.}</math>}}

-	<math>~~\log _{3}~~12~~-\log _{3}4~~=~~\log _{~~3}\~~frac{12}{~~4}=\~~log _{3}3~~=1</math>	+	Another way is to write <math>12 = 3\cdot 4</math> and use the logarithm law,
		+	<math>\lg (ab) = \lg a + \lg b\,</math>,

-		+	{{Displayed math\|\|<math>\begin{align}
-	~~Another way is to write~~	+	\log _{3}12 - \log _{3}4
-	~~<math>\text~~{~~12}=\text~~{~~3}\centerdot \text{4 }</~~math>	+	&= \log_{3}(3\cdot 4) - \log_{3} 4\\[5pt]
-	~~and use the logarithm law,~~	+	&= \log_{3}3 + \log _{3}4 - \log _{3}4\\[5pt]
-	~~<math>\lg \left( ab \right)=\lg a+\lg b</math>,~~	+	&= \log _{3}3\\[5pt]
-		+	&= 1\,\textrm{.}
-		+	\end{align}</math>}}
-	<math>\begin{align}	+
-	& \log _{3}12-\log _{3}4=\~~log _~~{3}~~\left~~( 3\~~centerdot~~ 4 ~~\right~~)-\~~log _~~{3}4 \\	+
-	& =\~~log _~~{3}3+\log _{3}4-\log _{3}4=\log _{3}3=1 \\	+
-	\end{align}</math>	+

2008-09-25T14:33:16Z

←Older revision		Revision as of 14:33, 25 September 2008
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-	{{~~NAVCONTENT_START~~}}	+	Using the logarithm law,
-	<~~center~~> ~~[[Image:3_3_3g.gif]]~~ </~~center~~>	+	<math>\lg a-\lg b=\lg \left( \frac{a}{b} \right)</math>, the expression can be calculated as
-	{{~~NAVCONTENT_STOP~~}}	+
		+
		+	<math>\log _{3}12-\log _{3}4=\log _{3}\frac{12}{4}=\log _{3}3=1</math>
		+
		+
		+	Another way is to write
		+	<math>\text{12}=\text{3}\centerdot \text{4 }</math>
		+	and use the logarithm law,
		+	<math>\lg \left( ab \right)=\lg a+\lg b</math>,
		+
		+
		+	<math>\begin{align}
		+	& \log _{3}12-\log _{3}4=\log _{3}\left( 3\centerdot 4 \right)-\log _{3}4 \\
		+	& =\log _{3}3+\log _{3}4-\log _{3}4=\log _{3}3=1 \\
		+	\end{align}</math>

2008-09-09T09:25:23Z

←Older revision

Revision as of 09:25, 9 September 2008

2008-08-21T06:59:49Z

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2008-04-02T09:47:21Z

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