Solution 3.3:3b - Revision history

Tek at 06:27, 2 October 2008

2008-10-02T06:27:57Z

←Older revision		Revision as of 06:27, 2 October 2008
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-	Because we are working with	+	Because we are working with <math>\log _{9}</math>, we express 1/3 as a power of 9,
-	<math>\log _{9}</math>, we express	+
-	~~<math>{~~1}/{3~~}\;</math>~~	+
-	as a power of	+
-	~~<math>\text{~~9~~}</math>~~,	+
-		+
-		+
-	~~<math>\frac{1}{3}=\frac{1}{\sqrt{9}}=\frac{1}{9^{{1}/{2}\;}}=9^{-{1}/{2}\;}</math>~~	+

		+	{{Displayed math\|\|<math>\frac{1}{3} = \frac{1}{\sqrt{9}} = \frac{1}{9^{1/2}} = 9^{-1/2}\,\textrm{.}</math>}}

	Using the logarithm laws, we get		Using the logarithm laws, we get

-		+	{{Displayed math\|\|<math>\log_9\frac{1}{3} = \log_9 9^{-1/2} = -\frac{1}{2}\cdot\log_9 9 = -\frac{1}{2}\cdot 1 = -\frac{1}{2}\,\textrm{.}</math>}}
-	<math>\~~log _{9}~~\frac{1}{3}=\~~log _{9}~~9^{-{1}/{2~~}\;~~}=-\frac{1}{2}\~~centerdot~~ \~~log _{9}~~9=-\frac{1}{2}\~~centerdot~~ 1=-\frac{1}{2}.</math>	+

2008-09-25T14:06:19Z

←Older revision		Revision as of 14:06, 25 September 2008
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-	{{~~NAVCONTENT_START~~}}	+	Because we are working with
-	<~~center~~> ~~[[Image:3_3_3b.gif]]~~ </~~center~~>	+	<math>\log _{9}</math>, we express
-	{{~~NAVCONTENT_STOP~~}}	+	<math>{1}/{3}\;</math>
		+	as a power of
		+	<math>\text{9}</math>,
		+
		+
		+	<math>\frac{1}{3}=\frac{1}{\sqrt{9}}=\frac{1}{9^{{1}/{2}\;}}=9^{-{1}/{2}\;}</math>
		+
		+
		+	Using the logarithm laws, we get
		+
		+
		+	<math>\log _{9}\frac{1}{3}=\log _{9}9^{-{1}/{2}\;}=-\frac{1}{2}\centerdot \log _{9}9=-\frac{1}{2}\centerdot 1=-\frac{1}{2}.</math>

2008-09-09T09:23:43Z

←Older revision

Revision as of 09:23, 9 September 2008

2008-08-21T06:58:59Z

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2008-04-02T09:35:46Z

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