Solution 3.3:3d - Revision history

Tek at 06:36, 2 October 2008

Tek — Thu, 02 Oct 2008 06:36:56 GMT

←Older revision		Revision as of 06:36, 2 October 2008
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-	We write the argument of	+	We write the argument of <math>\log_{3}</math> as a power of 3,
-	<math>\~~log _~~{3}</math>	+
-	as a power of	+
-	~~<math>\text{~~3~~}</math>~~,	+

		+	{{Displayed math\|\|<math>9\cdot 3^{1/3} = 3^2\cdot 3^{1/3} = 3^{2+1/3} = 3^{7/3}\,,</math>}}

-	~~<math>9\centerdot 3^{{1}/{3}\;}=3^{2}\centerdot 3^{{1}/{3}\;}=3^{2+\frac{1}{3}}=3^{\frac{7}{3}}</math>~~	+	and then simplify the expression with the logarithm laws

-		+	{{Displayed math\|\|<math>\log _3 (9\cdot 3^{1/3}) = \log_3 3^{7/3} = \frac{7}{3}\cdot \log_3 3 = \frac{7}{3}\cdot 1 = \frac{7}{3}\,\textrm{.}</math>}}
-	~~and then simplify the expression with the logarithm laws:~~	+
-		+
-		+
-	<math>\log ~~_{3}\left~~( 9\~~centerdot~~ 3^{{1}/{3}~~\;} \right~~)=\~~log _{3}~~3^~~{\frac~~{7}{3}}=\frac{7}{3}\~~centerdot~~ \~~log _{3}~~3=\frac{7}{3}\~~centerdot~~ 1=\frac{7}{3}.</math>	+

Ian — Thu, 25 Sep 2008 14:19:14 GMT

←Older revision		Revision as of 14:19, 25 September 2008
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-	{{~~NAVCONTENT_START}~~}	+	We write the argument of
-	<~~center~~> ~~[[Image:3_3_3d.gif]]~~ </~~center~~>	+	<math>\log _{3}</math>
-	{{~~NAVCONTENT_STOP~~}}	+	as a power of
		+	<math>\text{3}</math>,
		+
		+
		+	<math>9\centerdot 3^{{1}/{3}\;}=3^{2}\centerdot 3^{{1}/{3}\;}=3^{2+\frac{1}{3}}=3^{\frac{7}{3}}</math>
		+
		+
		+	and then simplify the expression with the logarithm laws:
		+
		+
		+	<math>\log _{3}\left( 9\centerdot 3^{{1}/{3}\;} \right)=\log _{3}3^{\frac{7}{3}}=\frac{7}{3}\centerdot \log _{3}3=\frac{7}{3}\centerdot 1=\frac{7}{3}.</math>

Tekbot — Tue, 09 Sep 2008 09:24:23 GMT

←Older revision

Revision as of 09:24, 9 September 2008

Tekbot — Thu, 21 Aug 2008 06:59:19 GMT

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

Oskar — Wed, 02 Apr 2008 09:40:11 GMT

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