Aus Online Mathematik Brückenkurs 1

Version vom 14:19, 25. Sep. 2008 (bearbeiten) Ian (Diskussion | Beiträge) ← Zum vorherigen Versionsunterschied		Version vom 06:36, 2. Okt. 2008 (bearbeiten) (rückgängig) Tek (Diskussion | Beiträge) K Zum nächsten Versionsunterschied →
Zeile 1:		Zeile 1:
-	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>	+

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Version vom 06:36, 2. Okt. 2008

We write the argument of \displaystyle \log_{3} as a power of 3,

Vorlage:Displayed math

and then simplify the expression with the logarithm laws

Vorlage:Displayed math