Aus Online Mathematik Brückenkurs 1

Version vom 14:06, 25. Sep. 2008 (bearbeiten) Ian (Diskussion | Beiträge) ← Zum vorherigen Versionsunterschied		Version vom 06:27, 2. Okt. 2008 (bearbeiten) (rückgängig) Tek (Diskussion | Beiträge) K Zum nächsten Versionsunterschied →
Zeile 1:		Zeile 1:
-	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>	+

---

Version vom 06:27, 2. Okt. 2008

Because we are working with \displaystyle \log _{9}, we express 1/3 as a power of 9,

Vorlage:Displayed math

Using the logarithm laws, we get

Vorlage:Displayed math