Lösung 3.3:3h - Versionsgeschichte

Markus.bez: Sprache und Formulierung

Markus.bez — Fri, 12 Jun 2009 10:58:00 GMT

Sprache und Formulierung

← Nächstältere Version		Version vom 10:58, 12. Jun. 2009
Zeile 1:		Zeile 1:
-	~~Nachdem~~ <math>a^{2}\sqrt{a} = a^{2}a^{1/2} = a^{2+1/2} = a^{5/2}</math>, ~~gibt~~ das Logarithmengesetz <math>b\lg a = \lg a^b</math>, den Ausdruck	+	Da <math>a^{2}\sqrt{a} = a^{2}a^{1/2} = a^{2+1/2} = a^{5/2}</math>, liefert uns das Logarithmengesetz <math>b\lg a = \lg a^b</math> den Ausdruck

	{{Abgesetzte Formel\|\|<math>\log_{a} \bigl(a^{2}\sqrt{a}\,\bigr) = \log_{a}a^{5/2} = \frac{5}{2}\cdot\log_{a}a = \frac{5}{2}\cdot 1 = \frac{5}{2}\,,</math>}}		{{Abgesetzte Formel\|\|<math>\log_{a} \bigl(a^{2}\sqrt{a}\,\bigr) = \log_{a}a^{5/2} = \frac{5}{2}\cdot\log_{a}a = \frac{5}{2}\cdot 1 = \frac{5}{2}\,,</math>}}

-	Wo wir <math>\log_{a}a = 1\,</math> ~~benutzt haben~~.	+	indem wir <math>\log_{a}a = 1\,</math> benutzen.

-	Hinweis: Wir müssen annehmen dass <math>a > 0</math> und <math>a\ne 1\,</math>.	+	Hinweis: Wir müssen annehmen, dass <math>a > 0</math> und <math>a\ne 1\,</math>.

Martin — Thu, 26 Mar 2009 20:30:41 GMT

← Nächstältere Version		Version vom 20:30, 26. Mär. 2009
Zeile 1:		Zeile 1:
-	~~Because~~ <math>a^{2}\sqrt{a} = a^{2}a^{1/2} = a^{2+1/2} = a^{5/2}</math>, ~~the logarithm law,~~ <math>b\lg a = \lg a^b</math>, ~~gives that~~	+	Nachdem <math>a^{2}\sqrt{a} = a^{2}a^{1/2} = a^{2+1/2} = a^{5/2}</math>, gibt das Logarithmengesetz <math>b\lg a = \lg a^b</math>, den Ausdruck

	{{Abgesetzte Formel\|\|<math>\log_{a} \bigl(a^{2}\sqrt{a}\,\bigr) = \log_{a}a^{5/2} = \frac{5}{2}\cdot\log_{a}a = \frac{5}{2}\cdot 1 = \frac{5}{2}\,,</math>}}		{{Abgesetzte Formel\|\|<math>\log_{a} \bigl(a^{2}\sqrt{a}\,\bigr) = \log_{a}a^{5/2} = \frac{5}{2}\cdot\log_{a}a = \frac{5}{2}\cdot 1 = \frac{5}{2}\,,</math>}}

-	~~where we have used that~~ <math>\log_{a}a = 1\,</math>.	+	Wo wir <math>\log_{a}a = 1\,</math> benutzt haben.

-		+	Hinweis: Wir müssen annehmen dass <math>a > 0</math> und <math>a\ne 1\,</math>.
-	~~Note~~: ~~In this exercise, we assume, implicitly, that~~ <math>a > 0</math> ~~and~~ <math>a\ne 1\,</math>.	+

Tekbot — Wed, 22 Oct 2008 14:36:04 GMT

hat „Solution 3.3:3h“ nach „Lösung 3.3:3h“ verschoben: Robot: moved page

← Nächstältere Version

Version vom 14:36, 22. Okt. 2008

Tekbot — Wed, 22 Oct 2008 08:43:38 GMT

Robot: Automated text replacement (-{{Displayed math +{{Abgesetzte Formel)

← Nächstältere Version		Version vom 08:43, 22. Okt. 2008
Zeile 1:		Zeile 1:
	Because <math>a^{2}\sqrt{a} = a^{2}a^{1/2} = a^{2+1/2} = a^{5/2}</math>, the logarithm law, <math>b\lg a = \lg a^b</math>, gives that		Because <math>a^{2}\sqrt{a} = a^{2}a^{1/2} = a^{2+1/2} = a^{5/2}</math>, the logarithm law, <math>b\lg a = \lg a^b</math>, gives that

-	{{~~Displayed math~~\|\|<math>\log_{a} \bigl(a^{2}\sqrt{a}\,\bigr) = \log_{a}a^{5/2} = \frac{5}{2}\cdot\log_{a}a = \frac{5}{2}\cdot 1 = \frac{5}{2}\,,</math>}}	+	{{Abgesetzte Formel\|\|<math>\log_{a} \bigl(a^{2}\sqrt{a}\,\bigr) = \log_{a}a^{5/2} = \frac{5}{2}\cdot\log_{a}a = \frac{5}{2}\cdot 1 = \frac{5}{2}\,,</math>}}

	where we have used that <math>\log_{a}a = 1\,</math>.		where we have used that <math>\log_{a}a = 1\,</math>.

Tek — Thu, 02 Oct 2008 07:04:04 GMT

← Nächstältere Version		Version vom 07:04, 2. Okt. 2008
Zeile 1:		Zeile 1:
-	Because	+	Because <math>a^{2}\sqrt{a} = a^{2}a^{1/2} = a^{2+1/2} = a^{5/2}</math>, the logarithm law, <math>b\lg a = \lg a^b</math>, gives that
-	<math>a^{2}\sqrt{a}=a^{2}a^~~{\frac~~{1}{2}}=a^{2+~~\frac{~~1}{2}}=a^~~{\frac~~{5}{2}}</math>, the logarithm law,	+
-	<math>b\lg a=\lg a^{b}</math>, gives that	+

		+	{{Displayed math\|\|<math>\log_{a} \bigl(a^{2}\sqrt{a}\,\bigr) = \log_{a}a^{5/2} = \frac{5}{2}\cdot\log_{a}a = \frac{5}{2}\cdot 1 = \frac{5}{2}\,,</math>}}

-	<math>\~~log _~~{a}a~~^{2}\sqrt{a}~~=~~\log _{a}a^{\frac{5}{2}}=\frac{5}{2}\centerdot \log _{a}a=\frac{5}{2}\centerdot~~ 1=\~~frac{5}{2}~~,</math>	+	where we have used that <math>\log_{a}a = 1\,</math>.


-	~~where we have used that~~	+	Note: In this exercise, we assume, implicitly, that <math>a > 0</math> and <math>a\ne 1\,</math>.
-	~~<math>\log _{a}a=1</math>.~~	+
-		+
-	~~NOTE~~: In this exercise, we assume, implicitly, that	+
-	<math>~~\text{~~a}>0~~\text{ }~~</math>	+
-	and	+
-	<math>~~\text{~~a}\ne ~~\text{~~1}</math>.	+

Ian — Thu, 25 Sep 2008 14:41:14 GMT

← Nächstältere Version		Version vom 14:41, 25. Sep. 2008
Zeile 1:		Zeile 1:
-	{{~~NAVCONTENT_START~~}}	+	Because
-	<~~center~~> ~~[[Image:3_3_3h.gif]]~~ </~~center~~>	+	<math>a^{2}\sqrt{a}=a^{2}a^{\frac{1}{2}}=a^{2+\frac{1}{2}}=a^{\frac{5}{2}}</math>, the logarithm law,
-	{{~~NAVCONTENT_STOP~~}}	+	<math>b\lg a=\lg a^{b}</math>, gives that
		+
		+
		+	<math>\log _{a}a^{2}\sqrt{a}=\log _{a}a^{\frac{5}{2}}=\frac{5}{2}\centerdot \log _{a}a=\frac{5}{2}\centerdot 1=\frac{5}{2},</math>
		+
		+
		+	where we have used that
		+	<math>\log _{a}a=1</math>.
		+
		+	NOTE: In this exercise, we assume, implicitly, that
		+	<math>\text{a}>0\text{ }</math>
		+	and
		+	<math>\text{a}\ne \text{1}</math>.

Tekbot — Tue, 09 Sep 2008 09:25:43 GMT

← Nächstältere Version

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

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

Oskar — Wed, 02 Apr 2008 09:48:07 GMT

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