Lösung 3.3:3a - Versionsgeschichte

Markus.bez: Sprache und Formulierung

2009-06-12T10:54:30Z

Sprache und Formulierung

← Nächstältere Version		Version vom 10:54, 12. Jun. 2009
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-	~~Indem wir~~ das Argument <math>8</math> ~~wie~~ <math>8 = 2\cdot 4 = 2\cdot 2\cdot 2 = 2^3</math> ~~schreiben, gibt das~~ Logarithmengesetz <math>\lg a^b = b\lg a</math>	+	Wir schreiben das Argument <math>8</math> als <math>8 = 2\cdot 4 = 2\cdot 2\cdot 2 = 2^3</math>. Das Logarithmengesetz <math>\lg a^b = b\lg a</math> liefert

-	{{Abgesetzte Formel\|\|<math>\log _{2}8 = \log _{2} 2^3 = 3\cdot\log _{2}2 = 3\cdot 1 = 3\,,</math>}}	+	{{Abgesetzte Formel\|\|<math>\log _{2}8 = \log _{2} 2^3 = 3\cdot\log _{2}2 = 3\cdot 1 = 3\,.</math>}}

-	wo wir <math>\log _{2}2 = 1\,</math> verwendet ~~haben~~.	+	Dabei haben wir <math>\log _{2}2 = 1\,</math> verwendet.

2009-03-26T19:39:04Z

← Nächstältere Version		Version vom 19:39, 26. Mär. 2009
Zeile 1:		Zeile 1:
-	~~By writing the argument~~ <math>8</math> as <math>8 = 2\cdot 4 = 2\cdot 2\cdot 2 = 2^3</math>, ~~the logarithm law,~~ <math>\lg a^b = b\lg a</math>~~, gives~~	+	Indem wir das Argument <math>8</math> wie <math>8 = 2\cdot 4 = 2\cdot 2\cdot 2 = 2^3</math> schreiben, gibt das Logarithmengesetz <math>\lg a^b = b\lg a</math>

	{{Abgesetzte Formel\|\|<math>\log _{2}8 = \log _{2} 2^3 = 3\cdot\log _{2}2 = 3\cdot 1 = 3\,,</math>}}		{{Abgesetzte Formel\|\|<math>\log _{2}8 = \log _{2} 2^3 = 3\cdot\log _{2}2 = 3\cdot 1 = 3\,,</math>}}

-	~~where we have used~~ <math>\log _{2}2 = 1\,</math>.	+	wo wir <math>\log _{2}2 = 1\,</math> verwendet haben.

2008-10-22T14:33:45Z

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

← Nächstältere Version

Version vom 14:33, 22. Okt. 2008

2008-10-22T08:42:38Z

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

← Nächstältere Version		Version vom 08:42, 22. Okt. 2008
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	By writing the argument <math>8</math> as <math>8 = 2\cdot 4 = 2\cdot 2\cdot 2 = 2^3</math>, the logarithm law, <math>\lg a^b = b\lg a</math>, gives		By writing the argument <math>8</math> as <math>8 = 2\cdot 4 = 2\cdot 2\cdot 2 = 2^3</math>, the logarithm law, <math>\lg a^b = b\lg a</math>, gives

-	{{~~Displayed math~~\|\|<math>\log _{2}8 = \log _{2} 2^3 = 3\cdot\log _{2}2 = 3\cdot 1 = 3\,,</math>}}	+	{{Abgesetzte Formel\|\|<math>\log _{2}8 = \log _{2} 2^3 = 3\cdot\log _{2}2 = 3\cdot 1 = 3\,,</math>}}

	where we have used <math>\log _{2}2 = 1\,</math>.		where we have used <math>\log _{2}2 = 1\,</math>.

2008-10-02T06:24:18Z

← Nächstältere Version		Version vom 06:24, 2. Okt. 2008
Zeile 1:		Zeile 1:
-	By writing the argument	+	By writing the argument <math>8</math> as <math>8 = 2\cdot 4 = 2\cdot 2\cdot 2 = 2^3</math>, the logarithm law, <math>\lg a^b = b\lg a</math>, gives
-	<math>~~\text{~~8}</math>	+
-	as	+
-	<math>8=2\~~centerdot~~ 4=2\~~centerdot~~ 2\~~centerdot~~ 2=2^{3}</math>, the logarithm law,	+
-	<math>\lg a^{b}=b\lg a</math>, gives	+

-		+	{{Displayed math\|\|<math>\log _{2}8 = \log _{2} 2^3 = 3\cdot\log _{2}2 = 3\cdot 1 = 3\,,</math>}}
-	<math>\log _{2}8=\log _{2}2^{3}=3\~~centerdot~~ \log _{2}2=3\~~centerdot~~ 1=3</math>	+

-		+	where we have used <math>\log _{2}2 = 1\,</math>.
-	where we have used	+
-	<math>\log _{2}2=1</math>.	+

2008-09-25T13:57:40Z

← Nächstältere Version		Version vom 13:57, 25. Sep. 2008
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-	{{~~NAVCONTENT_START}~~}	+	By writing the argument
-	<~~center~~> ~~[[Image:3_3_3a.gif]]~~ </~~center~~>	+	<math>\text{8}</math>
-	{{~~NAVCONTENT_STOP~~}}	+	as
		+	<math>8=2\centerdot 4=2\centerdot 2\centerdot 2=2^{3}</math>, the logarithm law,
		+	<math>\lg a^{b}=b\lg a</math>, gives
		+
		+
		+	<math>\log _{2}8=\log _{2}2^{3}=3\centerdot \log _{2}2=3\centerdot 1=3</math>
		+
		+
		+	where we have used
		+	<math>\log _{2}2=1</math>.

2008-09-09T09:23:23Z

← Nächstältere Version

Version vom 09:23, 9. Sep. 2008

2008-08-21T06:58:49Z

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

2008-04-02T09:34:12Z

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