Lösung 3.4:3b - Versionsgeschichte

Markus.bez: Sprache und Formulierung

2009-06-12T14:08:17Z

Sprache und Formulierung

← Nächstältere Version		Version vom 14:08, 12. Jun. 2009
Zeile 1:		Zeile 1:
-	Die Ausdrücke <math>\ln\bigl(x^2+3x\bigr)</math> und <math>\ln\bigl(3x^2-2x \bigr)</math> sind gleich wenn deren Argumente ~~Gleich~~ sind, also wenn	+	Die Ausdrücke <math>\ln\bigl(x^2+3x\bigr)</math> und <math>\ln\bigl(3x^2-2x \bigr)</math> sind gleich, wenn deren Argumente gleich sind, also wenn

	{{Abgesetzte Formel\|\|<math>x^2 + 3x = 3x^2 - 2x\,\textrm{.}</math>}}		{{Abgesetzte Formel\|\|<math>x^2 + 3x = 3x^2 - 2x\,\textrm{.}</math>}}

-	Hier müssen wir aber vorsichtig sein. Falls beide ~~seiten~~ negativ sind, ist dies keine gültige Lösung, nachdem der Logarithmus nicht für negative Argumente definiert ist. Daher müssen wir testen ob die Ausdrücke <math>x^2 + 3x</math> und <math>3x^2 - 2x</math> ~~positiv sind,~~ für unsere Lösungen ''x''.	+	Hier müssen wir aber vorsichtig sein: Falls beide Seiten negativ sind, ist dies keine gültige Lösung, nachdem der Logarithmus nicht für negative Argumente definiert ist. Daher müssen wir testen, ob die Ausdrücke <math>x^2 + 3x</math> und <math>3x^2 - 2x</math> für unsere Lösungen ''x'' positiv sind.

-	Wir ~~sammeln~~ alle Terme auf ~~einer~~ Seite, und erhalten so eine quadratische Gleichung.	+	Wir schreiben alle Terme auf eine Seite und erhalten so eine quadratische Gleichung:

	{{Abgesetzte Formel\|\|<math>2x^2-5x=0</math>}}		{{Abgesetzte Formel\|\|<math>2x^2-5x=0</math>}}
Zeile 13:		Zeile 13:
	{{Abgesetzte Formel\|\|<math>x(2x-5) = 0\,\textrm{.}</math>}}		{{Abgesetzte Formel\|\|<math>x(2x-5) = 0\,\textrm{.}</math>}}

-	~~und~~ sehen direkt dass die Lösungen <math>x=0</math>	+	Wir sehen direkt, dass die Lösungen <math>x=0</math>
	und <math>x=5/2\,</math> sind.		und <math>x=5/2\,</math> sind.

-	Wir sehen dass wenn <math>x=0</math> ~~ist~~ <math>x^2 + 3x = 3x^2 - 2x = 0</math>~~, und also~~ ist <math>x=0</math> keine Lösung. <math>x=5/2</math> ergibt aber <math>x^2 + 3x = 3x^2 - 2x = 55/4 > 0</math>, also ist <math>x=5/2</math> eine Lösung.	+	Wir sehen, dass, wenn <math>x=0</math>, dann <math>x^2 + 3x = 3x^2 - 2x = 0</math>. Also ist <math>x=0</math> keine Lösung. <math>x=5/2</math> ergibt aber <math>x^2 + 3x = 3x^2 - 2x = 55/4 > 0</math>, also ist <math>x=5/2</math> eine Lösung.

Martin um 17:25, 27. Mär. 2009

2009-03-27T17:25:49Z

← Nächstältere Version		Version vom 17:25, 27. Mär. 2009
Zeile 1:		Zeile 1:
-	~~The expressions~~ <math>\ln\bigl(x^2+3x\bigr)</math> ~~and~~ <math>\ln\bigl(3x^2-2x \bigr)</math> ~~are equal only if their arguments are equal~~, ~~i.e.~~	+	Die Ausdrücke <math>\ln\bigl(x^2+3x\bigr)</math> und <math>\ln\bigl(3x^2-2x \bigr)</math> sind gleich wenn deren Argumente Gleich sind, also wenn

	{{Abgesetzte Formel\|\|<math>x^2 + 3x = 3x^2 - 2x\,\textrm{.}</math>}}		{{Abgesetzte Formel\|\|<math>x^2 + 3x = 3x^2 - 2x\,\textrm{.}</math>}}

-	~~However~~, ~~we have to be careful! If we obtain a value for ''x'' which makes the arguments equal but negative or zero~~, ~~then it will not correspond to a genuine solution because ln is not defined for~~ negative ~~arguments~~. ~~At the end of the exercise, we must therefore check that~~ <math>x^2 + 3x</math> ~~and~~ <math>3x^2 - 2x</math> ~~really are positive for those solutions that we have calculated~~.	+	Hier müssen wir aber vorsichtig sein. Falls beide seiten negativ sind, ist dies keine gültige Lösung, nachdem der Logarithmus nicht für negative Argumente definiert ist. Daher müssen wir testen ob die Ausdrücke <math>x^2 + 3x</math> und <math>3x^2 - 2x</math> positiv sind, für unsere Lösungen ''x''.

-	~~If we move all the terms over to one side in the equation for the arguments~~, ~~we get the second-degree equation~~	+	Wir sammeln alle Terme auf einer Seite, und erhalten so eine quadratische Gleichung.

	{{Abgesetzte Formel\|\|<math>2x^2-5x=0</math>}}		{{Abgesetzte Formel\|\|<math>2x^2-5x=0</math>}}

-	~~and we see that both terms contain~~ ''x'', ~~which we can take out as a factor,~~	+	Nachdem alle Terme ''x'' enthalten, ziehen wir den Faktor ''x'' heraus

	{{Abgesetzte Formel\|\|<math>x(2x-5) = 0\,\textrm{.}</math>}}		{{Abgesetzte Formel\|\|<math>x(2x-5) = 0\,\textrm{.}</math>}}

-	~~From this factorized expression, we read off that the solutions are~~ <math>x=0</math>	+	und sehen direkt dass die Lösungen <math>x=0</math>
-	~~and~~ <math>x=5/2\,</math>.	+	und <math>x=5/2\,</math> sind.

-	~~A final check shows that when~~ <math>x=0</math> ~~then~~ <math>x^2 + 3x = 3x^2 - 2x = 0</math>, so <math>x=0</math> ~~is not a solution~~. ~~On the other hand, when~~ <math>x=5/2</math> ~~then~~ <math>x^2 + 3x = 3x^2 - 2x = 55/4 > 0</math>, so <math>x=5/2</math> ~~is a solution~~.	+	Wir sehen dass wenn <math>x=0</math> ist <math>x^2 + 3x = 3x^2 - 2x = 0</math>, und also ist <math>x=0</math> keine Lösung. <math>x=5/2</math> ergibt aber <math>x^2 + 3x = 3x^2 - 2x = 55/4 > 0</math>, also ist <math>x=5/2</math> eine Lösung.

Tekbot: hat „Solution 3.4:3b“ nach „Lösung 3.4:3b“ verschoben: Robot: moved page

2008-10-22T14:42:44Z

hat „Solution 3.4:3b“ nach „Lösung 3.4:3b“ verschoben: Robot: moved page

← Nächstältere Version

Version vom 14:42, 22. Okt. 2008

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

2008-10-22T08:46:48Z

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

← Nächstältere Version		Version vom 08:46, 22. Okt. 2008
Zeile 1:		Zeile 1:
	The expressions <math>\ln\bigl(x^2+3x\bigr)</math> and <math>\ln\bigl(3x^2-2x \bigr)</math> are equal only if their arguments are equal, i.e.		The expressions <math>\ln\bigl(x^2+3x\bigr)</math> and <math>\ln\bigl(3x^2-2x \bigr)</math> are equal only if their arguments are equal, i.e.

-	{{~~Displayed math~~\|\|<math>x^2 + 3x = 3x^2 - 2x\,\textrm{.}</math>}}	+	{{Abgesetzte Formel\|\|<math>x^2 + 3x = 3x^2 - 2x\,\textrm{.}</math>}}

	However, we have to be careful! If we obtain a value for ''x'' which makes the arguments equal but negative or zero, then it will not correspond to a genuine solution because ln is not defined for negative arguments. At the end of the exercise, we must therefore check that <math>x^2 + 3x</math> and <math>3x^2 - 2x</math> really are positive for those solutions that we have calculated.		However, we have to be careful! If we obtain a value for ''x'' which makes the arguments equal but negative or zero, then it will not correspond to a genuine solution because ln is not defined for negative arguments. At the end of the exercise, we must therefore check that <math>x^2 + 3x</math> and <math>3x^2 - 2x</math> really are positive for those solutions that we have calculated.
Zeile 7:		Zeile 7:
	If we move all the terms over to one side in the equation for the arguments, we get the second-degree equation		If we move all the terms over to one side in the equation for the arguments, we get the second-degree equation

-	{{~~Displayed math~~\|\|<math>2x^2-5x=0</math>}}	+	{{Abgesetzte Formel\|\|<math>2x^2-5x=0</math>}}

	and we see that both terms contain ''x'', which we can take out as a factor,		and we see that both terms contain ''x'', which we can take out as a factor,

-	{{~~Displayed math~~\|\|<math>x(2x-5) = 0\,\textrm{.}</math>}}	+	{{Abgesetzte Formel\|\|<math>x(2x-5) = 0\,\textrm{.}</math>}}

	From this factorized expression, we read off that the solutions are <math>x=0</math>		From this factorized expression, we read off that the solutions are <math>x=0</math>

Tek um 14:25, 2. Okt. 2008

2008-10-02T14:25:08Z

← Nächstältere Version		Version vom 14:25, 2. Okt. 2008
Zeile 1:		Zeile 1:
-	The expressions	+	The expressions <math>\ln\bigl(x^2+3x\bigr)</math> and <math>\ln\bigl(3x^2-2x \bigr)</math> are equal only if their arguments are equal, i.e.
-	<math>\~~text{~~ln}\~~left~~( x^~~{\text{~~2}}+\~~text{3}x \right~~)~~\text{ }~~</math>	+
-	and	+
-	<math>\~~text{~~ln}\~~left~~( ~~\text{3}x~~^~~{\text{~~2}}-\~~text{2}x \right~~)~~\text{ }~~</math>	+
-	are equal only if their arguments are equal, i.e.	+

		+	{{Displayed math\|\|<math>x^2 + 3x = 3x^2 - 2x\,\textrm{.}</math>}}

-	~~<math>\left( x^{\text{2}}+\text{3}x \right)=\left( \text{3}x^{\text{2}}-\text{2}x \right)\text{ }</math>~~	+	However, we have to be careful! If we obtain a value for ''x'' which makes the arguments equal but negative or zero, then it will not correspond to a genuine solution because ln is not defined for negative arguments. At the end of the exercise, we must therefore check that <math>x^2 + 3x</math> and <math>3x^2 - 2x</math> really are positive for those solutions that we have calculated.
-		+
-		+
-	~~however~~, we have to be careful! If we obtain a value for	+
-	~~<math>~~x~~</math>~~	+
-	which makes the arguments equal but negative or zero, then it will not correspond to a genuine solution because ln is not defined for negative arguments. At the end of the exercise, we must therefore check that	+
-	<math>x^~~{\text{~~2}}+~~\text{3}x\text{ }~~</math>	+
-	and	+
-	<math>~~\text{3}x~~^~~{\text{~~2}}-~~\text{2}x\text{ }~~</math>	+
-	really are positive for those solutions that we have calculated.	+

	If we move all the terms over to one side in the equation for the arguments, we get the second-degree equation		If we move all the terms over to one side in the equation for the arguments, we get the second-degree equation

		+	{{Displayed math\|\|<math>2x^2-5x=0</math>}}

-	~~<math>2x^{2}-5x=0</math>~~	+	and we see that both terms contain ''x'', which we can take out as a factor,
-		+
-		+
-	and we see that both terms contain x, which we can take out as a factor:	+
-		+
-		+
-	~~<math>x\left( 2x-5 \right)=0</math>~~	+

		+	{{Displayed math\|\|<math>x(2x-5) = 0\,\textrm{.}</math>}}

-	From this factorized expression, we read off that the solutions are	+	From this factorized expression, we read off that the solutions are <math>x=0</math>
-	<math>x=0~~\text{ }~~</math>	+	and <math>x=5/2\,</math>.
-	and	+
-	<math>x={5}/{2}\;</math>.	+

-	A final check shows that when	+	A final check shows that when <math>x=0</math> then <math>x^2 + 3x = 3x^2 - 2x = 0</math>, so <math>x=0</math> is not a solution. On the other hand, when <math>x=5/2</math> then <math>x^2 + 3x = 3x^2 - 2x = 55/4 > 0</math>, so <math>x=5/2</math> is a solution.
-	<math>x=0~~\text{ }~~</math>	+
-	then	+
-	<math>x^~~{\text{~~2}}+~~\text{3}x~~=~~\text{ 3}x~~^~~{\text{~~2}}-~~\text{2}x\text{ }~~=0</math>, so	+
-	<math>x=0~~\text{ }~~</math>	+
-	is not a solution. On the other hand, when	+
-	<math>x={5}/{2~~}\;~~</math>	+
-	then	+
-	<math>x^~~{\text{~~2}}+~~\text{3}x~~=~~\text{ 3}x~~^~~{\text{~~2}}-~~\text{2}x~~=~~\text{~~55}/~~\text{~~4}>0</math>, so	+
-	<math>x={5}/{2~~}\;~~</math>	+
-	is a solution.	+

Ian um 11:52, 26. Sep. 2008

2008-09-26T11:52:38Z

← Nächstältere Version		Version vom 11:52, 26. Sep. 2008
Zeile 1:		Zeile 1:
-	{{~~NAVCONTENT_START~~}}	+	The expressions
-	<~~center~~> ~~[[Image:3_4_3b-1~~(2).~~gif]]~~ </~~center~~>	+	<math>\text{ln}\left( x^{\text{2}}+\text{3}x \right)\text{ }</math>
-	{{~~NAVCONTENT_STOP~~}}	+	and
-	{{~~NAVCONTENT_START~~}}	+	<math>\text{ln}\left( \text{3}x^{\text{2}}-\text{2}x \right)\text{ }</math>
-	<~~center~~> ~~[[Image~~:~~3_4_3b~~-2(2).~~gif]]~~ </~~center~~>	+	are equal only if their arguments are equal, i.e.
-	{{~~NAVCONTENT_STOP~~}}	+
		+
		+	<math>\left( x^{\text{2}}+\text{3}x \right)=\left( \text{3}x^{\text{2}}-\text{2}x \right)\text{ }</math>
		+
		+
		+	however, we have to be careful! If we obtain a value for
		+	<math>x</math>
		+	which makes the arguments equal but negative or zero, then it will not correspond to a genuine solution because ln is not defined for negative arguments. At the end of the exercise, we must therefore check that
		+	<math>x^{\text{2}}+\text{3}x\text{ }</math>
		+	and
		+	<math>\text{3}x^{\text{2}}-\text{2}x\text{ }</math>
		+	really are positive for those solutions that we have calculated.
		+
		+	If we move all the terms over to one side in the equation for the arguments, we get the second-degree equation
		+
		+
		+	<math>2x^{2}-5x=0</math>
		+
		+
		+	and we see that both terms contain x, which we can take out as a factor:
		+
		+
		+	<math>x\left( 2x-5 \right)=0</math>
		+
		+
		+	From this factorized expression, we read off that the solutions are
		+	<math>x=0\text{ }</math>
		+	and
		+	<math>x={5}/{2}\;</math>.
		+
		+	A final check shows that when
		+	<math>x=0\text{ }</math>
		+	then
		+	<math>x^{\text{2}}+\text{3}x=\text{ 3}x^{\text{2}}-\text{2}x\text{ }=0</math>, so
		+	<math>x=0\text{ }</math>
		+	is not a solution. On the other hand, when
		+	<math>x={5}/{2}\;</math>
		+	then
		+	<math>x^{\text{2}}+\text{3}x=\text{ 3}x^{\text{2}}-\text{2}x=\text{55}/\text{4}>0</math>, so
		+	<math>x={5}/{2}\;</math>
		+	is a solution.

Tekbot: Lösning 3.4:3b moved to Solution 3.4:3b: Robot: moved page

2008-09-09T09:32:23Z

Lösning 3.4:3b moved to Solution 3.4:3b: Robot: moved page

← Nächstältere Version

Version vom 09:32, 9. Sep. 2008

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

2008-08-21T07:03:19Z

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

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2008-04-02T11:34:05Z

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