Lösung 3.4:2c - Versionsgeschichte

Markus.bez um 13:49, 12. Jun. 2009

2009-06-12T13:49:18Z

← Nächstältere Version		Version vom 13:49, 12. Jun. 2009
Zeile 11:		Zeile 11:
	\end{align}</math>}}		\end{align}</math>}}

-	Wir sammeln alle Terme auf einer Seite der Gleichung, und erhalten	+	Wir sammeln alle Terme auf einer Seite der Gleichung und erhalten

	{{Abgesetzte Formel\|\|<math>x^{2}-x\cdot \ln 2 + \ln 3 = 0</math>}}		{{Abgesetzte Formel\|\|<math>x^{2}-x\cdot \ln 2 + \ln 3 = 0</math>}}
Zeile 22:		Zeile 22:
	\end{align}</math>}}		\end{align}</math>}}

-	Nachdem <math>2 < e < 3</math> und daher <math>\ln 2 < 1 < \ln 3</math>, ist <math>\tfrac{1}{4}(\ln 2)^2 < \ln 3</math>, und die rechte Seite der Gleichung ist also negativ. Nachdem die linke Seite der Gleichung ~~eine Quadrate~~ ist, ist sie immer positiv. Also hat die Gleichung keine Lösung.	+	Nachdem <math>2 < e < 3</math> und daher <math>\ln 2 < 1 < \ln 3</math>, ist <math>\tfrac{1}{4}(\ln 2)^2 < \ln 3</math> und die rechte Seite der Gleichung ist also negativ. Nachdem die linke Seite der Gleichung ein Quadrat ist, ist sie immer positiv. Also hat die Gleichung keine Lösung.

Martin um 17:09, 27. Mär. 2009

2009-03-27T17:09:09Z

← Nächstältere Version		Version vom 17:09, 27. Mär. 2009
Zeile 1:		Zeile 1:
-	~~Regardless of what value~~ ''x'' ~~has, both sides of the equation are positive, because they are of the type "positive number raised to something"~~. ~~It is therefore possible to take the logarithm of both sides and, by using the log laws, to get ''x'' down from the exponents,~~	+	Beide Seiten der Gleichung sind positiv, unabhängig von ''x''. Daher können wir beide Seiten direkt logarithmieren. Durch die Logarithmusgesetze erhalten wir

	{{Abgesetzte Formel\|\|<math>\begin{align}		{{Abgesetzte Formel\|\|<math>\begin{align}
-	\text{~~LHS~~}	+	\text{Linke Seite}
	&= \ln\bigl( 3e^{x^{2}}\bigr)		&= \ln\bigl( 3e^{x^{2}}\bigr)
	= \ln 3 + \ln e^{x^2}		= \ln 3 + \ln e^{x^2}
	= \ln 3 + x^2\ln e		= \ln 3 + x^2\ln e
	= \ln 3 + x^2\,\textrm{,}\\[5pt]		= \ln 3 + x^2\,\textrm{,}\\[5pt]
-	\text{~~RHS~~}	+	\text{Rechte Seite}
	&= \ln 2^x = x\ln 2\,\textrm{.}		&= \ln 2^x = x\ln 2\,\textrm{.}
	\end{align}</math>}}		\end{align}</math>}}

-	~~If we collect the terms onto one side~~, ~~the equation becomes~~	+	Wir sammeln alle Terme auf einer Seite der Gleichung, und erhalten

	{{Abgesetzte Formel\|\|<math>x^{2}-x\cdot \ln 2 + \ln 3 = 0</math>}}		{{Abgesetzte Formel\|\|<math>x^{2}-x\cdot \ln 2 + \ln 3 = 0</math>}}

-	~~which is a standard second-order equation for which we complete the square~~	+	Dies ist eine quadratische Gleichung, die wir durch quadratische Ergänzung lösen

	{{Abgesetzte Formel\|\|<math>\begin{align}		{{Abgesetzte Formel\|\|<math>\begin{align}
Zeile 22:		Zeile 22:
	\end{align}</math>}}		\end{align}</math>}}

-	~~Now, we have to be cautious and remember that, because~~ <math>2 < e < 3</math> ~~and thus~~ <math>\ln 2 < 1 < \ln 3</math> ~~we have that~~ <math>\tfrac{1}{4}(\ln 2)^2 < \ln 3</math> ~~and the right-hand side is therefore negative~~. ~~Since the square on the left-hand side cannot be negative~~, ~~the equation has no solution~~.	+	Nachdem <math>2 < e < 3</math> und daher <math>\ln 2 < 1 < \ln 3</math>, ist <math>\tfrac{1}{4}(\ln 2)^2 < \ln 3</math>, und die rechte Seite der Gleichung ist also negativ. Nachdem die linke Seite der Gleichung eine Quadrate ist, ist sie immer positiv. Also hat die Gleichung keine Lösung.

Tekbot: hat „Solution 3.4:2c“ nach „Lösung 3.4:2c“ verschoben: Robot: moved page

2008-10-22T14:42:04Z

hat „Solution 3.4:2c“ nach „Lösung 3.4:2c“ verschoben: Robot: moved page

← Nächstältere Version

Version vom 14:42, 22. Okt. 2008

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

2008-10-22T08:46:28Z

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

← Nächstältere Version		Version vom 08:46, 22. Okt. 2008
Zeile 1:		Zeile 1:
	Regardless of what value ''x'' has, both sides of the equation are positive, because they are of the type "positive number raised to something". It is therefore possible to take the logarithm of both sides and, by using the log laws, to get ''x'' down from the exponents,		Regardless of what value ''x'' has, both sides of the equation are positive, because they are of the type "positive number raised to something". It is therefore possible to take the logarithm of both sides and, by using the log laws, to get ''x'' down from the exponents,

-	{{~~Displayed math~~\|\|<math>\begin{align}	+	{{Abgesetzte Formel\|\|<math>\begin{align}
	\text{LHS}		\text{LHS}
	&= \ln\bigl( 3e^{x^{2}}\bigr)		&= \ln\bigl( 3e^{x^{2}}\bigr)
Zeile 13:		Zeile 13:
	If we collect the terms onto one side, the equation becomes		If we collect the terms onto one side, the equation becomes

-	{{~~Displayed math~~\|\|<math>x^{2}-x\cdot \ln 2 + \ln 3 = 0</math>}}	+	{{Abgesetzte Formel\|\|<math>x^{2}-x\cdot \ln 2 + \ln 3 = 0</math>}}

	which is a standard second-order equation for which we complete the square		which is a standard second-order equation for which we complete the square

-	{{~~Displayed math~~\|\|<math>\begin{align}	+	{{Abgesetzte Formel\|\|<math>\begin{align}
	\Bigl(x-\frac{1}{2}\ln 2\Bigr)^{2} - \Bigl(\frac{1}{2}\ln 2\Bigr)^{2} + \ln 3 = 0\,\textrm{,}\\[5pt]		\Bigl(x-\frac{1}{2}\ln 2\Bigr)^{2} - \Bigl(\frac{1}{2}\ln 2\Bigr)^{2} + \ln 3 = 0\,\textrm{,}\\[5pt]
	\Bigl(x-\frac{1}{2}\ln 2\Bigr)^{2} = \Bigl(\frac{1}{2}\ln 2\Bigr)^{2} - \ln 3\,\textrm{.}		\Bigl(x-\frac{1}{2}\ln 2\Bigr)^{2} = \Bigl(\frac{1}{2}\ln 2\Bigr)^{2} - \ln 3\,\textrm{.}

Tek um 11:28, 2. Okt. 2008

2008-10-02T11:28:25Z

← Nächstältere Version		Version vom 11:28, 2. Okt. 2008
Zeile 1:		Zeile 1:
-	Regardless of what value	+	Regardless of what value ''x'' has, both sides of the equation are positive, because they are of the type "positive number raised to something". It is therefore possible to take the logarithm of both sides and, by using the log laws, to get ''x'' down from the exponents,
-	~~<math>~~x~~</math>~~	+
-	has, both sides of the equation are positive, because they are of the type "positive number raised to something". It is therefore possible to take the logarithm of both sides and, by using the log laws, to get	+
-	~~<math>~~x~~</math>~~	+
-	down from the exponents:	+
-		+
-	~~LHS~~	+
-	~~<math>=\ln \left( 3e^{x^{2}} \right)=\ln 3+\ln e^{x^{2}}=\ln 3+x^{2}\ln e=\ln 3+x^{2}</math>~~	+
-		+
-	~~RHS~~	+
-	~~<math>=\ln 2^{x}=x\ln 2</math>~~	+

		+	{{Displayed math\|\|<math>\begin{align}
		+	\text{LHS}
		+	&= \ln\bigl( 3e^{x^{2}}\bigr)
		+	= \ln 3 + \ln e^{x^2}
		+	= \ln 3 + x^2\ln e
		+	= \ln 3 + x^2\,\textrm{,}\\[5pt]
		+	\text{RHS}
		+	&= \ln 2^x = x\ln 2\,\textrm{.}
		+	\end{align}</math>}}

	If we collect the terms onto one side, the equation becomes		If we collect the terms onto one side, the equation becomes

		+	{{Displayed math\|\|<math>x^{2}-x\cdot \ln 2 + \ln 3 = 0</math>}}

-	~~<math>x^{2}-x\centerdot \ln 2+\ln 3=0</math>~~	+	which is a standard second-order equation for which we complete the square
-		+
-		+
-	which is a standard second-order equation for which we complete the square:	+
-		+

-	<math>\begin{align}	+	{{Displayed math\|\|<math>\begin{align}
-	& \~~left~~( x-\frac{1}{2}\ln 2 \~~right~~)^{2}-\~~left~~( \frac{1}{2}\ln 2 \~~right~~)^{2}+\ln 3=0 \\	+	\Bigl(x-\frac{1}{2}\ln 2\Bigr)^{2} - \Bigl(\frac{1}{2}\ln 2\Bigr)^{2} + \ln 3 = 0\,\textrm{,}\\[5pt]
-	& \~~left~~( x-\frac{1}{2}\ln 2 \~~right~~)^{2}=\~~left~~( \frac{1}{2}\ln 2 \~~right~~)^{2}-\ln 3 \\	+	\Bigl(x-\frac{1}{2}\ln 2\Bigr)^{2} = \Bigl(\frac{1}{2}\ln 2\Bigr)^{2} - \ln 3\,\textrm{.}
-	\end{align}</math>	+	\end{align}</math>}}

-	Now, we have to be cautious and remember that, because	+	Now, we have to be cautious and remember that, because <math>2 < e < 3</math> and thus <math>\ln 2 < 1 < \ln 3</math> we have that <math>\tfrac{1}{4}(\ln 2)^2 < \ln 3</math> and the right-hand side is therefore negative. Since the square on the left-hand side cannot be negative, the equation has no solution.
-	<math>~~\text{~~2}<e<~~\text{~~3}</math>	+
-	and thus	+
-	<math>\~~text{~~ln 2}<~~\text{~~1}<\~~text{~~ln 3}</math>	+
-	~~which means~~ that	+
-	<math>\~~frac~~{1}{4}~~\left~~( \ln 2 ~~\right~~)^{2}<\ln 3</math>	+
-	and the right-hand side is therefore negative. Since the square on the ~~right~~-hand side cannot be negative, the equation has no solution.	+

Ian um 11:28, 26. Sep. 2008

2008-09-26T11:28:18Z

← Nächstältere Version		Version vom 11:28, 26. Sep. 2008
Zeile 1:		Zeile 1:
-	{{~~NAVCONTENT_START~~}}	+	Regardless of what value
-	<~~center~~> ~~[[Image~~:~~3_4_2c.gif]]~~ </~~center~~>	+	<math>x</math>
-	{{~~NAVCONTENT_STOP~~}}	+	has, both sides of the equation are positive, because they are of the type "positive number raised to something". It is therefore possible to take the logarithm of both sides and, by using the log laws, to get
		+	<math>x</math>
		+	down from the exponents:
		+
		+	LHS
		+	<math>=\ln \left( 3e^{x^{2}} \right)=\ln 3+\ln e^{x^{2}}=\ln 3+x^{2}\ln e=\ln 3+x^{2}</math>
		+
		+	RHS
		+	<math>=\ln 2^{x}=x\ln 2</math>
		+
		+
		+	If we collect the terms onto one side, the equation becomes
		+
		+
		+	<math>x^{2}-x\centerdot \ln 2+\ln 3=0</math>
		+
		+
		+	which is a standard second-order equation for which we complete the square:
		+
		+
		+	<math>\begin{align}
		+	& \left( x-\frac{1}{2}\ln 2 \right)^{2}-\left( \frac{1}{2}\ln 2 \right)^{2}+\ln 3=0 \\
		+	& \left( x-\frac{1}{2}\ln 2 \right)^{2}=\left( \frac{1}{2}\ln 2 \right)^{2}-\ln 3 \\
		+	\end{align}</math>
		+
		+	Now, we have to be cautious and remember that, because
		+	<math>\text{2}<e<\text{3}</math>
		+	and thus
		+	<math>\text{ln 2}<\text{1}<\text{ln 3}</math>
		+	which means that
		+	<math>\frac{1}{4}\left( \ln 2 \right)^{2}<\ln 3</math>
		+	and the right-hand side is therefore negative. Since the square on the right-hand side cannot be negative, the equation has no solution.

Tekbot: Lösning 3.4:2c moved to Solution 3.4:2c: Robot: moved page

2008-09-09T09:31:43Z

Lösning 3.4:2c moved to Solution 3.4:2c: Robot: moved page

← Nächstältere Version

Version vom 09:31, 9. Sep. 2008

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

2008-08-21T07:02:59Z

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

← Nächstältere Version		Version vom 07:02, 21. Aug. 2008
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	{{NAVCONTENT_START}}		{{NAVCONTENT_START}}
-	<center> [[~~Bild~~:3_4_2c.gif]] </center>	+	<center> [[Image:3_4_2c.gif]] </center>
	{{NAVCONTENT_STOP}}		{{NAVCONTENT_STOP}}

Oskar um 11:31, 2. Apr. 2008

2008-04-02T11:31:09Z

← Nächstältere Version		Version vom 11:31, 2. Apr. 2008
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	{{NAVCONTENT_START}}		{{NAVCONTENT_START}}
-	<center> [[Bild:~~3_4_c(2)~~.gif]] </center>	+	<center> [[Bild:3_4_2c.gif]] </center>
	{{NAVCONTENT_STOP}}		{{NAVCONTENT_STOP}}

Oskar: Ny sida: {{NAVCONTENT_START}} Bild:3_4_c(2).gif {{NAVCONTENT_STOP}}

2008-04-02T11:30:32Z

Ny sida: {{NAVCONTENT_START}} <center> Bild:3_4_c(2).gif </center> {{NAVCONTENT_STOP}}

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