Lösung 3.4:1b - Versionsgeschichte

Markus.bez: Sprache und Formulierung

2009-06-12T13:44:37Z

Sprache und Formulierung

Martin um 16:09, 27. Mär. 2009

2009-03-27T16:09:08Z

← Nächstältere Version		Version vom 16:09, 27. Mär. 2009
Zeile 1:		Zeile 1:
-	~~In the equation~~, ~~both sides are positive because the factors~~ <math>e^{x}</math> ~~and~~ <math>3^{-x}</math> ~~are positive regardless of the value of~~ <math>x</math> ~~(a positive base raised to a number always gives a positive number)~~. ~~We can therefore take the natural logarithm of both sides,~~	+	Beide Seiten der Gleichung sind positiv, nachdem <math>e^{x}</math> und <math>3^{-x}</math> für alle <math>x</math> positiv sind. Daher können wir beide Seiten logarithmieren

	{{Abgesetzte Formel\|\|<math>\ln\bigl(13e^{x}\bigr) = \ln\bigl(2\cdot 3^{-x}\bigr)\,\textrm{.}</math>}}		{{Abgesetzte Formel\|\|<math>\ln\bigl(13e^{x}\bigr) = \ln\bigl(2\cdot 3^{-x}\bigr)\,\textrm{.}</math>}}

-	~~Using the log laws, we can divide up the products into several logarithmic terms,~~	+	Wir benutzen die Logarithmusgesetze und erhalten

	{{Abgesetzte Formel\|\|<math>\ln 13+\ln e^{x} =\ln 2+\ln 3^{-x},</math>}}		{{Abgesetzte Formel\|\|<math>\ln 13+\ln e^{x} =\ln 2+\ln 3^{-x},</math>}}

-	~~and using the law <math>\ln a^{b}=b\cdot \ln a</math>, we can get rid of <math>x</math> from the exponents~~	+	und

	{{Abgesetzte Formel\|\|<math>\ln 13 + x\ln e = \ln 2 + (-x)\ln 3\,\textrm{.}</math>}}		{{Abgesetzte Formel\|\|<math>\ln 13 + x\ln e = \ln 2 + (-x)\ln 3\,\textrm{.}</math>}}

-	~~Collect~~ <math>x</math> ~~on one side and the other terms on the other~~,	+	Wir hohlen alle <math>x</math>-Terme auf eine Seite,

	{{Abgesetzte Formel\|\|<math>x\ln e+x\ln 3=\ln 2-\ln 13\,\textrm{.}</math>}}		{{Abgesetzte Formel\|\|<math>x\ln e+x\ln 3=\ln 2-\ln 13\,\textrm{.}</math>}}

-	~~Take out <math>x</math> on the left-hand side and use~~ <math>\ln e=1</math>,	+
		+	und benutzen dass <math>\ln e=1</math>,

	{{Abgesetzte Formel\|\|<math>x( 1+\ln 3)=\ln 2-\ln 13\,\textrm{.}</math>}}		{{Abgesetzte Formel\|\|<math>x( 1+\ln 3)=\ln 2-\ln 13\,\textrm{.}</math>}}

-	~~Then, solve for~~ <math>x</math>,	+	Jetzt lösen wir die Gleichung für <math>x</math>,

	{{Abgesetzte Formel\|\|<math>x=\frac{\ln 2-\ln 13}{1+\ln 3}\,\textrm{.}</math>}}		{{Abgesetzte Formel\|\|<math>x=\frac{\ln 2-\ln 13}{1+\ln 3}\,\textrm{.}</math>}}


-	~~Note~~: ~~Because~~ <math>\ln 2 < \ln 13</math>, ~~we can write the answer as~~	+	Hinweis: Nachdem <math>\ln 2 < \ln 13</math>, können wir die Antwort wie

	{{Abgesetzte Formel\|\|<math>x=-\frac{\ln 13-\ln 2}{1+\ln 3}</math>}}		{{Abgesetzte Formel\|\|<math>x=-\frac{\ln 13-\ln 2}{1+\ln 3}</math>}}

-	~~to indicate that~~ <math>x</math> ~~is negative~~.	+	schreiben, um zu zeigen das <math>x</math> negativ ist.

Tekbot: hat „Solution 3.4:1b“ nach „Lösung 3.4:1b“ verschoben: Robot: moved page

2008-10-22T14:40:44Z

hat „Solution 3.4:1b“ nach „Lösung 3.4:1b“ verschoben: Robot: moved page

← Nächstältere Version

Version vom 14:40, 22. Okt. 2008

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

2008-10-22T08:45:48Z

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

Tek um 08:54, 2. Okt. 2008

2008-10-02T08:54:43Z

← Nächstältere Version		Version vom 08:54, 2. Okt. 2008
Zeile 1:		Zeile 1:
-	In the equation, both sides are positive because the factors	+	In the equation, both sides are positive because the factors <math>e^{x}</math> and <math>3^{-x}</math> are positive regardless of the value of <math>x</math> (a positive base raised to a number always gives a positive number). We can therefore take the natural logarithm of both sides,
-	<math>e^{x}</math>	+
-	and	+
-	<math>3^{-x}</math>	+
-	are positive regardless of the value of	+
-	<math>x</math>	+
-	(a positive base raised to a number always gives a positive number). We can therefore take the natural logarithm of both ~~numbers~~,	+

		+	{{Displayed math\|\|<math>\ln\bigl(13e^{x}\bigr) = \ln\bigl(2\cdot 3^{-x}\bigr)\,\textrm{.}</math>}}

-	~~<math>\ln \left( 13e^{x} \right)=\ln \left( 2\centerdot 3^{-x} \right)</math>~~	+	Using the log laws, we can divide up the products into several logarithmic terms,

		+	{{Displayed math\|\|<math>\ln 13+\ln e^{x} =\ln 2+\ln 3^{-x},</math>}}

-	~~Using~~ the ~~log~~ law, we can ~~divide up~~ the ~~products into several logarithmic terms,~~	+	and using the law <math>\ln a^{b}=b\cdot \ln a</math>, we can get rid of <math>x</math> from the exponents

		+	{{Displayed math\|\|<math>\ln 13 + x\ln e = \ln 2 + (-x)\ln 3\,\textrm{.}</math>}}

-	<math>~~\ln 13+\ln e^{~~x~~}=\ln 2+\ln 3^{-x}~~</math>	+	Collect <math>x</math> on one side and the other terms on the other,

		+	{{Displayed math\|\|<math>x\ln e+x\ln 3=\ln 2-\ln 13\,\textrm{.}</math>}}

-	~~and using the law~~	+	Take out <math>x</math> on the left-hand side and use <math>\ln e=1</math>,
-	<math>~~\ln a^{b}=b\centerdot \ln a~~</math>~~, we can get rid of~~	+
-	<math>x</math>	+
-	~~from the exponents:~~	+

		+	{{Displayed math\|\|<math>x( 1+\ln 3)=\ln 2-\ln 13\,\textrm{.}</math>}}

-	<math>~~\ln 13+~~x~~\ln e=\ln 2+\left( -x \right)\ln 3~~</math>	+	Then, solve for <math>x</math>,

		+	{{Displayed math\|\|<math>x=\frac{\ln 2-\ln 13}{1+\ln 3}\,\textrm{.}</math>}}

-	Collecting together
-	<math>x</math>
-	on one side and the other terms on the other,

		+	Note: Because <math>\ln 2 < \ln 13</math>, we can write the answer as

-	<math>x\~~ln e+x~~\ln 3=\ln 2-\ln 13</math>	+	{{Displayed math\|\|<math>x=-\frac{\ln 13-\ln 2}{1+\ln 3}</math>}}

-		+	to indicate that <math>x</math> is negative.
-	~~Take out~~	+
-	~~<math>x</math>~~	+
-	~~on the left-hand side and use~~	+
-	~~<math>\ln e=1</math>~~	+
-	:	+
-		+
-		+
-	~~<math>x\left( 1+\ln 3 \right)=\ln 2-\ln 13</math>~~	+
-		+
-		+
-	~~Then, solve for~~	+
-	~~<math>x</math>~~	+
-	:	+
-		+
-		+
-	~~<math>x=\frac{\ln 2-\ln 13}{1+\ln 3}</math>~~	+
-		+
-		+
-	~~NOTE: Because~~	+
-	~~<math>\ln 2<\ln 13</math>, we can write the answer as~~	+
-		+
-		+
-	~~<math>x=-\frac{\ln 13-\ln 2}{1+\ln 3}</math>~~	+
-		+
-		+
-	~~in order~~ to indicate that	+
-	<math>x</math>	+
-	is negative.	+

Ian um 12:55, 12. Sep. 2008

2008-09-12T12:55:06Z

← Nächstältere Version		Version vom 12:55, 12. Sep. 2008
Zeile 1:		Zeile 1:
-	{~~{NAVCONTENT_START}~~}	+	In the equation, both sides are positive because the factors
-	<~~center~~> ~~[[Image:3_4_1b~~-1(2).~~gif]]~~ </~~center~~>	+	<math>e^{x}</math>
-	{{~~NAVCONTENT_STOP}~~}	+	and
-	{~~{NAVCONTENT_START}~~}	+	<math>3^{-x}</math>
-	<~~center~~> ~~[[Image~~:~~3_4_1b-~~2(2)~~.gif]]~~ </~~center~~>	+	are positive regardless of the value of
-	{{~~NAVCONTENT_STOP~~}}	+	<math>x</math>
		+	(a positive base raised to a number always gives a positive number). We can therefore take the natural logarithm of both numbers,
		+
		+
		+	<math>\ln \left( 13e^{x} \right)=\ln \left( 2\centerdot 3^{-x} \right)</math>
		+
		+
		+	Using the log law, we can divide up the products into several logarithmic terms,
		+
		+
		+	<math>\ln 13+\ln e^{x}=\ln 2+\ln 3^{-x}</math>
		+
		+
		+	and using the law
		+	<math>\ln a^{b}=b\centerdot \ln a</math>, we can get rid of
		+	<math>x</math>
		+	from the exponents:
		+
		+
		+	<math>\ln 13+x\ln e=\ln 2+\left( -x \right)\ln 3</math>
		+
		+
		+	Collecting together
		+	<math>x</math>
		+	on one side and the other terms on the other,
		+
		+
		+	<math>x\ln e+x\ln 3=\ln 2-\ln 13</math>
		+
		+
		+	Take out
		+	<math>x</math>
		+	on the left-hand side and use
		+	<math>\ln e=1</math>
		+	:
		+
		+
		+	<math>x\left( 1+\ln 3 \right)=\ln 2-\ln 13</math>
		+
		+
		+	Then, solve for
		+	<math>x</math>
		+	:
		+
		+
		+	<math>x=\frac{\ln 2-\ln 13}{1+\ln 3}</math>
		+
		+
		+	NOTE: Because
		+	<math>\ln 2<\ln 13</math>, we can write the answer as
		+
		+
		+	<math>x=-\frac{\ln 13-\ln 2}{1+\ln 3}</math>
		+
		+
		+	in order to indicate that
		+	<math>x</math>
		+	is negative.

Tekbot: Lösning 3.4:1b moved to Solution 3.4:1b: Robot: moved page

2008-09-09T09:30:23Z

Lösning 3.4:1b moved to Solution 3.4:1b: Robot: moved page

← Nächstältere Version

Version vom 09:30, 9. Sep. 2008

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

2008-08-21T07:02:19Z

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

← Nächstältere Version		Version vom 07:02, 21. Aug. 2008
Zeile 1:		Zeile 1:
	{{NAVCONTENT_START}}		{{NAVCONTENT_START}}
-	<center> [[~~Bild~~:3_4_1b-1(2).gif]] </center>	+	<center> [[Image:3_4_1b-1(2).gif]] </center>
	{{NAVCONTENT_STOP}}		{{NAVCONTENT_STOP}}
	{{NAVCONTENT_START}}		{{NAVCONTENT_START}}
-	<center> [[~~Bild~~:3_4_1b-2(2).gif]] </center>	+	<center> [[Image:3_4_1b-2(2).gif]] </center>
	{{NAVCONTENT_STOP}}		{{NAVCONTENT_STOP}}

Oskar: Ny sida: {{NAVCONTENT_START}} Bild:3_4_1b-1(2).gif {{NAVCONTENT_STOP}} {{NAVCONTENT_START}} Bild:3_4_1b-2(2).gif {{NAVCONTENT_STOP}}

2008-04-02T10:28:35Z

Ny sida: {{NAVCONTENT_START}} <center> Bild:3_4_1b-1(2).gif </center> {{NAVCONTENT_STOP}} {{NAVCONTENT_START}} <center> Bild:3_4_1b-2(2).gif </center> {{NAVCONTENT_STOP}}

Neue Seite

{{NAVCONTENT_START}}
<center> [[Bild:3_4_1b-1(2).gif]] </center>
{{NAVCONTENT_STOP}}
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<center> [[Bild:3_4_1b-2(2).gif]] </center>
{{NAVCONTENT_STOP}}

← Nächstältere Version		Version vom 13:44, 12. Jun. 2009
Zeile 5:		Zeile 5:
	Wir benutzen die Logarithmusgesetze und erhalten		Wir benutzen die Logarithmusgesetze und erhalten

-	{{Abgesetzte Formel\|\|<math>\ln 13+\ln e^{x} =\ln 2+\ln 3^{-x},</math>}}	+	{{Abgesetzte Formel\|\|<math>\ln 13+\ln e^{x} =\ln 2+\ln 3^{-x}</math>}}

	und		und
Zeile 11:		Zeile 11:
	{{Abgesetzte Formel\|\|<math>\ln 13 + x\ln e = \ln 2 + (-x)\ln 3\,\textrm{.}</math>}}		{{Abgesetzte Formel\|\|<math>\ln 13 + x\ln e = \ln 2 + (-x)\ln 3\,\textrm{.}</math>}}

-	Wir ~~hohlen~~ alle <math>x</math>-Terme auf eine Seite,	+	Wir schreiben alle <math>x</math>-Terme auf eine Seite:

	{{Abgesetzte Formel\|\|<math>x\ln e+x\ln 3=\ln 2-\ln 13\,\textrm{.}</math>}}		{{Abgesetzte Formel\|\|<math>x\ln e+x\ln 3=\ln 2-\ln 13\,\textrm{.}</math>}}


-	~~und benutzen~~ dass <math>\ln e=1</math>,	+	Dann verwenden wir, dass <math>\ln e=1</math>:

	{{Abgesetzte Formel\|\|<math>x( 1+\ln 3)=\ln 2-\ln 13\,\textrm{.}</math>}}		{{Abgesetzte Formel\|\|<math>x( 1+\ln 3)=\ln 2-\ln 13\,\textrm{.}</math>}}

-	Jetzt lösen wir die Gleichung für <math>x</math>,	+	Jetzt lösen wir die Gleichung für <math>x</math>:

	{{Abgesetzte Formel\|\|<math>x=\frac{\ln 2-\ln 13}{1+\ln 3}\,\textrm{.}</math>}}		{{Abgesetzte Formel\|\|<math>x=\frac{\ln 2-\ln 13}{1+\ln 3}\,\textrm{.}</math>}}


-	Hinweis: Nachdem <math>\ln 2 < \ln 13</math>, können wir die Antwort ~~wie~~	+	Hinweis: Nachdem <math>\ln 2 < \ln 13</math>, können wir die Antwort als

	{{Abgesetzte Formel\|\|<math>x=-\frac{\ln 13-\ln 2}{1+\ln 3}</math>}}		{{Abgesetzte Formel\|\|<math>x=-\frac{\ln 13-\ln 2}{1+\ln 3}</math>}}

-	schreiben, um zu zeigen ~~das~~ <math>x</math> negativ ist.	+	schreiben, um zu zeigen, dass <math>x</math> negativ ist.

← Nächstältere Version		Version vom 08:45, 22. Okt. 2008
Zeile 1:		Zeile 1:
	In the equation, both sides are positive because the factors <math>e^{x}</math> and <math>3^{-x}</math> are positive regardless of the value of <math>x</math> (a positive base raised to a number always gives a positive number). We can therefore take the natural logarithm of both sides,		In the equation, both sides are positive because the factors <math>e^{x}</math> and <math>3^{-x}</math> are positive regardless of the value of <math>x</math> (a positive base raised to a number always gives a positive number). We can therefore take the natural logarithm of both sides,

-	{{~~Displayed math~~\|\|<math>\ln\bigl(13e^{x}\bigr) = \ln\bigl(2\cdot 3^{-x}\bigr)\,\textrm{.}</math>}}	+	{{Abgesetzte Formel\|\|<math>\ln\bigl(13e^{x}\bigr) = \ln\bigl(2\cdot 3^{-x}\bigr)\,\textrm{.}</math>}}

	Using the log laws, we can divide up the products into several logarithmic terms,		Using the log laws, we can divide up the products into several logarithmic terms,

-	{{~~Displayed math~~\|\|<math>\ln 13+\ln e^{x} =\ln 2+\ln 3^{-x},</math>}}	+	{{Abgesetzte Formel\|\|<math>\ln 13+\ln e^{x} =\ln 2+\ln 3^{-x},</math>}}

	and using the law <math>\ln a^{b}=b\cdot \ln a</math>, we can get rid of <math>x</math> from the exponents		and using the law <math>\ln a^{b}=b\cdot \ln a</math>, we can get rid of <math>x</math> from the exponents

-	{{~~Displayed math~~\|\|<math>\ln 13 + x\ln e = \ln 2 + (-x)\ln 3\,\textrm{.}</math>}}	+	{{Abgesetzte Formel\|\|<math>\ln 13 + x\ln e = \ln 2 + (-x)\ln 3\,\textrm{.}</math>}}

	Collect <math>x</math> on one side and the other terms on the other,		Collect <math>x</math> on one side and the other terms on the other,

-	{{~~Displayed math~~\|\|<math>x\ln e+x\ln 3=\ln 2-\ln 13\,\textrm{.}</math>}}	+	{{Abgesetzte Formel\|\|<math>x\ln e+x\ln 3=\ln 2-\ln 13\,\textrm{.}</math>}}

	Take out <math>x</math> on the left-hand side and use <math>\ln e=1</math>,		Take out <math>x</math> on the left-hand side and use <math>\ln e=1</math>,

-	{{~~Displayed math~~\|\|<math>x( 1+\ln 3)=\ln 2-\ln 13\,\textrm{.}</math>}}	+	{{Abgesetzte Formel\|\|<math>x( 1+\ln 3)=\ln 2-\ln 13\,\textrm{.}</math>}}

	Then, solve for <math>x</math>,		Then, solve for <math>x</math>,

-	{{~~Displayed math~~\|\|<math>x=\frac{\ln 2-\ln 13}{1+\ln 3}\,\textrm{.}</math>}}	+	{{Abgesetzte Formel\|\|<math>x=\frac{\ln 2-\ln 13}{1+\ln 3}\,\textrm{.}</math>}}


	Note: Because <math>\ln 2 < \ln 13</math>, we can write the answer as		Note: Because <math>\ln 2 < \ln 13</math>, we can write the answer as

-	{{~~Displayed math~~\|\|<math>x=-\frac{\ln 13-\ln 2}{1+\ln 3}</math>}}	+	{{Abgesetzte Formel\|\|<math>x=-\frac{\ln 13-\ln 2}{1+\ln 3}</math>}}

	to indicate that <math>x</math> is negative.		to indicate that <math>x</math> is negative.