Solution 3.4:3a - Revision history

Tek at 14:15, 2 October 2008

Tek — Thu, 02 Oct 2008 14:15:24 GMT

←Older revision		Revision as of 14:15, 2 October 2008
Line 1:		Line 1:
-	Both left- and right-hand sides are positive for all values of	+	Both left- and right-hand sides are positive for all values of ''x'' and this means that we can take the logarithm of both sides and get a more manageable equation,
-	~~<math>~~x~~</math>~~	+
-	and this means that we can take the logarithm of both sides and get a more manageable equation.	+
-		+
-	~~LHS~~	+
-	~~<math>=\ln 2^{-x^{2}}=-x^{2}\centerdot \ln 2~~,~~</math>~~	+
-		+
-	~~RHS~~	+
-	~~<math>=\ln \left( 2e^{2x} \right)=\ln 2+\ln e^{2x}=\ln 2+2x\centerdot \ln e=\ln 2+2x\centerdot 1</math>~~	+

		+	{{Displayed math\|\|<math>\begin{align}
		+	\text{LHS} &= \ln 2^{-x^{2}} = -x^{2}\cdot \ln 2\,,\\[5pt]
		+	\text{RHS} &= \ln \bigl(2e^{2x}\bigr) = \ln 2 + \ln e^{2x} = \ln 2 + 2x\cdot \ln e = \ln 2 + 2x\cdot 1\,\textrm{.}
		+	\end{align}</math>}}

	After a little rearranging, the equation becomes		After a little rearranging, the equation becomes

		+	{{Displayed math\|\|<math>x^{2}+\frac{2}{\ln 2}x+1=0\,\textrm{.}</math>}}

-	~~<math>x^{2}+\frac{2}{\ln 2}x+1=0</math>~~	+	We complete the square of the left-hand side,

-	~~We complete the square of the left~~-~~hand side~~	+	{{Displayed math\|\|<math>\Bigl(x+\frac{1}{\ln 2}\Bigr)^{2} - \Bigl(\frac{1}{\ln 2} \Bigr)^{2} + 1 = 0\,,</math>}}

		+	and move the constant terms over to the right-hand side,

-	<math>\~~left~~( x+\frac{1}{\ln 2} \~~right~~)^{2}-\~~left~~( \frac{1}{\ln 2} \~~right~~)^{2}+1=0</math>	+	{{Displayed math\|\|<math>\Bigl(x+\frac{1}{\ln 2}\Bigr)^{2} = \Bigl(\frac{1}{\ln 2} \Bigr)^{2} - 1\,\textrm{.}</math>}}

-		+	It can be difficult to see whether the right-hand side is positive or not, but if we remember that <math>e > 2</math> and that thus <math>\ln 2 < \ln e = 1\,</math>, we must have that <math>(1/\ln 2)^{2} > 1\,</math>, i.e. the right-hand side is positive.
-	~~and move the constant terms over to the right-hand side:~~	+
-		+
-		+
-	~~<math>\left( x+\frac{1}{\ln 2} \right)^{2}=\left( \frac{1}{\ln 2} \right)^{2}-1</math>~~	+
-		+
-		+
-	It can be difficult to see whether the right-hand side is positive or not, but if we remember that	+
-	<math>e>2</math>	+
-	and that thus	+
-	<math>\~~text{~~ln 2}<\ln e=\~~text{1}~~</math>, we must have that	+
-	<math>~~\left~~( {1}/{\ln 2~~}\; \right~~)^{2}>1</math>, i.e. the right-hand side is positive.	+

	The equation therefore has the solutions		The equation therefore has the solutions

-		+	{{Displayed math\|\|<math>x=-\frac{1}{\ln 2}\pm \sqrt{\Bigl(\frac{1}{\ln 2} \Bigr)^{2}-1}\,,</math>}}
-	<math>x=-\frac{1}{\ln 2}\pm \sqrt{\~~left~~( \frac{1}{\ln 2} \~~right~~)^{2}-1,}</math>	+
-		+

	which can also be written as		which can also be written as

-		+	{{Displayed math\|\|<math>x=\frac{-1\pm \sqrt{1-(\ln 2)^{2}}}{\ln 2}\,\textrm{.}</math>}}
-	<math>x=\frac{-1\pm \sqrt{1-~~\left~~( \ln 2 ~~\right~~)^{2}}}{\ln 2}</math>	+

Ian at 11:41, 26 September 2008

Ian — Fri, 26 Sep 2008 11:41:39 GMT

←Older revision		Revision as of 11:41, 26 September 2008
Line 1:		Line 1:
-	{{~~NAVCONTENT_START~~}}	+	Both left- and right-hand sides are positive for all values of
-	<~~center~~> ~~[[Image:3_4_3a~~-1(2)~~.gif]]~~ </~~center~~>	+	<math>x</math>
-	{{~~NAVCONTENT_STOP~~}}	+	and this means that we can take the logarithm of both sides and get a more manageable equation.
-	{{~~NAVCONTENT_START~~}}	+
-	<~~center~~> ~~[[Image:3_4_3a~~-2(2)~~.gif]]~~ </~~center~~>	+	LHS
-	{{~~NAVCONTENT_STOP~~}}	+	<math>=\ln 2^{-x^{2}}=-x^{2}\centerdot \ln 2,</math>
		+
		+	RHS
		+	<math>=\ln \left( 2e^{2x} \right)=\ln 2+\ln e^{2x}=\ln 2+2x\centerdot \ln e=\ln 2+2x\centerdot 1</math>
		+
		+
		+	After a little rearranging, the equation becomes
		+
		+
		+	<math>x^{2}+\frac{2}{\ln 2}x+1=0</math>
		+
		+	We complete the square of the left-hand side
		+
		+
		+	<math>\left( x+\frac{1}{\ln 2} \right)^{2}-\left( \frac{1}{\ln 2} \right)^{2}+1=0</math>
		+
		+
		+	and move the constant terms over to the right-hand side:
		+
		+
		+	<math>\left( x+\frac{1}{\ln 2} \right)^{2}=\left( \frac{1}{\ln 2} \right)^{2}-1</math>
		+
		+
		+	It can be difficult to see whether the right-hand side is positive or not, but if we remember that
		+	<math>e>2</math>
		+	and that thus
		+	<math>\text{ln 2}<\ln e=\text{1}</math>, we must have that
		+	<math>\left( {1}/{\ln 2}\; \right)^{2}>1</math>, i.e. the right-hand side is positive.
		+
		+	The equation therefore has the solutions
		+
		+
		+	<math>x=-\frac{1}{\ln 2}\pm \sqrt{\left( \frac{1}{\ln 2} \right)^{2}-1,}</math>
		+
		+
		+	which can also be written as
		+
		+
		+	<math>x=\frac{-1\pm \sqrt{1-\left( \ln 2 \right)^{2}}}{\ln 2}</math>

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

Tekbot — Tue, 09 Sep 2008 09:32:03 GMT

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

←Older revision

Revision as of 09:32, 9 September 2008

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

Tekbot — Thu, 21 Aug 2008 07:03:09 GMT

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

←Older revision		Revision as of 07:03, 21 August 2008
Line 1:		Line 1:
	{{NAVCONTENT_START}}		{{NAVCONTENT_START}}
-	<center> [[~~Bild~~:3_4_3a-1(2).gif]] </center>	+	<center> [[Image:3_4_3a-1(2).gif]] </center>
	{{NAVCONTENT_STOP}}		{{NAVCONTENT_STOP}}
	{{NAVCONTENT_START}}		{{NAVCONTENT_START}}
-	<center> [[~~Bild~~:3_4_3a-2(2).gif]] </center>	+	<center> [[Image:3_4_3a-2(2).gif]] </center>
	{{NAVCONTENT_STOP}}		{{NAVCONTENT_STOP}}

Oskar: Ny sida: {{NAVCONTENT_START}} Bild:3_4_3a-1(2).gif {{NAVCONTENT_STOP}} {{NAVCONTENT_START}} Bild:3_4_3a-2(2).gif {{NAVCONTENT_STOP}}

Oskar — Wed, 02 Apr 2008 11:32:45 GMT

Ny sida: {{NAVCONTENT_START}} <center> Bild:3_4_3a-1(2).gif </center> {{NAVCONTENT_STOP}} {{NAVCONTENT_START}} <center> Bild:3_4_3a-2(2).gif </center> {{NAVCONTENT_STOP}}

New page

{{NAVCONTENT_START}}
<center> [[Bild:3_4_3a-1(2).gif]] </center>
{{NAVCONTENT_STOP}}
{{NAVCONTENT_START}}
<center> [[Bild:3_4_3a-2(2).gif]] </center>
{{NAVCONTENT_STOP}}