Solution 3.4:3b - Revision history

Tek at 14:25, 2 October 2008

2008-10-02T14:25:08Z

←Older revision		Revision as of 14:25, 2 October 2008
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-	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 at 11:52, 26 September 2008

2008-09-26T11:52:38Z

←Older revision		Revision as of 11:52, 26 September 2008
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-	{{~~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

←Older revision

Revision as of 09:32, 9 September 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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