Solution 3.4:3c - Revision history

Tek at 14:34, 2 October 2008

Tek — Thu, 02 Oct 2008 14:34:12 GMT

←Older revision		Revision as of 14:34, 2 October 2008
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	With the log laws, we can write the left-hand side as one logarithmic expression,		With the log laws, we can write the left-hand side as one logarithmic expression,

		+	{{Displayed math\|\|<math>\ln x+\ln (x+4) = \ln (x(x+4))\,,</math>}}

-	<math>\ln x+\ln ~~\left~~( x+4 ~~\right~~)~~=\ln \left(~~ x~~\left(~~ x+4 \~~right) \right)~~</math>	+	but this rewriting presupposes that the expressions <math>\ln x</math> and <math>\ln (x+4)</math> are defined, i.e. <math>x > 0</math> and <math>x+4 > 0\,</math>. Therefore, if we choose to continue with the equation

		+	{{Displayed math\|\|<math>\ln (x(x+4)) = \ln (2x+3)</math>}}

-	~~but this rewriting presupposes that the expressions~~	+	we must remember to permit only solutions that satisfy <math>x > 0</math> (the condition <math>x+\text{4}>0</math> is then automatically satisfied).
-	~~<math>\text{ln }x\text{ }</math>~~	+
-	~~and~~	+
-	~~<math>\text{ln}\left( x+\text{4} \right)</math>~~	+
-	~~are defined, i.e.~~	+
-	~~<math>x>0</math>~~	+
-	~~and~~	+
-	~~<math>x+\text{4}>0</math>. Therefore, if we choose to continue with the equation~~	+
-		+
-		+
-	~~<math>\ln \left( x\left( x+4 \right) \right)=\ln \left( 2x+3 \right)</math>~~	+
-		+
-		+
-	we must remember to permit only solutions that satisfy	+
-	<math>x>0</math>	+
-	(the condition	+
-	<math>x+\text{4}>0</math>	+
-	is then automatically satisfied).	+

	The equation rewritten in this way is, in turn, only satisfied if the arguments		The equation rewritten in this way is, in turn, only satisfied if the arguments
-	<math>x~~\left~~( x+~~\text{~~4~~} \right~~)~~\text{ }~~</math>	+	<math>x(x+4)</math> and <math>2x+3</math> are equal to each other and positive, i.e.
-	and	+
-	<math>~~\text{2}x~~+~~\text{~~3}</math>	+
-	are equal to each other and positive, i.e.	+
-		+
-		+
-	~~<math>x\left( x+\text{4} \right)=\text{2}x+\text{3}</math>~~	+
-		+
-		+
-	~~We rewrite this equation as~~	+
-	~~<math>x^{\text{2}}-\text{2}x-\text{3}=0</math>~~	+
-	~~and completing the square gives~~	+

		+	{{Displayed math\|\|<math>x(x+4) = 2x+3\,\textrm{.}</math>}}

-	<math>~~\begin{align}~~	+	We rewrite this equation as <math>x^2+2x-3=0</math> and completing the square gives
-	~~& \left(~~ x~~+1 \right)~~^{2~~}-1^{2}~~-3=0 \\	+
-	~~& \left( x+1 \right)^{2}=4 \\~~	+
-	~~\end{align}~~</math>	+

		+	{{Displayed math\|\|<math>\begin{align}
		+	(x+1)^2-1^2-3 &= 0\,,\\
		+	(x+1)^2=4\,,
		+	\end{align}</math>}}

-	which means that	+	which means that <math>x=-1\pm 2</math>, i.e. <math>x=-3</math> and <math>x=1\,</math>.
-	<math>x=-~~\text{~~1}\pm ~~\text{~~2}</math>, i.e.	+
-	<math>x=-~~\text{~~3}</math>	+
-	and	+
-	<math>x=~~\text{~~1}</math>.	+

-	Because	+	Because <math>x=-3</math> is negative, we neglect it, whilst for <math>x=1</math> we have both that <math>x > 0</math> and <math>x(x+4) = 2x+3 > 0\,</math>. Therefore, the answer is <math>x=1\,</math>.
-	<math>x=-~~\text{~~3}</math>	+
-	is negative, we neglect it, whilst for	+
-	<math>x=~~\text{~~1}</math>	+
-	we have both that	+
-	<math>x>0~~\text{ }~~</math>	+
-	and	+
-	<math>x~~\left~~( x+~~\text{~~4~~} \right~~)=~~\text{2}x~~+~~\text{~~3}>0</math>. Therefore, the answer is	+
-	<math>x=~~\text{~~1}</math>.	+

Ian at 12:05, 26 September 2008

Ian — Fri, 26 Sep 2008 12:05:06 GMT

←Older revision		Revision as of 12:05, 26 September 2008
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-	{{~~NAVCONTENT_START}~~}	+	With the log laws, we can write the left-hand side as one logarithmic expression,
-	<~~center~~> ~~[[Image:3_4_3c-1~~(2).~~gif]]~~ </~~center~~>	+
-	{{~~NAVCONTENT_STOP~~}}	+
-	{{~~NAVCONTENT_START~~}}	+	<math>\ln x+\ln \left( x+4 \right)=\ln \left( x\left( x+4 \right) \right)</math>
-	<~~center~~> ~~[[Image:3_4_3c~~-2(2).~~gif]]~~ </~~center~~>	+
-	{{~~NAVCONTENT_STOP~~}}	+
		+	but this rewriting presupposes that the expressions
		+	<math>\text{ln }x\text{ }</math>
		+	and
		+	<math>\text{ln}\left( x+\text{4} \right)</math>
		+	are defined, i.e.
		+	<math>x>0</math>
		+	and
		+	<math>x+\text{4}>0</math>. Therefore, if we choose to continue with the equation
		+
		+
		+	<math>\ln \left( x\left( x+4 \right) \right)=\ln \left( 2x+3 \right)</math>
		+
		+
		+	we must remember to permit only solutions that satisfy
		+	<math>x>0</math>
		+	(the condition
		+	<math>x+\text{4}>0</math>
		+	is then automatically satisfied).
		+
		+	The equation rewritten in this way is, in turn, only satisfied if the arguments
		+	<math>x\left( x+\text{4} \right)\text{ }</math>
		+	and
		+	<math>\text{2}x+\text{3}</math>
		+	are equal to each other and positive, i.e.
		+
		+
		+	<math>x\left( x+\text{4} \right)=\text{2}x+\text{3}</math>
		+
		+
		+	We rewrite this equation as
		+	<math>x^{\text{2}}-\text{2}x-\text{3}=0</math>
		+	and completing the square gives
		+
		+
		+	<math>\begin{align}
		+	& \left( x+1 \right)^{2}-1^{2}-3=0 \\
		+	& \left( x+1 \right)^{2}=4 \\
		+	\end{align}</math>
		+
		+
		+	which means that
		+	<math>x=-\text{1}\pm \text{2}</math>, i.e.
		+	<math>x=-\text{3}</math>
		+	and
		+	<math>x=\text{1}</math>.
		+
		+	Because
		+	<math>x=-\text{3}</math>
		+	is negative, we neglect it, whilst for
		+	<math>x=\text{1}</math>
		+	we have both that
		+	<math>x>0\text{ }</math>
		+	and
		+	<math>x\left( x+\text{4} \right)=\text{2}x+\text{3}>0</math>. Therefore, the answer is
		+	<math>x=\text{1}</math>.

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

Tekbot — Tue, 09 Sep 2008 09:32:43 GMT

Lösning 3.4:3c moved to Solution 3.4:3c: 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 07:03:29 GMT

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

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Oskar — Wed, 02 Apr 2008 11:35:44 GMT

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