Solution 3.4:2b - Revision history

Tek at 10:18, 7 October 2008

Tek — Tue, 07 Oct 2008 10:18:25 GMT

←Older revision		Revision as of 10:18, 7 October 2008
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	If we write the equation as		If we write the equation as

		+	{{Displayed math\|\|<math>\bigl(e^{x}\bigr)^{2} + e^{x} = 4</math>}}

		+	we see that <math>x</math> appears only in the combination <math>e^{x}</math> and it is therefore appropriate to treat <math>e^{x}</math> as a new unknown in the equation and then, when we have obtained the value of <math>e^{x}</math>, we can calculate the corresponding value of <math>x</math> by simply taking the logarithm.

-	~~<math>\left( e^{x} \right)^{2}+e^{x}=4</math>~~	+	For clarity, we set <math>t=e^{x}</math>, so that the equation can be written as
-		+
-		+
-	~~we see that~~	+
-	~~<math>x</math>~~	+
-	~~appears only in the combination~~	+
-	~~<math>e^{x}</math>~~	+
-	~~and it is therefore appropriate to treat~~	+
-	~~<math>e^{x}</math>~~	+
-	~~as a new unknown in the equation and then, when we have obtained the value of~~	+
-	~~<math>e^{x}</math>, we can calculate the corresponding value of~~	+
-	~~<math>x</math>~~	+
-	~~by simply taking the logarithm.~~	+
-		+
-	For clarity, we set	+
-	<math>t=e^{x}</math>, so that the equation can be written as	+
-		+
-		+
-	~~<math>t^{2}+t=4</math>~~	+

		+	{{Displayed math\|\|<math>t^{2}+t=4</math>}}

	and we solve this second-degree equation by completing the square,		and we solve this second-degree equation by completing the square,

-		+	{{Displayed math\|\|<math>t^{2}+t = \Bigl( t+\frac{1}{2} \Bigr)^{2}-\Bigl( \frac{1}{2} \Bigr)^{2} = \Bigl( t+\frac{1}{2} \Bigr)^{2} - \frac{1}{4}\,,</math>}}
-	<math>t^{2}+t=\~~left~~( t+\frac{1}{2} \~~right~~)^{2}-\~~left~~( \frac{1}{2} \~~right~~)^{2}=\~~left~~( t+\frac{1}{2} \~~right~~)^{2}-\frac{1}{4}</math>	+
-		+

	which gives		which gives

		+	{{Displayed math\|\|<math>\Bigl(t+\frac{1}{2}\Bigr)^{2} - \frac{1}{4} = 4\quad \Leftrightarrow \quad t = -\frac{1}{2}\pm \frac{\sqrt{17}}{2}\,\textrm{.}</math>}}

-	~~<math>\left( t+\frac{1}{2} \right)^{2}-\frac{1}{4}=4\quad \Leftrightarrow \quad t=-\frac{1}{2}\pm \frac{\sqrt{17}}{2}</math>~~	+	These two roots give us two possible values for <math>e^{x}</math>,
-		+
-		+
-	These two roots give us two possible values for	+
-	<math>e^{x}</math>,	+
-		+
-		+
-	~~<math>e^{x}=-\frac{1}{2}-\frac{\sqrt{17}}{2}</math>~~	+
-	or	+
-	~~<math>e^{x}=-\frac{1}{2}+\frac{\sqrt{17}}{2}</math>~~	+
-		+
-		+
-	~~In the first case, the right-hand side is negative and because "~~	+
-	~~<math>e</math>~~	+
-	~~raised to anything" can never be negative, there is no~~	+
-	~~<math>x</math>~~	+
-	~~that can satisfy this equality. The other case, on the other hand, has a positive right-hand side (because~~	+
-	~~<math>\sqrt{17}>1</math>~~	+
-	~~) and we can take the logarithm of both sides to obtain~~	+
-		+
-		+
-	~~<math>x=\ln \left( \frac{\sqrt{17}}{2}-\frac{1}{2} \right)</math>~~	+

		+	{{Displayed math\|\|<math>e^{x}=-\frac{1}{2}-\frac{\sqrt{17}}{2}\qquad\text{or}\qquad e^{x} = -\frac{1}{2}+\frac{\sqrt{17}}{2}\,\textrm{.}</math>}}

-	~~NOTE: It~~ is ~~a little tricky to check the answer~~ to the ~~original equation~~, ~~so we can be satisfied with substituting~~	+	In the first case, the right-hand side is negative and because "''e'' raised to anything" can never be negative, there is no ''x'' that can satisfy this equality. The other case, on the other hand, has a positive right-hand side (because
-	<math>~~t=\frac{~~\sqrt{17}~~}{2}-\frac{~~1~~}{2}~~</math>	+	<math>\sqrt{17}>1</math>) and we can take the logarithm of both sides to obtain
-	~~into~~ the ~~equation~~	+
-	~~<math>t^{\text{2}}+t=\text{4}</math>,~~	+

-	~~LHS~~	+	{{Displayed math\|\|<math>x=\ln \Bigl(\frac{\sqrt{17}}{2}-\frac{1}{2}\Bigr)\,\textrm{.}</math>}}
-	<math>=</math>	+

-
-	<math>\begin{align}
-	& =\left( \frac{\sqrt{17}}{2}-\frac{1}{2} \right)^{2}+\left( \frac{\sqrt{17}}{2}-\frac{1}{2} \right)=\frac{17}{4}-2\centerdot \frac{1}{2}\centerdot \frac{\sqrt{17}}{2}+\frac{1}{4}+\frac{\sqrt{17}}{2}-\frac{1}{2} \\
-	& =\frac{17}{4}+\frac{1}{4}-\frac{1}{2}=\frac{17+1-2}{4}=\frac{16}{4}=4= \\
-	\end{align}</math>

		+	Note: It is a little tricky to check the answer to the original equation, so we can be satisfied with substituting <math>t=\sqrt{17}/2-1/2</math> into the equation <math>t^2+t=4</math>,

-	<math>=</math>	+	{{Displayed math\|\|<math>\begin{align}
-	~~RHS.~~	+	\text{LHS}
		+	&= \Bigl(\frac{\sqrt{17}}{2}-\frac{1}{2}\Bigr)^2 + \Bigl(\frac{\sqrt{17}}{2}-\frac{1}{2}\Bigr)\\[5pt]
		+	&= \frac{17}{4}-2\cdot \frac{1}{2}\cdot \frac{\sqrt{17}}{2}+\frac{1}{4}+\frac{\sqrt{17}}{2}-\frac{1}{2}\\[5pt]
		+	&= \frac{17}{4}+\frac{1}{4}-\frac{1}{2}\\[5pt]
		+	&= \frac{17+1-2}{4}\\[5pt]
		+	&=\frac{16}{4}\\[5pt]
		+	&= 4\\[5pt]
		+	&= \text{RHS.}
		+	\end{align}</math>}}

Ian at 11:44, 6 October 2008

Ian — Mon, 06 Oct 2008 11:44:11 GMT

←Older revision		Revision as of 11:44, 6 October 2008
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-	{{~~NAVCONTENT_START~~}}	+	If we write the equation as
-	<~~center~~> ~~[[Image:3_4_2b~~-1(2)~~.gif]]~~ </~~center~~>	+
-	{{~~NAVCONTENT_STOP~~}}	+
-	{{~~NAVCONTENT_START~~}}	+
-	<~~center~~> ~~[[Image:3_4_2b~~-2(2)~~.gif]]~~ </~~center~~>	+	<math>\left( e^{x} \right)^{2}+e^{x}=4</math>
-	{{~~NAVCONTENT_STOP~~}}	+
		+
		+	we see that
		+	<math>x</math>
		+	appears only in the combination
		+	<math>e^{x}</math>
		+	and it is therefore appropriate to treat
		+	<math>e^{x}</math>
		+	as a new unknown in the equation and then, when we have obtained the value of
		+	<math>e^{x}</math>, we can calculate the corresponding value of
		+	<math>x</math>
		+	by simply taking the logarithm.
		+
		+	For clarity, we set
		+	<math>t=e^{x}</math>, so that the equation can be written as
		+
		+
		+	<math>t^{2}+t=4</math>
		+
		+
		+	and we solve this second-degree equation by completing the square,
		+
		+
		+	<math>t^{2}+t=\left( t+\frac{1}{2} \right)^{2}-\left( \frac{1}{2} \right)^{2}=\left( t+\frac{1}{2} \right)^{2}-\frac{1}{4}</math>
		+
		+
		+	which gives
		+
		+
		+	<math>\left( t+\frac{1}{2} \right)^{2}-\frac{1}{4}=4\quad \Leftrightarrow \quad t=-\frac{1}{2}\pm \frac{\sqrt{17}}{2}</math>
		+
		+
		+	These two roots give us two possible values for
		+	<math>e^{x}</math>,
		+
		+
		+	<math>e^{x}=-\frac{1}{2}-\frac{\sqrt{17}}{2}</math>
		+	or
		+	<math>e^{x}=-\frac{1}{2}+\frac{\sqrt{17}}{2}</math>
		+
		+
		+	In the first case, the right-hand side is negative and because "
		+	<math>e</math>
		+	raised to anything" can never be negative, there is no
		+	<math>x</math>
		+	that can satisfy this equality. The other case, on the other hand, has a positive right-hand side (because
		+	<math>\sqrt{17}>1</math>
		+	) and we can take the logarithm of both sides to obtain
		+
		+
		+	<math>x=\ln \left( \frac{\sqrt{17}}{2}-\frac{1}{2} \right)</math>
		+
		+
		+	NOTE: It is a little tricky to check the answer to the original equation, so we can be satisfied with substituting
		+	<math>t=\frac{\sqrt{17}}{2}-\frac{1}{2}</math>
		+	into the equation
		+	<math>t^{\text{2}}+t=\text{4}</math>,
		+
		+	LHS
		+	<math>=</math>
		+
		+
		+	<math>\begin{align}
		+	& =\left( \frac{\sqrt{17}}{2}-\frac{1}{2} \right)^{2}+\left( \frac{\sqrt{17}}{2}-\frac{1}{2} \right)=\frac{17}{4}-2\centerdot \frac{1}{2}\centerdot \frac{\sqrt{17}}{2}+\frac{1}{4}+\frac{\sqrt{17}}{2}-\frac{1}{2} \\
		+	& =\frac{17}{4}+\frac{1}{4}-\frac{1}{2}=\frac{17+1-2}{4}=\frac{16}{4}=4= \\
		+	\end{align}</math>
		+
		+
		+	<math>=</math>
		+	RHS.

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

Tekbot — Tue, 09 Sep 2008 09:31:23 GMT

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

←Older revision

Revision as of 09:31, 9 September 2008

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

Tekbot — Thu, 21 Aug 2008 07:02:49 GMT

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

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Oskar — Wed, 02 Apr 2008 11:29:21 GMT

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