Solution 3.4:2c - Revision history

Tek at 11:28, 2 October 2008

Tek — Thu, 02 Oct 2008 11:28:25 GMT

←Older revision		Revision as of 11:28, 2 October 2008
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-	Regardless of what value	+	Regardless of what value ''x'' has, both sides of the equation are positive, because they are of the type "positive number raised to something". It is therefore possible to take the logarithm of both sides and, by using the log laws, to get ''x'' down from the exponents,
-	~~<math>~~x~~</math>~~	+
-	has, both sides of the equation are positive, because they are of the type "positive number raised to something". It is therefore possible to take the logarithm of both sides and, by using the log laws, to get	+
-	~~<math>~~x~~</math>~~	+
-	down from the exponents:	+
-		+
-	~~LHS~~	+
-	~~<math>=\ln \left( 3e^{x^{2}} \right)=\ln 3+\ln e^{x^{2}}=\ln 3+x^{2}\ln e=\ln 3+x^{2}</math>~~	+
-		+
-	~~RHS~~	+
-	~~<math>=\ln 2^{x}=x\ln 2</math>~~	+

		+	{{Displayed math\|\|<math>\begin{align}
		+	\text{LHS}
		+	&= \ln\bigl( 3e^{x^{2}}\bigr)
		+	= \ln 3 + \ln e^{x^2}
		+	= \ln 3 + x^2\ln e
		+	= \ln 3 + x^2\,\textrm{,}\\[5pt]
		+	\text{RHS}
		+	&= \ln 2^x = x\ln 2\,\textrm{.}
		+	\end{align}</math>}}

	If we collect the terms onto one side, the equation becomes		If we collect the terms onto one side, the equation becomes

		+	{{Displayed math\|\|<math>x^{2}-x\cdot \ln 2 + \ln 3 = 0</math>}}

-	~~<math>x^{2}-x\centerdot \ln 2+\ln 3=0</math>~~	+	which is a standard second-order equation for which we complete the square
-		+
-		+
-	which is a standard second-order equation for which we complete the square:	+
-		+

-	<math>\begin{align}	+	{{Displayed math\|\|<math>\begin{align}
-	& \~~left~~( x-\frac{1}{2}\ln 2 \~~right~~)^{2}-\~~left~~( \frac{1}{2}\ln 2 \~~right~~)^{2}+\ln 3=0 \\	+	\Bigl(x-\frac{1}{2}\ln 2\Bigr)^{2} - \Bigl(\frac{1}{2}\ln 2\Bigr)^{2} + \ln 3 = 0\,\textrm{,}\\[5pt]
-	& \~~left~~( x-\frac{1}{2}\ln 2 \~~right~~)^{2}=\~~left~~( \frac{1}{2}\ln 2 \~~right~~)^{2}-\ln 3 \\	+	\Bigl(x-\frac{1}{2}\ln 2\Bigr)^{2} = \Bigl(\frac{1}{2}\ln 2\Bigr)^{2} - \ln 3\,\textrm{.}
-	\end{align}</math>	+	\end{align}</math>}}

-	Now, we have to be cautious and remember that, because	+	Now, we have to be cautious and remember that, because <math>2 < e < 3</math> and thus <math>\ln 2 < 1 < \ln 3</math> we have that <math>\tfrac{1}{4}(\ln 2)^2 < \ln 3</math> and the right-hand side is therefore negative. Since the square on the left-hand side cannot be negative, the equation has no solution.
-	<math>~~\text{~~2}<e<~~\text{~~3}</math>	+
-	and thus	+
-	<math>\~~text{~~ln 2}<~~\text{~~1}<\~~text{~~ln 3}</math>	+
-	~~which means~~ that	+
-	<math>\~~frac~~{1}{4}~~\left~~( \ln 2 ~~\right~~)^{2}<\ln 3</math>	+
-	and the right-hand side is therefore negative. Since the square on the ~~right~~-hand side cannot be negative, the equation has no solution.	+

Ian at 11:28, 26 September 2008

Ian — Fri, 26 Sep 2008 11:28:18 GMT

←Older revision		Revision as of 11:28, 26 September 2008
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-	{{~~NAVCONTENT_START~~}}	+	Regardless of what value
-	<~~center~~> ~~[[Image~~:~~3_4_2c.gif]]~~ </~~center~~>	+	<math>x</math>
-	{{~~NAVCONTENT_STOP~~}}	+	has, both sides of the equation are positive, because they are of the type "positive number raised to something". It is therefore possible to take the logarithm of both sides and, by using the log laws, to get
		+	<math>x</math>
		+	down from the exponents:
		+
		+	LHS
		+	<math>=\ln \left( 3e^{x^{2}} \right)=\ln 3+\ln e^{x^{2}}=\ln 3+x^{2}\ln e=\ln 3+x^{2}</math>
		+
		+	RHS
		+	<math>=\ln 2^{x}=x\ln 2</math>
		+
		+
		+	If we collect the terms onto one side, the equation becomes
		+
		+
		+	<math>x^{2}-x\centerdot \ln 2+\ln 3=0</math>
		+
		+
		+	which is a standard second-order equation for which we complete the square:
		+
		+
		+	<math>\begin{align}
		+	& \left( x-\frac{1}{2}\ln 2 \right)^{2}-\left( \frac{1}{2}\ln 2 \right)^{2}+\ln 3=0 \\
		+	& \left( x-\frac{1}{2}\ln 2 \right)^{2}=\left( \frac{1}{2}\ln 2 \right)^{2}-\ln 3 \\
		+	\end{align}</math>
		+
		+	Now, we have to be cautious and remember that, because
		+	<math>\text{2}<e<\text{3}</math>
		+	and thus
		+	<math>\text{ln 2}<\text{1}<\text{ln 3}</math>
		+	which means that
		+	<math>\frac{1}{4}\left( \ln 2 \right)^{2}<\ln 3</math>
		+	and the right-hand side is therefore negative. Since the square on the right-hand side cannot be negative, the equation has no solution.

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

Tekbot — Tue, 09 Sep 2008 09:31:43 GMT

Lösning 3.4:2c moved to Solution 3.4:2c: 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 07:02:59 GMT

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

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Oskar at 11:31, 2 April 2008

Oskar — Wed, 02 Apr 2008 11:31:09 GMT

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Oskar — Wed, 02 Apr 2008 11:30:32 GMT

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