Solution 3.4:2a - Revision history

Tek at 11:15, 2 October 2008

2008-10-02T11:15:43Z

←Older revision		Revision as of 11:15, 2 October 2008
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-	The left-hand side is "	+	The left-hand side is "2 raised to something", and therefore a positive number regardless of whatever value the exponent has. We can therefore take the log of both sides,
-	~~<math>\text{~~2~~}</math>~~	+
-	raised to something", and therefore a positive number regardless of whatever value the exponent has. We can therefore take the log of both sides,	+

		+	{{Displayed math\|\|<math>\ln 2^{x^2-2} = \ln 1\,,</math>}}

-	<math>\ln 2^{x^{2}-2~~}=\ln 1~~</math>	+	and use the log law <math>\ln a^b = b\cdot \ln a</math> to get the exponent <math>x^2-2</math> as a factor on the left-hand side,

-	~~and use the log law~~	+	{{Displayed math\|\|<math>\bigl(x^2-2\bigr)\ln 2 = \ln 1\,\textrm{.}</math>}}
-	~~<math>\lg a^~~{~~b}=b\centerdot \lg a</~~math>	+
-	~~to get the exponent~~	+
-	<math>x^~~{\text{~~2}}-\~~text~~{2 }</math>	+
-	~~as a factor on the left-hand side~~	+

		+	Because <math>e^{0}=1</math>, so <math>\ln 1 = 0</math>, giving

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

		+	This means that ''x'' must satisfy the second-degree equation

-	~~Because~~	+	{{Displayed math\|\|<math>x^2-2 = 0\,\textrm{.}</math>}}
-	<math>e^{0~~}=1</math>~~, so	+
-	~~<math>~~\~~text~~{~~ln 1~~}=0</math>~~, giving:~~	+

		+	Taking the root gives <math>x=-\sqrt{2}</math> or <math>x=\sqrt{2}\,</math>.

-	<math>\left( x^{\text{2}}-\text{2 } \right)\ln 2=0</math>

-		+	Note: The exercise is taken from a Finnish upper-secondary final examination from March 2007.
-	~~This means that~~	+
-	~~<math>x</math>~~	+
-	~~must satisfy the second-degree equation~~	+
-		+
-		+
-	~~<math>\left( x^{\text{2}}-\text{2 } \right)=0</math>~~	+
-		+
-		+
-	~~Taking the root gives~~	+
-	~~<math>x=-\sqrt{2}</math>~~	+
-	or	+
-	~~<math>x=\sqrt{2}.</math>~~	+
-		+
-		+
-	~~NOTE~~: ~~the~~ exercise is taken from a Finnish upper-secondary final examination from March 2007.	+

Ian at 10:11, 26 September 2008

2008-09-26T10:11:01Z

←Older revision		Revision as of 10:11, 26 September 2008
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-	{{~~NAVCONTENT_START~~}}	+	The left-hand side is "
-	<~~center~~> ~~[[Image~~:~~3_4_2a.gif]]~~ </~~center~~>	+	<math>\text{2}</math>
-	{{~~NAVCONTENT_STOP~~}}	+	raised to something", and therefore a positive number regardless of whatever value the exponent has. We can therefore take the log of both sides,
		+
		+
		+	<math>\ln 2^{x^{2}-2}=\ln 1</math>
		+
		+	and use the log law
		+	<math>\lg a^{b}=b\centerdot \lg a</math>
		+	to get the exponent
		+	<math>x^{\text{2}}-\text{2 }</math>
		+	as a factor on the left-hand side
		+
		+
		+	<math>\left( x^{\text{2}}-\text{2 } \right)\ln 2=\ln 1</math>
		+
		+
		+	Because
		+	<math>e^{0}=1</math>, so
		+	<math>\text{ln 1}=0</math>, giving:
		+
		+
		+	<math>\left( x^{\text{2}}-\text{2 } \right)\ln 2=0</math>
		+
		+
		+	This means that
		+	<math>x</math>
		+	must satisfy the second-degree equation
		+
		+
		+	<math>\left( x^{\text{2}}-\text{2 } \right)=0</math>
		+
		+
		+	Taking the root gives
		+	<math>x=-\sqrt{2}</math>
		+	or
		+	<math>x=\sqrt{2}.</math>
		+
		+
		+	NOTE: the exercise is taken from a Finnish upper-secondary final examination from March 2007.

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

2008-09-09T09:31:03Z

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

←Older revision

Revision as of 09:31, 9 September 2008

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

2008-08-21T07:02:39Z

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

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2008-04-02T10:30:51Z

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