Solution 3.4:1b - Revision history

Tek at 08:54, 2 October 2008

Tek — Thu, 02 Oct 2008 08:54:43 GMT

←Older revision		Revision as of 08:54, 2 October 2008
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-	In the equation, both sides are positive because the factors	+	In the equation, both sides are positive because the factors <math>e^{x}</math> and <math>3^{-x}</math> are positive regardless of the value of <math>x</math> (a positive base raised to a number always gives a positive number). We can therefore take the natural logarithm of both sides,
-	<math>e^{x}</math>	+
-	and	+
-	<math>3^{-x}</math>	+
-	are positive regardless of the value of	+
-	<math>x</math>	+
-	(a positive base raised to a number always gives a positive number). We can therefore take the natural logarithm of both ~~numbers~~,	+

		+	{{Displayed math\|\|<math>\ln\bigl(13e^{x}\bigr) = \ln\bigl(2\cdot 3^{-x}\bigr)\,\textrm{.}</math>}}

-	~~<math>\ln \left( 13e^{x} \right)=\ln \left( 2\centerdot 3^{-x} \right)</math>~~	+	Using the log laws, we can divide up the products into several logarithmic terms,

		+	{{Displayed math\|\|<math>\ln 13+\ln e^{x} =\ln 2+\ln 3^{-x},</math>}}

-	~~Using~~ the ~~log~~ law, we can ~~divide up~~ the ~~products into several logarithmic terms,~~	+	and using the law <math>\ln a^{b}=b\cdot \ln a</math>, we can get rid of <math>x</math> from the exponents

		+	{{Displayed math\|\|<math>\ln 13 + x\ln e = \ln 2 + (-x)\ln 3\,\textrm{.}</math>}}

-	<math>~~\ln 13+\ln e^{~~x~~}=\ln 2+\ln 3^{-x}~~</math>	+	Collect <math>x</math> on one side and the other terms on the other,

		+	{{Displayed math\|\|<math>x\ln e+x\ln 3=\ln 2-\ln 13\,\textrm{.}</math>}}

-	~~and using the law~~	+	Take out <math>x</math> on the left-hand side and use <math>\ln e=1</math>,
-	<math>~~\ln a^{b}=b\centerdot \ln a~~</math>~~, we can get rid of~~	+
-	<math>x</math>	+
-	~~from the exponents:~~	+

		+	{{Displayed math\|\|<math>x( 1+\ln 3)=\ln 2-\ln 13\,\textrm{.}</math>}}

-	<math>~~\ln 13+~~x~~\ln e=\ln 2+\left( -x \right)\ln 3~~</math>	+	Then, solve for <math>x</math>,

		+	{{Displayed math\|\|<math>x=\frac{\ln 2-\ln 13}{1+\ln 3}\,\textrm{.}</math>}}

-	Collecting together
-	<math>x</math>
-	on one side and the other terms on the other,

		+	Note: Because <math>\ln 2 < \ln 13</math>, we can write the answer as

-	<math>x\~~ln e+x~~\ln 3=\ln 2-\ln 13</math>	+	{{Displayed math\|\|<math>x=-\frac{\ln 13-\ln 2}{1+\ln 3}</math>}}

-		+	to indicate that <math>x</math> is negative.
-	~~Take out~~	+
-	~~<math>x</math>~~	+
-	~~on the left-hand side and use~~	+
-	~~<math>\ln e=1</math>~~	+
-	:	+
-		+
-		+
-	~~<math>x\left( 1+\ln 3 \right)=\ln 2-\ln 13</math>~~	+
-		+
-		+
-	~~Then, solve for~~	+
-	~~<math>x</math>~~	+
-	:	+
-		+
-		+
-	~~<math>x=\frac{\ln 2-\ln 13}{1+\ln 3}</math>~~	+
-		+
-		+
-	~~NOTE: Because~~	+
-	~~<math>\ln 2<\ln 13</math>, we can write the answer as~~	+
-		+
-		+
-	~~<math>x=-\frac{\ln 13-\ln 2}{1+\ln 3}</math>~~	+
-		+
-		+
-	~~in order~~ to indicate that	+
-	<math>x</math>	+
-	is negative.	+

Ian at 12:55, 12 September 2008

Ian — Fri, 12 Sep 2008 12:55:06 GMT

←Older revision		Revision as of 12:55, 12 September 2008
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-	{~~{NAVCONTENT_START}~~}	+	In the equation, both sides are positive because the factors
-	<~~center~~> ~~[[Image:3_4_1b~~-1(2).~~gif]]~~ </~~center~~>	+	<math>e^{x}</math>
-	{{~~NAVCONTENT_STOP}~~}	+	and
-	{~~{NAVCONTENT_START}~~}	+	<math>3^{-x}</math>
-	<~~center~~> ~~[[Image~~:~~3_4_1b-~~2(2)~~.gif]]~~ </~~center~~>	+	are positive regardless of the value of
-	{{~~NAVCONTENT_STOP~~}}	+	<math>x</math>
		+	(a positive base raised to a number always gives a positive number). We can therefore take the natural logarithm of both numbers,
		+
		+
		+	<math>\ln \left( 13e^{x} \right)=\ln \left( 2\centerdot 3^{-x} \right)</math>
		+
		+
		+	Using the log law, we can divide up the products into several logarithmic terms,
		+
		+
		+	<math>\ln 13+\ln e^{x}=\ln 2+\ln 3^{-x}</math>
		+
		+
		+	and using the law
		+	<math>\ln a^{b}=b\centerdot \ln a</math>, we can get rid of
		+	<math>x</math>
		+	from the exponents:
		+
		+
		+	<math>\ln 13+x\ln e=\ln 2+\left( -x \right)\ln 3</math>
		+
		+
		+	Collecting together
		+	<math>x</math>
		+	on one side and the other terms on the other,
		+
		+
		+	<math>x\ln e+x\ln 3=\ln 2-\ln 13</math>
		+
		+
		+	Take out
		+	<math>x</math>
		+	on the left-hand side and use
		+	<math>\ln e=1</math>
		+	:
		+
		+
		+	<math>x\left( 1+\ln 3 \right)=\ln 2-\ln 13</math>
		+
		+
		+	Then, solve for
		+	<math>x</math>
		+	:
		+
		+
		+	<math>x=\frac{\ln 2-\ln 13}{1+\ln 3}</math>
		+
		+
		+	NOTE: Because
		+	<math>\ln 2<\ln 13</math>, we can write the answer as
		+
		+
		+	<math>x=-\frac{\ln 13-\ln 2}{1+\ln 3}</math>
		+
		+
		+	in order to indicate that
		+	<math>x</math>
		+	is negative.

Tekbot: Lösning 3.4:1b moved to Solution 3.4:1b: Robot: moved page

Tekbot — Tue, 09 Sep 2008 09:30:23 GMT

Lösning 3.4:1b moved to Solution 3.4:1b: Robot: moved page

←Older revision

Revision as of 09:30, 9 September 2008

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

Tekbot — Thu, 21 Aug 2008 07:02:19 GMT

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

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Oskar — Wed, 02 Apr 2008 10:28:35 GMT

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