Solution 3.3:6c - Revision history

Tek at 08:40, 2 October 2008

Tek — Thu, 02 Oct 2008 08:40:16 GMT

Ian at 09:49, 26 September 2008

Ian — Fri, 26 Sep 2008 09:49:57 GMT

←Older revision		Revision as of 09:49, 26 September 2008
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-	{{~~NAVCONTENT_START~~}}	+	Before we even start thinking about transforming
-	<~~center~~> ~~[[Image:3_3_6c-1~~(2)~~.gif]]~~ </~~center~~>	+	<math>\log _{2}</math>
-	{{~~NAVCONTENT_STOP~~}}	+	and
-	{{~~NAVCONTENT_START~~}}	+	<math>\log _{3}</math>
-	<~~center~~> ~~[[Image:3_3_6c-~~2(2).~~gif]]~~ </~~center~~>	+	to ln, we use the log laws
-	{{~~NAVCONTENT_STOP~~}}	+
-	[[~~Image:3_3_6_c.gif\|center~~]]	+
		+	<math>\lg a^{b}=b\centerdot \lg a</math>
		+
		+
		+	<math>\lg \left( a\centerdot b \right)=\lg a+\lg b</math>
		+
		+
		+	to simplify the expression
		+
		+
		+	<math>\begin{align}
		+	& \log _{3}\log _{2}3^{118}=\log _{3}\left( 118\centerdot \log _{2}3 \right) \\
		+	& =\log _{3}118+\log _{3}\log _{2}3 \\
		+	\end{align}</math>
		+
		+
		+	With help of the relation
		+
		+
		+	<math>2^{\log _{2}x}=x</math>
		+	and
		+	<math>3^{\log _{3}x}=x</math>
		+
		+
		+	and taking the natural logarithm , we can express
		+	<math>\log _{2}</math>
		+	and
		+	<math>\log _{3}</math>
		+	using ln,
		+
		+
		+	<math>\log _{2}x=\frac{\ln x}{\ln 2}</math>
		+	and
		+	<math>\log _{3}x=\frac{\ln x}{\ln 3}</math>
		+
		+
		+
		+	The two terms log3 118 and log3 log2 3 can therefore be written as
		+
		+
		+	<math>\log _{3}118=\frac{\ln 118}{\ln 3}</math>
		+	and
		+	<math>\log _{3}\log _{2}3=\log _{3}\left( \frac{\ln 3}{\ln 2} \right),</math>
		+
		+
		+	where we can simplify the last expression further with the logarithm law, log a/b = log a – log b, and then transform
		+	<math>\log _{3}</math>
		+	to ln,
		+
		+
		+	<math>\begin{align}
		+	& \log _{3}\left( \frac{\ln 3}{\ln 2} \right)=\log _{3}\ln 3-\log _{3}\ln 2 \\
		+	& =\frac{\ln \ln 3}{\ln 3}-\frac{\ln 2}{\ln 3} \\
		+	\end{align}</math>
		+
		+
		+	In all, we thus obtain
		+
		+
		+	<math>\log _{3}\log _{2}3^{118}=\frac{\ln 118}{\ln 3}+\frac{\ln \ln 3}{\ln 3}-\frac{\ln 2}{\ln 3}</math>
		+
		+
		+	Input into the calculator gives
		+
		+
		+	<math>\log _{3}\log _{2}3^{118}\approx 4.762</math>
		+
		+
		+
		+	NOTE: the button sequence on a calculator will be:
		+
		+
		+
		+	<math>\begin{align}
		+	& \left[ 1 \right]\quad \left[ 1 \right]\quad \left[ 8 \right]\quad \left[ \text{LN} \right]\quad \left[ \div \right]\quad \left[ 3 \right]\quad \left[ \text{LN} \right]\quad \left[ + \right] \\
		+	& \left[ 3 \right]\quad \left[ \text{LN} \right]\quad \left[ \text{LN} \right]\quad \left[ \div \right]\quad \left[ 3 \right]\quad \left[ \text{LN} \right]\quad \left[ - \right]\quad \left[ 2 \right] \\
		+	& \left[ \text{LN} \right]\quad \left[ \text{LN} \right]\quad \left[ \div \right]\quad \left[ 3 \right]\quad \left[ \text{LN} \right]\quad \left[ = \right] \\
		+	& \quad \\
		+	\end{align}</math>

Tekbot: Lösning 3.3:6c moved to Solution 3.3:6c: Robot: moved page

Tekbot — Tue, 09 Sep 2008 09:29:43 GMT

Lösning 3.3:6c moved to Solution 3.3:6c: Robot: moved page

←Older revision

Revision as of 09:29, 9 September 2008

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

Tekbot — Thu, 21 Aug 2008 07:01:59 GMT

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

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Madde at 12:16, 15 August 2008

Madde — Fri, 15 Aug 2008 12:16:35 GMT

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Oskar — Wed, 02 Apr 2008 09:58:04 GMT

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