From Förberedande kurs i matematik 1

Revision as of 09:29, 9 September 2008 (edit) Tekbot (Talk | contribs) m (Lösning 3.3:6c moved to Solution 3.3:6c: Robot: moved page) ← Previous diff		Revision as of 09:49, 26 September 2008 (edit) (undo) Ian (Talk | contribs) Next diff →
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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>

---

Revision as of 09:49, 26 September 2008

Before we even start thinking about transforming \displaystyle \log _{2} and \displaystyle \log _{3} to ln, we use the log laws

\displaystyle \lg a^{b}=b\centerdot \lg a

\displaystyle \lg \left( a\centerdot b \right)=\lg a+\lg b

to simplify the expression

\displaystyle \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}

With help of the relation

\displaystyle 2^{\log _{2}x}=x and \displaystyle 3^{\log _{3}x}=x

and taking the natural logarithm , we can express \displaystyle \log _{2} and \displaystyle \log _{3} using ln,

\displaystyle \log _{2}x=\frac{\ln x}{\ln 2} and \displaystyle \log _{3}x=\frac{\ln x}{\ln 3}

The two terms log3 118 and log3 log2 3 can therefore be written as

\displaystyle \log _{3}118=\frac{\ln 118}{\ln 3} and \displaystyle \log _{3}\log _{2}3=\log _{3}\left( \frac{\ln 3}{\ln 2} \right),

where we can simplify the last expression further with the logarithm law, log a/b = log a – log b, and then transform \displaystyle \log _{3} to ln,

\displaystyle \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}

In all, we thus obtain

\displaystyle \log _{3}\log _{2}3^{118}=\frac{\ln 118}{\ln 3}+\frac{\ln \ln 3}{\ln 3}-\frac{\ln 2}{\ln 3}

Input into the calculator gives

\displaystyle \log _{3}\log _{2}3^{118}\approx 4.762

NOTE: the button sequence on a calculator will be:

\displaystyle \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}