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Solution 3.3:6c

From Förberedande kurs i matematik 1

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m (Lösning 3.3:6c moved to Solution 3.3:6c: Robot: moved page)
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{{NAVCONTENT_START}}
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Before we even start thinking about transforming
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<center> [[Image:3_3_6c-1(2).gif]] </center>
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<math>\log _{2}</math>
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and
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<math>\log _{3}</math>
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<center> [[Image:3_3_6c-2(2).gif]] </center>
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to ln, we use the log laws
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{{NAVCONTENT_STOP}}
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[[Image:3_3_6_c.gif|center]]
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<math>\lg a^{b}=b\centerdot \lg a</math>
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<math>\lg \left( a\centerdot b \right)=\lg a+\lg b</math>
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to simplify the expression
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<math>\begin{align}
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& \log _{3}\log _{2}3^{118}=\log _{3}\left( 118\centerdot \log _{2}3 \right) \\
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& =\log _{3}118+\log _{3}\log _{2}3 \\
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\end{align}</math>
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With help of the relation
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<math>2^{\log _{2}x}=x</math>
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and
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<math>3^{\log _{3}x}=x</math>
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and taking the natural logarithm , we can express
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<math>\log _{2}</math>
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and
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<math>\log _{3}</math>
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using ln,
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<math>\log _{2}x=\frac{\ln x}{\ln 2}</math>
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and
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<math>\log _{3}x=\frac{\ln x}{\ln 3}</math>
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The two terms log3 118 and log3 log2 3 can therefore be written as
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<math>\log _{3}118=\frac{\ln 118}{\ln 3}</math>
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and
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<math>\log _{3}\log _{2}3=\log _{3}\left( \frac{\ln 3}{\ln 2} \right),</math>
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where we can simplify the last expression further with the logarithm law, log a/b = log a – log b, and then transform
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<math>\log _{3}</math>
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to ln,
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<math>\begin{align}
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& \log _{3}\left( \frac{\ln 3}{\ln 2} \right)=\log _{3}\ln 3-\log _{3}\ln 2 \\
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& =\frac{\ln \ln 3}{\ln 3}-\frac{\ln 2}{\ln 3} \\
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\end{align}</math>
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In all, we thus obtain
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<math>\log _{3}\log _{2}3^{118}=\frac{\ln 118}{\ln 3}+\frac{\ln \ln 3}{\ln 3}-\frac{\ln 2}{\ln 3}</math>
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Input into the calculator gives
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<math>\log _{3}\log _{2}3^{118}\approx 4.762</math>
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NOTE: the button sequence on a calculator will be:
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<math>\begin{align}
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& \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] \\
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& \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] \\
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& \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] \\
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& \quad \\
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\end{align}</math>

Revision as of 09:49, 26 September 2008

Before we even start thinking about transforming log2 and log3 to ln, we use the log laws


lgab=blga


lgab=lga+lgb 


to simplify the expression


log3log23118=log3118log23=log3118+log3log23 


With help of the relation


2log2x=x and 3log3x=x


and taking the natural logarithm , we can express log2 and log3 using ln,


log2x=lnxln2 and log3x=lnxln3


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


log3118=ln3ln118 and log3log23=log3ln2ln3 


where we can simplify the last expression further with the logarithm law, log a/b = log a – log b, and then transform log3 to ln,


log3ln2ln3=log3ln3log3ln2=ln3lnln3ln3ln2 


In all, we thus obtain


log3log23118=ln3ln118+ln3lnln3ln3ln2


Input into the calculator gives


log3log231184762


NOTE: the button sequence on a calculator will be:


118LN3LN+3LNLN3LN2LNLN3LN=