From Förberedande kurs i matematik 1
Before we even start thinking about transforming log2 and log3 to ln, we use the log laws
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| logablog(a b)=bloga =loga+logb |
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to simplify the expression
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| log3log23118=log3(118log23)=log3118+log3log23. |
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With help of the relation 2log2x=x and 3log3x=x and taking the natural logarithm , we can express log2 and log3 using ln,
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| log2x=lnxln2 and log3x=lnxln3.
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The two terms log3118 and log3log23 can therefore be written as
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| log3118=ln3ln118 and log3log23=log3ln2ln3
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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 log3 to ln,
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| log3ln2ln3=log3ln3−log3ln2=ln3lnln3−ln3lnln2. |
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In all, we thus obtain
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| \displaystyle \log_{3}\log_{2}3^{118} = \frac{\ln 118}{\ln 3} + \frac{\ln \ln 3}{\ln 3} - \frac{\ln\ln 2}{\ln 3}\,\textrm{.}
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Input into the calculator gives
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| \displaystyle \log_{3}\log_{2}3^{118}\approx 4\textrm{.}762\,\textrm{.}
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Note: The button sequence on the calculator will be:
