Question 1038775
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Notation: *[tex \Large \log_b(x)] in plain text is log_b(x).  Most folks assume base 10 if the base is omitted, so log(x) is generally interpreted as log_10(x).  So your note that the logs are base 10 is appropriate because sometimes *[tex \Large \log] is used to mean *[tex \Large \ln] which is *[tex \Large \log_e].  Also, get into the habit of putting the argument of the log in parentheses -- eliminates confusion.


So 10log_10(m) + 2log_10(n) - log_10 and (All logs base 10) 10log(m) + 2log(n) - log(p) would both have been interpreted as:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 10\log_{10}(m)\ +\ 2\log_{10}(n)\ -\ \log_{10}(p)]


Rules:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ a\cdot\log_b(x)\ =\ \log_b(x^a)]


The sum of the logs is the log of the product.


The difference of the logs is the log of the quotient.


So:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 10\log_{10}(m)\ +\ 2\log_{10}(n)\ -\ \log_{10}(p)\ =\ \log_{10}\left(\frac{m^{10}n^2}{p}\right)]


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
<img src="http://c0rk.blogs.com/gr0undzer0/darwin-fish.jpg">
*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \  

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