Question 1030367
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The log of the product is the sum of the logs, and *[tex \Large \log_b(a^n)\ =\ n\log_b(a)], and *[tex \Large \log_b(b)\ =\ 1]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  \log_b(14b^2)\ =\ \log_b(14)\ +\ \log_b(b^2)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  =\ \log_b(2)\ +\ \log_b(7)\ +\ 2\log_b(b)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  =\ \log_b(2)\ +\ \log_b(7)\ +\ 2]


But since


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log_b(4)\ =\ \log_b(2^2)\ =\ 2\log_b(2)]


and


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log_b(4)\ =\ 1.39]


then it follows that


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log_b(2)\ =\ \frac{1.39}{2}]


Therefore:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  \log_b(14b^2)\ =\ \frac{1.39}{2}\ +\ 1.95\ +\ 2]


You can do your own arithmetic. 


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it

*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \  

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