Question 461820
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Ok, you are also going to need:


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


Also note that *[tex \Large 25\ =\ 5^2,\ \ ] *[tex \Large 27\ =\ 3^3,\ \ ] and *[tex \Large 8\ =\ 2^3]


The trick is to change all three of your log functions to the same base, but it doesn't matter which base.  I'm going to use natural logs, just because, well, it just seems natural.


Starting with


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log_2(25)\ \cdot\ \log_3(27)\ \cdot\ \log_5(8)]


Do the base conversion to *[tex \Large e]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{\ln(25)}{\ln(2)}\ \cdot\ \frac{\ln(27)}{\ln(3)}\ \cdot\ \frac{\ln(8)}{\ln(5)}]


Make the substitutions alluded to earlier:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{\ln(5^2)}{\ln(2)}\ \cdot\ \frac{\ln(3^3)}{\ln(3)}\ \cdot\ \frac{\ln(2^3)}{\ln(5)}]


Then use


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


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{2\ln(5)}{\ln(2)}\ \cdot\ \frac{3\ln(3)}{\ln(3)}\ \cdot\ \frac{3\ln(2)}{\ln(5)}]


Eliminating common factors leaves you with:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2\ \cdot\ 3\ \cdot\ 3\ =\ 18]


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