Question 674455
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You need to convert so that the two logs are the same base.   You can convert the base 2 log to base 6, or the base 6 log to base 2, or convert both to some other base -- in the end it won't matter because you will always get the same answer.


Conversion formula:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log_b(x) = \frac{\log_c(x)}{\log_c(b)}]


Let's convert the base 6 log to base 2.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log_6(8) = \frac{\log_2(8)}{\log_2(6)}]


Now let's do the original multiplication:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log_2(6)\,\cdot\,\frac{\log_2(8)}{\log_2(6)}]


Which simplifies to:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log_2(8)]


But wait!  There's more!  Now what would you pay?  Oh, sorry, this isn't a Ginzu Knives infomercial, is it...but there is more for all that.


Note that *[tex \LARGE 8\ =\ 2^3], so:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log_2\left(2^3\right)]


But, remember:


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


So


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 3\log_2\left(2\right)]


But then:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log_b(b)\ =\ 1]


So


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log_2(6)\,\cdot\,\log_6(8)\ =\ 3]


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
<font face="Math1" size="+2">Egw to Beta kai to Sigma</font>
My calculator said it, I believe it, that settles it
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