Question 598270
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Nope.  Your numbers are wrong.  It is impossible for


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log_6(5)\ >\ \log_6(8)]


In fact, since *[tex \LARGE 8\ >\ 6],


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log_6(8)\ >\ 1]


With all that in mind, I checked on a hunch:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log_6(8)\ \approx\ 1.161]


Note that


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 8\ =\ 2^3]


so


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log_6(8)\ =\ \log_6\left(2^3\right)\ =\ 3\,\cdot\,\log_6(2)]


so


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log_6(2)\ \approx\ \frac{1.161}{3}\ =\ 0.387]


Since the the log of the product is the sum of the logs and since 10 is equal to 5 times 2:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log_6(10)\ =\ \log_6(5)\ +\ \log_6(2)]


All that is left is to plug in the values and do the arithmetic.


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