Question 763915
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You need to use the change of base formula:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log_a(x)\ =\ \frac{\log_b(x)}{\log_b(a)}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log_8(x)\ =\ \frac{\log_2(x)}{\log_2(8)}]


But then, since *[tex \LARGE y = \log_b(x) \ \ \Leftrightarrow\ \ b^y = x]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y = \log_2(8) \ \ \Leftrightarrow\ \ 2^y = 8]


hence *[tex \LARGE y\ =\ 3], so


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log_8(x)\ =\ \frac{\log_2(x)}{3}]


Similarly


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log_{64}(x)\ =\ \frac{\log_2(x)}{\log_2(64)}\ =\ \frac{\log_2(x)}{6}]


Substituting


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log_2(x)\ +\ \frac{\log_2(x)}{3}\ +\ \frac{\log_2(x)}{6}\ =\ 3]


Factor out *[tex \LARGE \log_2(x)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log_2(x)\left(1\ +\ \frac{1}{3}\ +\ \frac{1}{6}\right)\ =\ 3]


A little arithmetic, and


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log_2(x)\ =\ 2]


But since *[tex \LARGE y = \log_b(x) \ \ \Leftrightarrow\ \ b^y = x]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log_2(x)\ =\ 2\ \Rightarrow\ x\ =\ 4]


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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