Question 413995
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Let *[tex \Large x\ =\ \log_a(27)]


Let *[tex \Large y\ =\ \log_b(4)]


Then


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ +\ y\ =\ 5]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ -\ y\ =\ 1]


Using elimination


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ 3]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ =\ 2]


Hence:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log_a(27)\ =\ 3\ \Leftrightarrow\ a^3\ =\ 27\ \Rightarrow\ a\ =\ 3]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log_b(4)\ =\ 2\ \Leftrightarrow\ b^2\ =\ 4\ \Rightarrow\ b\ =\ \pm2]


But the base of a logarithm function is restricted to the positive reals, hence


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ b\ =\ 2]


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