Question 1178302
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log_b(x)\ =\ y\ \Leftrightarrow\ b^y\ =\ x]


and 


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


Therefore, if


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log_x(y^4)\ =\ m^3]


Then


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log_x(y)\ =\ \frac{m^3}{4}]


and 


Eq 1: *[tex \LARGE \ \ \ \ \ \ \ \ \ \ \ x^{\frac{m^3}{4}}\ =\ y]


Also if


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log_y(x)\ =\ \frac{4}{m^2}]


Then


Eq 2: *[tex \LARGE \ \ \ \ \ \ \ \ \ \ y^{\frac{4}{m^2}\ =\ x]


Substituting the equivalent expression for *[tex \Large x] from Eq 2 into Eq 1:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \(y^{\frac{4}{m^2}\)^{\frac{m^3}{4}}\ =\ y]


Simplifying:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y^m\ =\ y]


Therefore:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ m\ =\ 1]

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

From <https://www.algebra.com/cgi-bin/upload-illustration.mpl> 
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