Question 702045
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ =\ 3^{2x\,-\,1}]


Take the base 3 log of both sides:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log_3(y)\ =\ \log_3\left(3^{2x\,-\,1}\right)]


Apply *[tex \LARGE \ \ \ \log_b\left(x^n\right)\ =\ n\,\log_b(x)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log_3(y)\ =\ \left(2x\,-\,1\right)\log_3(3)]


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


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2x\ -\ 1\ =\ \log_3(y)]


The rest is just straight algebra:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ \frac{1}{2}\left(\log_3(y)\ +\ 1\right)] 


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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