Question 302788
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 3.5^{3x\,+\,1}\ =\ 65.4]


Take the log of both sides.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log{3.5^{3x\,+\,1}}\ =\ \log{65.4}]


So far, so good.


Use the Power Priciple


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (3x\,+\,1)\log{3.5}\ =\ \log{65.4}]


Still good, but here is where I would take this in a new direction.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 3x\,+\,1\ =\ \frac{\log{65.4}}{\log{3.5}}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 3x\ =\ \frac{\log{65.4}}{\log{3.5}}\ -\ 1]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ \frac{\frac{\log{\65.4}}{\log{3.5}}\ -\ 1}{3}]



The rest is just calculator work.



John
*[tex \LARGE e^{i\pi} + 1 = 0]
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