Question 637078
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 1.35\left(1\ +\ g\right)^4\ =\ 2.28]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(1\ +\ g\right)^4\ =\ \frac{2.28}{1.35} ]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \ln\left(\left(1\ +\ g\right)^4\right)\ =\ \ln\left(\frac{2.28}{1.35}\right)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 4\,\cdot\,\ln\left(\left(1\ +\ g\right)\right)\ =\ \ln\left(\frac{2.28}{1.35}\right)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \ln\left(\left(1\ +\ g\right)\right)\ =\ \frac{\ln\left(\frac{2.28}{1.35}\right)}{4}]


For neatness sake, let *[tex \LARGE u\ =\ \frac{\ln\left(\frac{2.28}{1.35}\right)}{4}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \ln\left(\left(1\ +\ g\right)\right)\ =\ u]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 1\ +\ g\ =\ e^u]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ g\ =\ e^u\ -\ 1]


You can do your own arithmetic.


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