Question 910920
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2.91e^{-1.7x}\ =\ 7.83]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ e^{-1.7x}\ =\ \frac{7.83}{2.91}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \ln\left(e^{-1.7x}\right)\ =\ \ln\left(\frac{7.83}{2.91}\right)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ -1.7x\cdot \ln\left(e\right)\ =\ \ln\left(\frac{7.83}{2.91}\right)]


but since *[tex \Large \log_b(b)\ =\ 1],


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ -1.7x\ =\ \ln\left(\frac{7.83}{2.91}\right)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ -\frac{\ln\left(\frac{7.83}{2.91}\right)}{1.7}]


Use your calculator to find a decimal approximation if you want.


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