Question 1015192
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ =\ y_0e^{kt}]


Substitute the values you know:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 20\ =\ 30e^{10k}]


So


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ e^{10k}\ =\ \frac{20}{30}]


Take the log of both sides


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \ln\left(e^{10k}\right)\ =\ \ln\left(\frac{2}{3}\right)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ k\ =\ \frac{\ln(2)\ -\ \ln(3)}{10}]


Now that you know the value of *[tex \Large k], solve for *[tex \Large t] in


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 1\ =\ 2e^{kt}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ e^{kt}\ =\ \frac{1}{2}]


Take the log


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ kt\ =\ \ln(1)\ -\ \ln(2)\ =\ -\ln(2)]


Substitute value of *[tex \Large k] and solve for *[tex \Large t]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ t\ =\ \frac{10\left(-\ln(2)\right)}{\ln(2)\ -\ \ln(3)}]


The rest is just calculator work.


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

*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \  

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