Question 1030370
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For continuous compounding,


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ A\ =\ Pe^{rt}]


where *[tex \Large e] is the base of the natural logarithms.


So


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ e^{rt}\ =\ \frac{A}{P}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \ln\left(e^{rt}\right)\ =\ \ln\left(\frac{A}{P}\right)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ rt\,\cdot\,\ln(e)\ =\ \ln(A)\ -\ \ln(P)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ rt\ =\ \ln(A)\ -\ \ln(P)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ t\ =\ \frac{\ln(A)\ -\ \ln(P)}{r}]


Plug in your numbers and do the arithmetic.


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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