Question 1122744
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{-dA(t)}{dt}\ =\ kA(t)]


Negative because # atoms is decreasing


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{dA(t)}{A(t)}\ =\ -kdt]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \int^{A(t)}_{A_o}\,\frac{dA}{A}\ =\ -\int_0^t\,kdt]


Where *[tex \Large A_o] is the quantity at time *[tex \Large t\ =\ 0]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \ln\,|A(t)|\ -\ \ln|A_o|\ =\ -kt]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \ln\,\|\frac{A(t)}{A_o}\|\ =\ -kt]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ e^{\large{\ln\,\|\frac{A(t)}{A_o}\|}}\ =\ e^{-kt}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{A(t)}{A_o}\ =\ e^{-kt}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ A(t)\ =\ A_oe^{-kt}]


If 20% has decayed in 10 years then the ratio of the initial amount to the ending amount is 0.80.  Therefore:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 0.8\ =\ e^{-10k}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \ln(0.8)\ =\ \ln\(e^{-10k}\)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ k\ =\ -\frac{\ln(0.8)}{10}]


So, for the half-life:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 0.5\ =\ e^{\frac{\ln(0.8)}{10}t}]


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