Question 1163965
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A radioactive substance decays at a rate proportional to the amount remaining.


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


When half-life is reached, *[tex \Large \frac{A(t)}{A_o}\ =\ \frac{1}{2}]


Hence the name "half-life"


Since the half-life of this particular substance is 17.5 days:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ e^{17.5\lambda}\ =\ \frac{1}{2}]


Take the natural log of both sides


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 17.5\lambda\ =\ \ln\(\frac{1}{2}\)\ =\ \ln(1)\ -\ \ln(2)\ =\ -\ln(2)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \lambda\ =\ \frac{-\ln(2)}{17.5}]


You can do any additional arithmetic required.


I sincerely hope you are not surprised that the answer is a negative number.


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