Question 1163332
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Radioactive decay conforms to the exponential model *[tex \Large A(t)\ =\ A_oe^{kt}] where *[tex \Large A_o] is the initial amount, *[tex \Large k] is the growth/decay constant, and *[tex \Large t] is the elapsed time.


Since the initial amount was 13 mg we have the initial expression:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ A(t)\ =\ 13e^{kt}]


But then we are given:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ A(24)\ =\ 13e^{24k}\ =\ 6\text{ mg}]


Solving for the constant of decay:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ e^{24k}\ =\ \frac{6}{13}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \ln\(e^{24k}\)\ =\ \ln\(\frac{6}{13}\)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ k\ =\ \frac{\ln(6)\ -\ \ln(13)}{24}]


Once you have done the arithmetic to find the value of *[tex \Large k] you can write the full function for your particular situation:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ A(t)\ =\ 13e^{kt}]


Insert the value of the constant and then set the function equal to 2 mg and solve for *[tex \Large t] minutes.
								
								
John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
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