The substance has a half-life of 3 years, which means it decays with rate k such that
[tex]\dfrac12 = e^{3k} \implies \ln\left(\dfrac12\right) = 3k \implies k = \dfrac13 \ln\left(\dfrac12\right) = -\dfrac{\ln(2)}3[/tex]
a) Starting at 200 mg, the amount of substance y remaining after t years have passed is then
[tex]y = 200 e^{kt} = \boxed{200 e^{(-\ln(2)/3) \, t}}[/tex]
b) After t = 10 years, you'd be left with
[tex]y = 200 e^{-10\ln(2)/3} = 200 e^{\ln(2)^{-10/3}} = 200 \cdot 2^{-10/3} \approx \boxed{19.843}[/tex]
mg of the substance.
c) The substance takes T years to decay to 10% of its initial amount such that
[tex]0.10 = e^{(-\ln(2)/3) \, T}[/tex]
Solve for T :
[tex]\ln(0.10) = -\dfrac{\ln(2)}3 T[/tex]
[tex]T = -\dfrac{3\ln(0.10)}{\ln(2)} = -3 \log_2(0.10) \approx \boxed{9.966}[/tex]
years.
