In the formula A(t) = A0ekt, A is the amount of radioactive material remaining from an

Question

initial amount A0 at a given time t, and k is a negative constant determined by the nature
of the material. An artifact is discovered at a certain site. If it has 72% of the carbon-14 it
originally contained, what is the approximate age of the artifact? (carbon-14 decays at the
rate of 0.0125% annually.) (Round to the nearest year.)

Answer 1

You are solving

.72 = 1(e^(-.0125t)
ln .72= ln (e^(-.0125t))
-.0125t = ln .72
t = ln.72/-.0125 = appr. 26.28 years

Answer 2

radioactive=decay
convert % to decimal.
72%=0.72, 0.0125%=0.000125
A=Aoe^-kt
0.72A=Aoe^-0.000125t
0.72=e^-0.000125t
ln 0.72=ln e^-0.000125t
ln 0.72=-0.000125t
ln 0.72/-0.000125=t
t=2628