algebra 2
posted by sky on .
In the formula A(t) = A0ekt, A is the amount of radioactive material remaining from an
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 carbon14 it
originally contained, what is the approximate age of the artifact? (carbon14 decays at the
rate of 0.0125% annually.) (Round to the nearest year.)

You are solving
.72 = 1(e^(.0125t)
ln .72= ln (e^(.0125t))
.0125t = ln .72
t = ln.72/.0125 = appr. 26.28 years 
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