Posted by **sky** on Wednesday, July 13, 2011 at 12:43am.

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 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.)

- algebra 2 -
**Reiny**, Wednesday, July 13, 2011 at 9:03am
You are solving

.72 = 1(e^(-.0125t)

ln .72= ln (e^(-.0125t))

-.0125t = ln .72

t = ln.72/-.0125 = appr. 26.28 years

- algebra 2 -
**Inu**, Thursday, June 20, 2013 at 12:25am
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

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