# Math Help - Radioactive Decay

In the formula A=Ie^kt, A is the amount of radioactive material remaining from an initial amount I at a given time t, and k is the negative constant determined by the nature of the material. An artifact is discovered at a certain site. If it has 52% 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)

2. Originally Posted by lollipopgang
In the formula A=Ie^kt, A is the amount of radioactive material remaining from an initial amount I at a given time t, and k is the negative constant determined by the nature of the material. An artifact is discovered at a certain site. If it has 52% 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)

A = I e^(kt) ----------(i)
A/I = e^(kt)
ln(A/I) = ln[e^(kt)]
ln(A/I) = (kt)ln(e)
ln(A/I) = kt ------------(ii)

(ii) and (i) are the same.

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"(carbon-14 decays at the rate of 0.0125% annually)"

So, if, say, one year ago, the carbon-14 content is x, then at present, after the 0.0125% decay, the new carbon-14 content is
((100 -0.0125)% =) 99.9875% of x.

Plugging those into (ii),
ln[(99.9875% of x) /x] = k(1) ------one year past, remember.
ln(99.9875%) = k
k = ln(0.999875) = -0.000125008

(Umm...., really? That is -0.0125%)

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" An artifact is discovered at a certain site. If it has 52% of the carbon-14 it originally contained, what is the approximate age of the artifact? "

If the artifact had y-amount of carbon-14 initially t-years ago, now it has 52% of y.

ln(A/I) = kt
ln[(52% of y) /y] = (-0.000125008)t
ln(0.52) = (-0.000125008)t
t = ln(0.52) /(-0.000125008)
t = 5231 years