The​ half-life of​ carbon-14 is 5600 years. If a piece of charcoal made from the wood of a tree shows only 68​% of the​ carbon-14 expected in living​ matter, when did the tree​ die?

Answer 1

0.68 = (0.5)^t/5600

ln 0.68 = (t / 5600) ln 0.5

-0.3856 = (t / 5600) (-0.6931)

3115.80 = t

Answer 2

Let A(t) = amount remaining t years after the tree died and let P be the amount of C-14 when the tree died/

A(t) = Pe^rt

0.5P = Pe^5600r

0.5 = e^5600r

ln(0.5) = 5600r

So, r = -0.0001238

A(t) = Pe^-0.0001238t

0.68P = Pe^-0.0001238t

ln(0.68) = -0.0001238t

t ≈ 3115 years ago