Knowing that Carbon 14 decays in half approximately every 57

Knowing that Carbon 14 decays in half approximately every 5,730 years, suppose you find a female skeleton that has only 15% of its Carbon-14 remaining. Approximately how old is the skeleton?

Solution

decay constant = 0.693/half life = 0.693/5730 =0.0001209

half life is 5730 years

1/2 Ao = Ao e^(5730k)

k = ln(1/2) / 5730 = .0001209

equation to represent exponential decay:

A = Aoe^(-kt)

15% remaining: 0.15Ao = Aoe^(-0.0001209t)

0.15 = e^(-0.0001209t)

taking log of both sides:

ln(0.15) = -0.0001209t

t = 15691.65 years

So, skeleton is 15691.65 years old