Given that: N=N0e^(-kt) where N is amount of C-14 at time t N0 is the initial amount k is the constant t is the time taken given that the amount found was 30% percent of the initial amount, the age of the mammal will be found as follows: let N=x, N0=0.3x, k=0.0001, t=? Plug in the formula we get: 0.3x=xe^(-0.0001t) x will cancel and we shall remain with: 0.3=e^(-0.0001t) introducing natural logs we get: ln 0.3=-0.0001t hence t=ln0.3/(-0.0001) t= 12039.7 years
