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
