Calculus of a Single Variable, Chapter 6, 6.2, Section 6.2, Problem 38

It is impossible to predict when a particular atom will decay. However, it is equally likely to decay at any instant in time. Therefore, given a sample of a particular radioisotope, the number of decay events −dN expected to occur in a small interval of time dt is proportional to the number of atoms present N, i.e.
-(dN)/(dt)propto N
For different atoms different decay constants apply.
-(dN)/(dt)=\lambda N
The above differential equation is easily solved by separation of variables.
N=N_0e^(-lambda t)
where N_0 is the number of undecayed atoms at time t=0.
We can now calculate decay constant lambda for carbon-14 using the given half-life.
N_0/2=N_0e^(-lambda 5715)
e^(-5715lambda)=1/2
-5715lambda=ln(1/2)
lambda=-(ln(1/2))/5715
lambda=1.21 times 10^-4
Note that the above constant is usually measured in seconds rather than years.
Now we can return to the problem at hand. Since the charcoal contains only 15% (0.15N_0 ) of the original carbon-14, we have
0.15N_0=N_0e^(-1.21times10^-4t)
Now we solve for t.
e^(-1.21times10^-4t=0.15)
1.21times10^-4=-ln 0.15
t=-(ln0.15)/(1.21times10^-4)
t=15678.68
According to our calculation the tree was burned approximately 15679 years ago.