Let [tex]U(t)[/tex] denote the amount of uranium-238 in the reactor at time [tex]t[/tex]. As conversion to plutonium-239 occurs, the amount of uranium will decrease, so the conversion rate is negative. Because the rate is proportional to the current amount of uranium, we have
[tex]\dfrac{\mathrm dU}{\mathrm dt}=-kU[/tex]
where [tex]k>0[/tex] is constant. Separating variables and integrating both sides gives
[tex]\dfrac{\mathrm dU}U=-k\,\mathrm dt\implies\ln|U|=-kt+C\implies U=Ce^{-kt}[/tex]
Suppose we start some amount [tex]u[/tex]. This means that at time [tex]t=0[/tex] we have [tex]U(0)=u[/tex], so that
[tex]u=Ce^{-0k}\implies C=u\implies U=ue^{-kt}[/tex]
We're given that after 10 years, 99.97% of the original amount of uranium remains. This means (if [tex]t[/tex] is taken to be in years) for some starting amount [tex]u[/tex],
[tex]0.9997u=ue^{-10k}\implies k=-\dfrac{\ln(0.9997)}{10}[/tex]
The half-life is the time [tex]t_{1/2}[/tex] it takes for the starting amount [tex]u[/tex] to decay to half, [tex]0.5u[/tex]:
[tex]0.5u=ue^{-kt_{1/2}}\implies t_{1/2}=-\dfrac{\ln(0.5)}k=\dfrac{10\ln2}{\ln(0.9997)}[/tex]
or about 23,101 years. Notice that it doesn't matter what the actual starting amount is, the half-life is independent of that.
