SOLUTION
We will use the formula given
[tex]\begin{gathered} N=N_0e^{-kt} \\ here\text{ N = 30\% of N}_0 \\ N=0.3N_0 \\ k=0.0001 \end{gathered}[/tex]Plugging in we have
[tex]\begin{gathered} N=N_0e^{-kt} \\ 0.3N_0=N_0e^{-0.0001t} \\ e^{-0.0001t}=\frac{0.3N_0}{N_0} \\ e^{-0.000t}=0.3 \end{gathered}[/tex]Taking Ln of both sides we have
[tex]\begin{gathered} ln(e^{-0.000t})=ln0.3 \\ -0.0001t=ln0.3 \\ t=\frac{ln0.3}{-0.0001} \\ t=12,039.728043 \end{gathered}[/tex]Hence the answer is 12,040 years to the nearest year
