1) Considering that the initial amount of Carbon 14 is 40% (0.4) and each year "t" is given in whole numbers. And we were told the amount of k we can insert into the equation:
[tex]\begin{gathered} N=N_0e^{-kt} \\ 0.6=\left(1\right)e^{-0.0001*t} \\ e^{\left\{-0.0001t\right\}}=0.6 \\ lne^{\left\{-0.0001t\right\}}=ln\left(0.6\right) \\ -0.0001t=ln\left(0.6\right) \\ t=5108.25623 \\ t\approx5108years \end{gathered}[/tex]Note that the initial value is 1 and the last one is 0.6 (40% less).