SOLUTION
The formula to apply is
[tex]\begin{gathered} A=A_oe^{-\lambda t} \\ Where\text{ A = amount of substance remaining = 0.5, after half decayed} \\ A_o=1 \\ \lambda=0.051 \\ t=\text{ time in years } \end{gathered}[/tex]
Putting in the values into the formula, we have
[tex]\begin{gathered} 0.5=1\times e^{-0.051t} \\ 0.5=e^{-0.051t} \\ Taking\text{ ln of both sides, we have} \\ ln0.5=-0.05t \\ t=\frac{ln0.5}{-0.051} \\ t=13.59112 \end{gathered}[/tex]
Hence the answer is 13.6 years to 1 d.p
