[tex]8.0\:hours[/tex]
1) In this problem, we must remind that the final amount on the left is half the initial one. Thus, we can plug into that formula the initial amount as 1 and the final 0.5
2) So, let's plug them into the formula:
[tex]\begin{gathered} y=y_0e^{-0.086t} \\ \\ 0.5=1\cdot e^{-0.086t} \\ \\ e^{-0.086t}=0.5 \\ \\ \ln \left(e^{-0.086t}\right)=\ln \left(0.5\right) \\ \\ -0.086t=\ln \left(0.5\right) \\ \\ \frac{-0.086t}{-0.086}=\frac{\ln \left(0.5\right)}{-0.086} \\ \\ t=-\frac{\ln \left(0.5\right)}{0.086} \\ \\ t\approx8.0 \end{gathered}[/tex]
Thus, the half-life is 8 hours
