Answer
6.02
Work Step by Step
Using the properties of logarithms, we find:
$$e^{-0.2t}=\frac{3}{10} \\ \ln \left(e^{-0.2t}\right)=\ln \left(\frac{3}{10}\right) \\ -0.2t=\ln \left(\frac{3}{10}\right) \\ t=-5\ln \left(\frac{3}{10}\right) \\ t=6.02$$