Chapter 9 - Section 9.3 - Logarithmic Functions and Models - Exercises - Page 658: 54b

$37$ years

Given $Q(t)\approx Q_{0}e^{-0.0248t}$, (from part a.) we want to find t for which $Q(t)=\displaystyle \frac{2}{5}Q_{0}$ $\displaystyle \frac{2}{5}Q_{0}=Q_{0}e^{-0.0248t}$ $\displaystyle \frac{2}{5}=e^{-0.139t}\qquad/\ln(...)$ $\displaystyle \ln(\frac{2}{5})=-0.0248t$ $t=\displaystyle \frac{\ln(\frac{2}{5})}{-0.0248}\approx 37$ years