Chapter 7 Exponential and Logarithmic Functions - 7.6 Solve Exponential and Logarithmic Equations - 7.6 Exercises - Problem Solving - Page 522: 61

5.9 weeks

Given: $h(t)=\frac{256}{1+13e^{-0.65t}}$ Substituting $h=200$, we have: $h(t)=\frac{256}{1+13e^{-0.65t}}\\ 1+13e^{-0.65t}=\frac{256}{200}\\13e^{-0.65t}=\frac{7}{25}\\ e^{-0.65t}=\frac{7}{325}\\\ln (e^{-0.65t})=\ln(\frac{7}{325})\\-0.65t=\ln(\frac{7}{325})\\t\approx5.9045\approx5.9$