Chapter 5 - 5.5 - Exponential and Logarithmic Models - 5.5 Exercises - Page 405: 27

$y=5 e^{\frac{-\ln 5}{4}x} =5e^{-0.4x}$

We have $y=ae^{bx}$ $ae^{b(0)}=5$ and $ae^{4b}=1$ Further, we have $ae^{0}=5$ $ \implies a=5$ and $5e^{4b} =1 \implies e^{4b} =\dfrac{1}{5}$ Take the $\log$ on each side. $\ln e^{4b} =\ln \dfrac{1}{5}$ or, $4b \approx -1.6094$ or, $ b \approx -0.4$ Now, the exponential decay model is: $y=5 e^{\frac{-\ln 5}{4}x} =5e^{-0.4x}$