Answer
$\approx 3.2$ hours
Work Step by Step
An exponential model has the form $y=Ab^{t}$.
We are given that $y=\displaystyle \frac{3}{4}A$ when $t=1$ hour.
Substitute and solve for k:
$\displaystyle \frac{3}{4}A=Ab^{1}$
$b=\displaystyle \frac{1}{4}, $
Given A=200, $y=200(0.75)^{t}$.
We want the time t it takes for $y $ to become 80:
$80=200(0.75)^{t}\qquad/\div 200$
$0.4=(0.75)^{t}\qquad/\log(...)$
$\log 0.4=t\log 0.75\qquad/\log 0.75$
$t=\displaystyle \frac{\log 0.4}{\log 0.75}\approx 3.2$ hours
