Answer
$\approx 3.89$ days
Work Step by Step
An exponential model has the form $y=Ab^{t}.$
We are given two (t,y) points: (0,1), and (2, 0.7).
From (0,1) $\Rightarrow 1=Ab^{0} \Rightarrow A=1.$
From (2, 0.7)$ \Rightarrow 0.7=1\cdot b^{2} \Rightarrow b=(0.7)^{1/2}$
So, $y= 1(0.7)^{(1/2)t}=(0.7)^{t/2}$
Now, to find t when $y=\displaystyle \frac{1}{2}$ (half of the initial quantity).
$0.5=(0.7)^{t/2}\qquad/\log(...)$
$\displaystyle \log(0.5)=\frac{t}{2}\log(0.7)\qquad/\times\frac{2}{\log(0.7)}$
$t=\displaystyle \frac{2\log(0.5)}{\log(0.7)}\approx 3.89$ days.
