Answer
$\approx 0.903$ days
Work Step by Step
An exponential model has the form $y=Ab^{t}.$
The initial quantity (after the colleague stole half) is $A=1$ g.
We are given that for t=3, $y=0.1.$ We find b:
$0.1=1\cdot b^{3}$
$b=(0.1)^{1/3}$
So, $y=(0.1)^{(1/3)t}=(0.1)^{t/3}$
Now, to find t when $y=\displaystyle \frac{1}{2}$ (half of the initial quantity).
$0.5=(0.1)^{t/3}\qquad/\log(...)$
$\displaystyle \log(0.5)=\frac{t}{3}\log(0.1)\qquad/\times\frac{3}{\log(0.1)}$
$t=\displaystyle \frac{3\log(0.5)}{\log(0.1)}\approx 0.903$ days
