Answer
$ k=\frac{1}{2\ln (4)}=0.3607$.
$F(6)=0.298=\frac{298}{1000}$.
Work Step by Step
The given model is
$F(t)=1-k\ln (t+1)$
Given values are
Time $t=3\;hours$.
The fraction of people after $3\;hours$ is $f(3)=\frac{1}{2}$
Substitute all the values into the given model.
$\Rightarrow F(3)=1-k\ln (3+1)$
$\Rightarrow \frac{1}{2}=1-k\ln (4)$
Add $k\ln (4)-\frac{1}{2}$ to both sides.
$\Rightarrow \frac{1}{2}+k\ln (4)-\frac{1}{2}=1-k\ln (4)+k\ln (4)-\frac{1}{2}$
Simplify.
$\Rightarrow k\ln (4)=1-\frac{1}{2}$
$\Rightarrow k\ln (4)=\frac{1}{2}$
Divide both sides by $\ln (4)$.
$\Rightarrow k=\frac{1}{2\ln (4)}$
Simplify.
$\Rightarrow k=0.3607$
Now Substitute $t=6$ and $k=\frac{1}{2\ln (4)}$ into the given model.
$\Rightarrow F(6)=1-\frac{1}{2\ln (4)}\ln (6+1)$
Simplify.
$\Rightarrow F(6)=1-\frac{1}{2\ln (4)}\ln (7)$
Use calculator and round to three decimal places.
$\Rightarrow F(6)=0.298$
Fraction with a denominator of $1000$.
$\Rightarrow F(6)=\frac{298}{1000}$.
