Question 898920
"At noon an injection..." gives the data to find k in the <i>first</i> equation.
{{{C/Co=e^(-kt)}}}
{{{ln(C/Co)=-kt*ln(e)}}}
{{{ln(C/Co)=-kt}}}
{{{k=-(1/t)ln(C/Co)}}}
{{{highlight_green(k=(1/t)ln(Co/C))}}}
The variable choices are slightly different than yours.
The data in the rest of that sentence and the next one put into this formula makes {{{k=(1/3)ln(2)}}}.
{{{highlight_green(k=0.231)}}}.


Knowing the value for k allows you to find the value for T, the time interval between effective and harmful.  Now, using YOUR choice of variables, C2=5*C and C1=C in this case for some constant C.  This gives {{{T=(1/k)ln(5)}}},
{{{T=(1/0.231)ln(5)}}}
{{{highlight(T=6.97)}}}, not sure if to the nearest hour might be best or not; but this could be stated T is 6 hours and 58 minutes.