Question 933433
The question description seems impractical.  Is  14.9 hours too short a time to purchase the expected 480 miligrams?  


{{{A=Ie^(-kt)}}}
{{{(1/2)=1*e^(-kt)}}}
{{{ln(1/2)=ln(1)-kt*ln(e)}}}
{{{-kt=ln(1/2)}}}
{{{k=-(1/t)ln(1/2)}}}
{{{k=-(1/t)(-ln(2))}}}
{{{k=ln(2)/t}}}
{{{highlight_green(k=ln(2)/14.9)}}}.


The model becomes {{{highlight(A=480*e^(-0.4652t))}}}.


You need a solution for t.
{{{ln(A)=ln(480)+ln(e^(-0.04652t))}}}
{{{ln(A)-ln(480)=-0.04652t}}}
{{{ln(480)-ln(A)=0.04652t}}}
{{{highlight_green(t=(1/0.04652)ln(480/A))}}}
Let A=3.75 and evaluate t.


t=104 hours.
---  
4 days 8 hours 30 minutes


CHECK:  This should be 7 periods of halflife time.
14.9*7=104.3, checks.