Question 719989
{{{Q=Q[0]e^(kt)}}} Basic Exponential Decay Model.


Half-life is 4.  {{{(1/2)Q[0]=Q[0]e^(k*4)}}}
{{{1/2=e^(4k)}}}
{{{ln(1/2)=4k*1}}}
{{{k=(1/4)ln(1/2)}}}
{{{highlight(k=-0.173)}}}.


Given is Q=2500 at t=0, so the model for this example is {{{highlight(Q=2500e^(-0.173t))}}}.