Question 713839
Using time 0 as 15 years ago, {{{A[0]=18500}}}, and A=5600.  Use these to find k.


{{{5600=18500*e^(k*15)}}}
{{{(56/185)=e^(k*15)}}}
{{{ln(56/185)=k*15*1}}}
{{{k=(1/15)ln(56/185)}}}
{{{highlight(k=-0.0797)}}}


a.  The decay equation to model the exponential DECAY (not growth) is A=A[0]e^(-0.0797t).  You choice of A[0] depends on when you want to take time point for zero time.  You can choose "now" as time=0, and then {{{A[0]=5600}}}.  For now, current time decay equation, then you can say {{{highlight(A=5600*e^(-0.0797t))}}}.


b.  For twenty years from now, to find how many birds living, (a prediction), use t=20.  Compute A.  Use the equation that was just found.