Question 1026385
p for population
x for time in hours
{{{p=8700*e^(kx)}}}


Solve the model for k.
{{{ln(p)=ln(8700)+kx*ln(e)}}}
{{{kx*1=ln(p)-ln(8700)}}}
{{{k=(1/x)(ln(p)-ln(8700))}}}


Substitute the point  (1,10000).
{{{k=(1/1)(ln(10000)-ln(8700))}}}
{{{k=ln(10000)-ln(8700)}}}
{{{k=0.13926}}}-----which you might want only to three significant figures, or maybe just two.


If you want p after 2 hours, you could simply use   {{{10000(10000/8700)}}} and it SHOULD be the same as  {{{8700*e^(0.139*2)}}};  just try it /them and see.  The model more specifically is {{{highlight(p=8700*e^(0.139*x))}}}.


Find x for how long to double.
Take logs of both sides and solve for x.
More simply,  {{{2=1*e^(0.13926*x)}}}---------start from this form.