Question 993293
{{{p=a*b^t}}} and you have some information for doubling time.  Twenty minutes is  {{{1/3}}} hour.  Try to use just that much and see what you can find.


Try taking a as the initial population when t=0.
{{{system(t=1/3,p=2,a=1)}}},
.
{{{2=1*b^(1/3)}}}

{{{b=root(3,2)}}}, or b=1.26 approximately.
The model might work as {{{p=a*(1.26)^t}}}.


The next part of the description is the given point  (1.5, 50000), and you want to know a, or value of p when t=0 instead of t=1.5.
{{{a*(1.26)^(3/2)=50000}}}


{{{a=50000/(1.26)^(3/2)}}}
and for convenience, recall where the "1.26" came from:
{{{a=50000/(root(3,2))^(3/2)}}}
...the rendering does not look properly aligned but that is an equation, formula for a =  fifty thousand over (cube root of two) to the three-halves power...
{{{a=50000/(2^(1/3))^(3/2)}}}

{{{a=50000/(2^((1/3)(3/2)))}}}

{{{a=50000/(2^(1/2))}}}

{{{highlight(a=50000/sqrt(2))}}}, simple radical form, although denominator is not rationalized.


This would be about {{{highlight(a=35400)}}}.