Question 1027424
p, the initial population
y, the amount of bacteria
x, the amount of time, in minutes


You might consider a continuous exponential growth model; and use k, a constant.

{{{highlight_green(y=pe^(kx))}}}


Find a formula for k.
{{{ln(y)=ln(pe^(kx))}}}
{{{ln(y)=ln(p)+ln(e^(kx))}}}
{{{ln(p)+kx*ln(e)=ln(y)}}}
{{{kx*1=ln(y)-ln(p)}}}-------* you may use this one again later.
{{{highlight_green(k=(1/x)ln(y/p))}}}
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The description tells you that for {{{x=17}}}, {{{y/p=2}}}.  Also that population will double every 17 minutes.
{{{k=(1/17)ln(2/1)}}}
{{{k=0.04077}}}
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Your more specific model is {{{highlight(y=1000*e^(0.04077x))}}}.
You do the rest.    *