Question 885246
You may be able to do this using hours instead of minutes.



I = initial count of bacteria
p = count of bacteria after time t
{{{highlight_green(p=Ie^(kt))}}}, exponential growth formula


Interesting that we do not know the count of bacteria at time t=0.


{{{ln(p)=ln(Ie^(kt))}}}
{{{ln(p)=ln(I)+ln(e^(kt))}}}
{{{ln(p)=ln(I)+kt*1}}}
{{{highlight_green(ln(p)=ln(I)+kt)}}}, a LINEAR equation.  The vertical axis intercept is {{{ln(I)}}} and the slope is {{{k}}}.  The vertical axis variable is {{{ln(p)}}} and is a function of time {{{t}}}.


You have two data points to help in getting vertical intercept and slope.
20 & 600;
30 & 1600.
The way you must use them first is this way:
20 & ln(600)=6.3969
30 & ln(1600)=7.3778
Meaning the two points for the LINEAR form of this description are these:
(20, 6.3969)  and  (30, 7.3778).
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k or SLOPE
{{{k=(7.3778-6.3969)/(30-20)}}}
{{{k=0.9809/10}}}
{{{k=0.09809}}}
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Vertical Intercept
{{{ln(p)=ln(I)+kt}}}
{{{ln(I)=ln(p)-kt}}}
Pick either point to finish finding ln(I).
{{{ln(I)=7.3778-0.09809*30}}}
{{{ln(I)=4.4351}}}
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Antilog, base e, for 4.4351 will be I.
{{{I=84}}},   more sensible than saying 84.36...
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Build back the exponential growth equation using the now found values:
{{{highlight(p=84*e^(0.09809t))}}}.
You can answer whatever questions you want for this fictional or factual bacteria culture.


<i>Find the doubling period.  </i>
Just let I=1 and p=2.  Now find t.


<i>Find the population after 90 minutes.  </i>
Obviously, just let t=90...


<i>When will the population reach 11000</i>
Let p=11000 and solve for t.  There are clues how to do these i some of the steps I showed.