Asked by Annie
Suppose an E. coli culture is growing exponentially at 37 ◦C. After 20 minutes at that temperature, there are 1.28×10^7 E. coli cells. After 60 minutes, there are 2.4×10^7 cells. How long does it take for the culture to have double the amount of cells that it had started with?
I do not know where to begin. I tried using the equation P(x)=Po e^kt but I don't know what to do because there would be two values for t.
I do not know where to begin. I tried using the equation P(x)=Po e^kt but I don't know what to do because there would be two values for t.
Answers
Answered by
bobpursley
You can do it that way, but I think simpler math will do..
amountfinal=amoutintial*(e)^.692 time/doubletime
amountfinal/amountintial=e^.692(t/doublettime)
so you are given two points...
2.4E7/1.28E7 = e^.692*40/timedouble
I am assuming the "after 60minutes"means 40 min after the first reading.
take the ln of each side
ln( 2.4/1.28)=.692*40/timedouble
solve for time to double.
amountfinal=amoutintial*(e)^.692 time/doubletime
amountfinal/amountintial=e^.692(t/doublettime)
so you are given two points...
2.4E7/1.28E7 = e^.692*40/timedouble
I am assuming the "after 60minutes"means 40 min after the first reading.
take the ln of each side
ln( 2.4/1.28)=.692*40/timedouble
solve for time to double.
Answered by
Annie
Can I ask why you are using 0.692 in the exponent?
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