Asked by Hannah
A colony of bacteria increases according to the law of unihabited growth.
a) If the number of bacteria doubles in 5 hours, find the function that gives the number of cells in the culture.
I have the answer to this one. It is N=No e ln2/5 (t)
b) If there are 10,000 cells initially, how long will it take for there to be 25,000?
This is the part I do not know how to do. Would I just pug ib 10,000 and 25,000 in the equation that I got for part a?
a) If the number of bacteria doubles in 5 hours, find the function that gives the number of cells in the culture.
I have the answer to this one. It is N=No e ln2/5 (t)
b) If there are 10,000 cells initially, how long will it take for there to be 25,000?
This is the part I do not know how to do. Would I just pug ib 10,000 and 25,000 in the equation that I got for part a?
Answers
Answered by
Reiny
Your equation should have been
N = N<sub>0</sub> e^((ln2/5)t) , where t is in hours
another version could have been
N = N<sub>0</sub> (2)^(t/5) , where t is hours
b) 25000 = 10000 e^((ln2/5)t)
2.5 = e^((ln2/5)t)
(ln2/5)t = ln2.5
t = 5ln2.5/ln2 = 6.61 hours
using the other form of the equation
2.5 = 2^(t/5)
ln 2.5 = (t/5) ln2
t/5 = ln2.5/ln2
t = 5ln2.5/ln2 = 6.61
notice I ended up with the same final calculation.
N = N<sub>0</sub> e^((ln2/5)t) , where t is in hours
another version could have been
N = N<sub>0</sub> (2)^(t/5) , where t is hours
b) 25000 = 10000 e^((ln2/5)t)
2.5 = e^((ln2/5)t)
(ln2/5)t = ln2.5
t = 5ln2.5/ln2 = 6.61 hours
using the other form of the equation
2.5 = 2^(t/5)
ln 2.5 = (t/5) ln2
t/5 = ln2.5/ln2
t = 5ln2.5/ln2 = 6.61
notice I ended up with the same final calculation.
Answered by
Hannah
Thank You
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