# Logarithm's and Such

There are initially 200 bacteria in a culture. After five minutes, the population has grown to 4080 bacteria. Estimate the doubling period.

I DID THIS:

$$4080=200(k)^5\\\frac{4080}{200}=k^5\\k=1.82$$

IS THIS CORRECT?

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Hootenanny
Staff Emeritus
Gold Member
I'm not quite sure what your doing here, but when solving exponential growth problems one should generally start with a differential equation such as;

$$\frac{dN}{dt} = kN$$

Where N is the number of bacteria. Do you know the solution to this differential?

no your using derivative crap...not that way

Hootenanny
Staff Emeritus
Gold Member
thomasrules said:
no your using derivative crap...not that way
How do you propose to solve it?

damnit how do u start a new line in latex

Hootenanny
Staff Emeritus
Gold Member
Type a double backslash like this: \\

I DID!!:

[.tex]4080=200(k)^5\\\frac{4080}{200}=k^5\\k=1.82[.tex]

i'm aware of the .

ok:
4080=200(k)^5

4080/200}=k^5

k=1.82

Hootenanny
Staff Emeritus
Gold Member
thomasrules said:
ok:
4080=200(k)^5

4080/200}=k^5

k=1.82
This is not correct.

$$\frac{dN}{dt} = kN$$. The solution of this equation is $$N_{0}e^{kt}$$. We know that $$N(0) = 200$$. So $$N= 200e^{kt}$$.

$$4080 = 200e^{5k}$$.

$$k = .6031$$

To find the doubling time, look at $$t = 1$$.

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Y look at 1 minute. It's not answering the question for doubling time. If you are starting with 200, then double would be 400. and the question is what is the time for that. so it would be 400=200e^(0.6031*t). solve for t.

yes, my fault. you are correct

HallsofIvy
Homework Helper
thomasrules said:
There are initially 200 bacteria in a culture. After five minutes, the population has grown to 4080 bacteria. Estimate the doubling period.

I DID THIS:

$$4080=200(k)^5\\\frac{4080}{200}=k^5 \\ k=1.82$$

IS THIS CORRECT?
Yes, that's a correct calculation but I don't see any point in doing it. The problem did not ask for "k".
If T is the time (in minutes) to double, then the population would be multiplied by 2 every T minutes: in t minutes, you will have doubled t/T times: $P= P_02^{\frac{t}{T}}$. If, after 5 minutes the population has increased from 200 to 4080, you want to find T such that
$$2^{\frac5}{T}}= 4080/200= 20.4$$
Take the logarithm of both sides to solve for T.