If a bacteria culture starts with 8000 bateria and doubles every 25 minutes, how many minutes will it take the population to reach 43000?

i need help setting up this problem, formula: y = Ce^(kt)

so far i have 43000 = 8000e^(25)k

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- Jan 25th 2007, 05:26 PMvietexponential growth
If a bacteria culture starts with 8000 bateria and doubles every 25 minutes, how many minutes will it take the population to reach 43000?

i need help setting up this problem, formula: y = Ce^(kt)

so far i have 43000 = 8000e^(25)k - Jan 25th 2007, 05:44 PMviet
ok here what i gotten sofar:

43000 = 8000e^(25t)

(43/8) = e^(25t)

ln(43/8) = 25t

(ln(43/8)) / 25 = t

so t = 0.67270 approx.

not sure what im doing wrong but the answer is wrong :confused: - Jan 25th 2007, 05:53 PMthedoge
you are in my calc II class too?

No calcs=an adjustment. Kind of stupid to require that when it is obvious you need them for the homework.

i think i finally realize i need to get a tutor. i do everything exactly like i think I should, but my answers are hit or miss. it's so frustrating not knowing the one error you are making.

I don't know about this problem either. I thought I had it but my answers aren't working.

BTW, I am doing it exactly like you. - Jan 25th 2007, 06:08 PMviet
yeah.. this class sucks. she doesnt allow notes on the test, so im going to be screwed cuz i dont have a good memory.

- Jan 25th 2007, 06:11 PMthedoge
Yeah, same. As bad as our last teacher taught in class at least he compensated for it on his tests. This teacher is fairly good, but she is a harder teacher. I always feel like I have this stuff down until webwork or a quiz comes, then I think I don't know it at all.

- Jan 25th 2007, 07:03 PMThePerfectHacker
You guys in the same class?

What chances are this of happening :rolleyes: - Jan 25th 2007, 07:08 PMthedoge
Yep. We had the same teacher last semester as well as this semester. There is actually a third person in the same situation that comes here :)

for future reference, the general formula to solve this is

Pf=Po*e^(kt)

k=ln(2)/(doublingtime)

solve for t - Jan 27th 2007, 11:08 AMearboth
Hello, Viet,

according to your problem you have exactly 16000 bacteria after 25 minutes. Now you can calculate the constant k:

$\displaystyle 16000=8000 \cdot e^{k \cdot 25} \Longleftrightarrow 25k=\ln\left(\frac{16000}{8000} \right) \Longleftrightarrow k \approx 0.027726$. Solve for t:

With these values you can solve the 2nd equation:

$\displaystyle 43000=8000 \cdot e^{0.027726 \cdot t} \Longleftrightarrow 0.027726 \cdot t=\ln\left(\frac{43}{8} \right)$

$\displaystyle t \approx 60.66 \text{ minutes}$

EB