1. ## exponential 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

2. 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

3. 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.

BTW, I am doing it exactly like you.

4. yeah.. this class sucks. she doesnt allow notes on the test, so im going to be screwed cuz i dont have a good memory.

5. 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.

6. You guys in the same class?
What chances are this of happening

7. 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

8. Originally Posted by viet
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
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