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
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.
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.
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
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