# exponential growth

• January 25th 2007, 05:26 PM
viet
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
• January 25th 2007, 05:44 PM
viet
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:
• January 25th 2007, 05:53 PM
thedoge
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.
• January 25th 2007, 06:08 PM
viet
yeah.. this class sucks. she doesnt allow notes on the test, so im going to be screwed cuz i dont have a good memory.
• January 25th 2007, 06:11 PM
thedoge
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.
• January 25th 2007, 07:03 PM
ThePerfectHacker
You guys in the same class?
What chances are this of happening :rolleyes:
• January 25th 2007, 07:08 PM
thedoge
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
• January 27th 2007, 11:08 AM
earboth
Quote:

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:

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

With these values you can solve the 2nd equation:

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

$t \approx 60.66 \text{ minutes}$

EB