anonymous
  • anonymous
how do i set this up?! a bacteria culture initially contains 100 dells and grows at a rate proportional to its size. after an hour the population has incresed to 420. a)find an expression for the number of bacteria after t hours. b)find the number of bacteria after 3 hours c)find the rate of growth after 3 hours d)when will the population reach 10,000
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
i have trouble converting it to numbers. here what i have so far and i working on more. i use the P(t)=P(0)e^kt
anonymous
  • anonymous
so P(0)=100 P(3)=420
anonymous
  • anonymous
i set the equation P(3)=100e^3k=420 i solved and got k=ln42/3

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anonymous
  • anonymous
is a [100e^[(ln42/3)(t)]
anonymous
  • anonymous
b) i just plug in 3 for t to the equation in a) ?
anonymous
  • anonymous
for d) i put equation a) equal to 10,000 right?
anonymous
  • anonymous
im stuck on c) though. how do you do that?
anonymous
  • anonymous
I guess that you can derive the expression you got before to obtain dP/dt. And then you sustitute the value of 3 on it. .
anonymous
  • anonymous
i got it down to 100(42^t/3) i dont how to dervie?
anonymous
  • anonymous
I think your initial conditions are wrong. You wrote P(3) = 420, and on the problem says P(1) = 420
anonymous
  • anonymous
\[420 = 100e^{k(1)}\]
anonymous
  • anonymous
so, \[k = \ln4.2\]
anonymous
  • anonymous
ok crap .yeah your right! good catch
anonymous
  • anonymous
\[P = 100e^{\ln(4.2)t}\]
anonymous
  • anonymous
if you derive you get: \[\frac{d}{dt}100e^{(\ln 4.2)k}= 100(\ln4.2)e^{(\ln 4.2)k}\]
anonymous
  • anonymous
sorry its t instead of k
anonymous
  • anonymous
im trying to get waht you got for the derviative right now. its not coming out for me.
anonymous
  • anonymous
what did you get?
anonymous
  • anonymous
instead of ln4.2 i get 4.2k
anonymous
  • anonymous
remember that \[\frac{d}{dx}e^{ax} = ae^x\]
anonymous
  • anonymous
Sorry i missed an "a" \[\frac{d}{dx}e^{ax}=ae^{ax}\]
anonymous
  • anonymous
oh. ok. got it now. thanks
anonymous
  • anonymous
You're welcome.

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