Diff E Q's Question

CatchThis2

Junior Member
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Feb 6, 2010
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A bacteria culture starts with 120 bacteria and grows at a rate propotional to its size. After 6 hours there are 720 bacteria.
(a) Find the population after t hours
y(t) = (function of t)
(b) Find the population after 5 hours.
y(5)=
(c) When will the population reach 1750 ?
T =

This is what I have so far. N(t)=No e^kt
N(6)=No e^kt
N(6)= 120No

T=6ln120/ln6

K=ln120/ln6


Not sure if this is the right work or if I am totally lost.
 
you first need to calculate k given the initial and final number and time (120,720 and 6 respectively)

solved for k algebraically, k= ln(Nt/No)/t

k = ln(720/120)/6

take that value of k and solve for your different value of time or final number or whatever else. Let me know if this helps.
 
tutor_joel said:
solved for k algebraically, k= ln(Nt/No)/t

k = ln(720/120)/6

Joel meant to divide by 6 (not multiply), as shown above in red. Perhaps, he suffered a TIA, doing the exercise in his head. That happens to me alot. :wink:
 
nope, You need to determine the argument of the log and then divide by 6. The Ln doesn't distribute like that. Plug it into your calculator as I have written it, or simplify first.

k=Ln(6)/6

k= 0.2986
 
To find the last part of the problem I should set
e=ln6/6=t

Then e ln6/6=t

(6/ln6)ln e(ln6/6)=lnt(6/ln6)

t=6lnt/6

Does this work look right and does the answer seem correct.
 
Sorry, no.

You are not understanding the fundamentals of this problem. The equation N(t) = No*e^kt has a time variable, where No is the initial number, N(t) is the number after a given time. solving for k gives you the rate of growth over a given time frame. Now that you know the rate, you set up a new equation with t = 5. You're unknown is then N(t), which is the number of bugs after time t at rate k.

Part A:
N(t) = 120e^(0.2986t)

Part B:
k = Ln(6)/6 = 0.2986
t = 5
No = 120

N(5) = 120e^(0.2986*5)

in part c your unknown is the time, since you're given N(t)
 
How do I go about finding When will the population reach 1750 ?


I think this could be the answer ? T= 5ln6/ln6
 
I'm not sure what you're doing here. Why don't you try to show more steps next time so that I can understand where you're making mistakes.

You have to use the equation from part A again and solve the equation for t.

N(t) = 120e^(0.2986t)

Find t for the population to reach 1750 means they give you N(t) = 1750. So,

1750 = 120e^(0.2986t)

1750/120 = e^0.2986t

Ln(1750/120) = 0.2986t

Ln(1750/120)/0.2986 = t

and there it is. Can you follow the steps?
 
Joel,

Thank you for breaking down the steps for me. I am a visual learner so seeing step by step how you got the answer really helps. Thanks for your time and patience. Have a good weekend!
 
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