Exponential growth equation - C4 Watch

jonnburton
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I've been trying to work out a question on this topic but haven't been able to get started properly. Could anyone tell me how to approach this?

The number, N, of bacteria increases at a rate proportional to the number present at any instant. After timing for 3 hours, the number is 5000 and after 5 hours, the number is 10 000. Find the initial number of bacteria.

I know the equation is  N = N_0 e^{kt}

and that

 N_0e^{3k} = 5000, and  N_0e^{5k} = 10000

But I can't work out how to get values for N_0 and k.

It also seems clear that  ln|\frac{5000}{n_0} = 3k so  \frac{1}{3} ln\frac{5000}{N_0} = k

But beyond that I get nowhere except going round in circles saying  e^k = \sqrt{\frac{5000}{N_0}} and  e^{3k} = \frac{5000}{N_0}

It seems like the kind of thing that could be solved simultaneously, but I'm not sure and don't really know how do do it..!
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J_W-x
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(Original post by jonnburton)
I've been trying to work out a question on this topic but haven't been able to get started properly. Could anyone tell me how to approach this?

The number, N, of bacteria increases at a rate proportional to the number present at any instant. After timing for 3 hours, the number is 5000 and after 5 hours, the number is 10 000. Find the initial number of bacteria.

I know the equation is  N = N_0 e^{kt}

and that

 N_0e^{3k} = 5000, and  N_0e^{5k} = 10000

But I can't work out how to get values for N_0 and k.

It also seems clear that  ln|\frac{5000}{n_0} = 3k so  \frac{1}{3} ln\frac{5000}{N_0} = k

But beyond that I get nowhere except going round in circles saying  e^k = \sqrt{\frac{5000}{N_0}} and  e^{3k} = \frac{5000}{N_0}

It seems like the kind of thing that could be solved simultaneously, but I'm not sure and don't really know how do do it..!
From your second line can you not say
 2NoE^(3k)=NoE^(5K)
my latex is awful, sorry.
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bembem
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5000=N0e^3k and 10000=N0e^5k

Equating and rearranging for N0

5000/e^3k = 10000/e^5k

Rearranging gives

e^2k = 2, now you can solve this for k.

Then using a given time t, a given number N and your value of k you can find N0 at these parameters.

You can also do the problem to simultaneously solve in terms of k first but I didnt want to try that in my head.
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jonnburton
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Ah, I completely missed that! I'll try it... Thanks JWX!

(Original post by J_W-x)
From your second line can you not say
 2NoE^(3k)=NoE^(5K)
my latex is awful, sorry.
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J_W-x
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(Original post by jonnburton)
Ah, I completely missed that! I'll try it... Thanks JWX!
It's alright , sometimes it's good to be lazy, look for the easiest way of solving it possible without getting into messy ln business is what I always do first :rolleyes:
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