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Worked Examples · Example 13

Q.In a certain culture of bacteria the rate of increase is proportional to the number present. It is found that there are 10,000 bacteria at the end of 3 hours and 40,000 bacteria at the end of 5 hours. How many bacteria were present in the beginning?

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Exponential growth N=N0ektN=N_0e^{kt}: from N(3)=10,000N(3)=10{,}000 and N(5)=40,000N(5)=40{,}000, ek=2e^k=2 and N0=1,250N_0=1{,}250.

dNdt=kN  ⇒  N=N0ekt\dfrac{dN}{dt}=kN\;\Rightarrow\;N=N_0e^{kt}, where N0N_0 = initial count, kk = growth constant, tt = time (hours).

Given: N(3)=10,000N(3)=10{,}000, N(5)=40,000N(5)=40{,}000.

  1. Model: N=N0ektN=N_0e^{kt}.
  2. Divide the two data equations: N(5)N(3)=N0e5kN0e3k=e2k=40,00010,000=4\dfrac{N(5)}{N(3)}=\dfrac{N_0e^{5k}}{N_0e^{3k}}=e^{2k}=\dfrac{40{,}000}{10{,}000}=4.
  3. So e2k=4⇒ek=2e^{2k}=4\Rightarrow e^{k}=2 (hence k=log⁡2k=\log 2). …

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