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Exercise 10.8 · Q1

Q.The rate of increase in the number of bacteria in a certain bacteria culture is proportional to the number present. Given that the number triples in 55 hours, find how many bacteria will be present after 1010 hours?

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✓ Free question

Set up the exponential growth model, fix the growth factor over 55 hours from the given tripling, then square that factor to jump to 1010 hours.

Step 1. Set up the model. dNdt=kN ⟹ N(t)=N0ekt\dfrac{dN}{dt}=kN\ \Longrightarrow\ N(t)=N_0e^{kt}, where N0N_0 is the count at t=0t=0.

Step 2. Use "triples in 5 hours". N(5)=3N0 ⟹ N0e5k=3N0 ⟹ e5k=3N(5)=3N_0\ \Longrightarrow\ N_0e^{5k}=3N_0\ \Longrightarrow\ e^{5k}=3.

Step 3. Find N(10)N(10). N(10)=N0e10k=N0(e5k)2=N0(3)2=9N0N(10)=N_0e^{10k}=N_0\left(e^{5k}\right)^2=N_0(3)^2=9N_0.

✓Final answer

N(10)=9N0N(10)=9N_0 — the bacteria count is 99 times the original number after 1010 hours.

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