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Exercise 5 · Q11

Q.Use the exponential growth model to show that the time it takes for a population to double (i.e., from an initial number A to 2A) is given by t=ln⁡2kt=\frac{\ln 2}{k}

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Starting from the exponential growth solution P=AektP=Ae^{kt}, imposing P=2AP=2A and solving for tt yields the doubling time t=log⁡2kt=\frac{\log 2}{k}.

dPdt=kP\dfrac{dP}{dt}=kP with solution P(t)=AektP(t)=Ae^{kt} — exponential growth, where AA = initial number (at t=0t=0), P(t)P(t) = number at time tt, k>0k>0 = growth constant, tt = time.

  1. State the model. The growth rate is proportional to the current amount: dPdt=kP\dfrac{dP}{dt}=kP.
  2. Solve the differential equation. Separate variables: dPP=k dt⇒∫dPP=∫k dt⇒log⁡P=kt+C\dfrac{dP}{P}=k\,dt\Rightarrow \int\dfrac{dP}{P}=\int k\,dt\Rightarrow \log P=kt+C.
  3. Apply the initial condition P(0)=AP(0)=A: log⁡A=C\log A=C, so log⁡P=kt+log⁡A⇒log⁡PA=kt⇒P=Aekt\log P=kt+\log A\Rightarrow \log\dfrac{P}{A}=kt\Rightarrow P=Ae^{kt}. …

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