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Worked Examples · Example 2

Q.A scientist running an experiment finds that a particular bacterial colony doubles its population every 20 hours. The experiment starts with 200 bacteria cells. The number of cells is expected to be given by the formula b=200(2)t20b = 200(\sqrt{2})^{\frac{t}{20}} where tt is the number of hours for which the experiment is running. Find the number of hours after which there will be 500 bacteria cells.

Puducherry CbseNCERTSubjective· 3mImportance★★★★★est
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Set the population formula equal to 500, isolate the exponential term, and solve for tt using logarithms.

[!FORMULA]

b=200(2)t/20b=200(\sqrt2)^{t/20}, where bb = number of bacteria cells and tt = elapsed time in hours. If (2)x=k(\sqrt2)^{x}=k then x=log⁡klog⁡2x=\dfrac{\log k}{\log\sqrt2}.

  1. Set b=500b=500: 500=200(2)t/20500=200(\sqrt2)^{t/20}.
  2. Divide both sides by 200: (2)t/20=500200=2.5(\sqrt2)^{t/20}=\dfrac{500}{200}=2.5.
  3. Take natural logs of both sides: t20log⁡(2)=log⁡(2.5)\dfrac{t}{20}\log(\sqrt2)=\log(2.5), i.e. t20⋅12ln⁡2=log⁡(2.5)\dfrac{t}{20}\cdot\dfrac12\ln2=\log(2.5).
  4. Solve for tt: t=20log⁡(2.5)12ln⁡2=40log⁡(2.5)ln⁡2t=\dfrac{20\log(2.5)}{\tfrac12\ln2}=\dfrac{40\log(2.5)}{\ln2}. …

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