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Exercise 5 · Q3

Q.In a certain culture of bacteria, the number of bacteria increased 5 times in 10 hours. How long did it take for the number of bacteria to double?

Lakshadweep CbseNCERTSubjective· 3mImportance★★★★★
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✓ Free question

Growth is exponential; a 5-fold rise in 10 hours fixes k=log⁡510k=\frac{\log 5}{10}, so the doubling time is t=10log⁡2log⁡5≈4.31t=\frac{10\log 2}{\log 5}\approx 4.31 hours.

N(t)=N0ektN(t)=N_0e^{kt} — exponential growth model, where N0N_0 = initial number of bacteria, N(t)N(t) = number at time tt, kk = growth constant, tt = time in hours.

  1. Set up from the given data. The count becomes 5 times in 10 hours: 5N0=N0e10k⇒e10k=55N_0=N_0e^{10k}\Rightarrow e^{10k}=5.
  2. Find kk. Taking natural logs: 10k=log⁡5⇒k=log⁡510=1.609410=0.16094 h−110k=\log 5\Rightarrow k=\dfrac{\log 5}{10}=\dfrac{1.6094}{10}=0.16094\ \text{h}^{-1}.
  3. Doubling condition. We need N=2N0N=2N_0: 2N0=N0ekt⇒ekt=2⇒kt=log⁡2⇒t=log⁡2k2N_0=N_0e^{kt}\Rightarrow e^{kt}=2\Rightarrow kt=\log 2\Rightarrow t=\dfrac{\log 2}{k}.
  4. Substitute kk. t=log⁡2log⁡5/10=10log⁡2log⁡5=10×0.69311.6094=6.9311.6094=4.31 ht=\dfrac{\log 2}{\log 5/10}=\dfrac{10\log 2}{\log 5}=\dfrac{10\times 0.6931}{1.6094}=\dfrac{6.931}{1.6094}=4.31\ \text{h}.
  5. Check. e0.16094×4.31=e0.6936≈2.00e^{0.16094\times 4.31}=e^{0.6936}\approx 2.00 ✓ (doubles as required).
✓Final answer

The bacteria take about 4.31 hours (≈ 4 hours 18 minutes) to double.

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