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Cell Doubling Time Calculator

Calculate doubling time (how long it takes a cell population to double) during exponential growth. Use N = N₀·2ⁿ and n = log(N/N₀) / log(2). Includes quick picks, steps, a mini growth-curve visual, and a doublings table.

Background

In exponential growth, each “doubling” multiplies the population by 2. If your population starts at N₀ and ends at N after time t, you can compute the number of doublings (generations) n, then: doubling time = t/n. This model is commonly used for cells, bacteria, or any population growing approximately exponentially.

Enter values

Tip: Best for exponential (log-phase) growth.

We’ll keep answers in your chosen units.

Since n = log(N/N₀)/log(2), the base doesn’t change the result. This setting is just to match your class notation.

Any count units work (cells/mL, CFU/mL, total cells) as long as N and N₀ match.

Enter a number in your selected time units.

Time per doubling. Same units as your time selection.

Options

Chips prefill values and calculate immediately.

Result

No results yet. Enter values and click Calculate.

How to use this calculator

  • Pick a mode (doubling time, final population, time, or doublings).
  • Enter the values you know (usually N₀, N, and t).
  • Click Calculate to get the answer, plus a visual + table (optional).

How this calculator works

  • Exponential growth model: N = N₀·2ⁿ
  • Doublings: n = log(N/N₀) / log(2)
  • Doubling time: DT = t/n
  • Predict N from DT: N = N₀·2^(t/DT)

Formula & Equation Used

Exponential growth model: N = N₀·2ⁿ

Doublings (generations): n = log(N/N₀) / log(2)

Doubling time: DT = t/n

Predict final population (from DT): N = N₀·2^(t/DT)

Solve elapsed time (from DT): t = n·DT

Example Problem & Step-by-Step Solution

Example 1 — Find doubling time

A cell culture grows from N₀ = 1.0×10⁵ to N = 8.0×10⁵ in t = 24 hr. Find the number of doublings n and the doubling time DT.

  1. Compute ratio: N/N₀ = 8.0
  2. Doublings: n = log(8)/log(2) = 3
  3. Doubling time: DT = t/n = 24/3 = 8 hr

Because 8× is exactly three doublings (1→2→4→8), the answer is perfectly clean.

Example 2 — Predict final population

Starting at N₀ = 5.0×10⁴, the doubling time is DT = 12 hr. After t = 3 days, what is N?

  1. Convert time: 3 days = 72 hr
  2. Doublings: n = t/DT = 72/12 = 6
  3. Final population: N = N₀·2⁶ = 5.0×10⁴·64 = 3.2×10⁶

Example 3 — Solve elapsed time

A population grows from N₀ = 1.0×10³ to N = 1.0×10⁶. If the doubling time is DT = 30 min, how long did it grow?

  1. Doublings: n = log(N/N₀)/log(2) = log(10³)/log(2) ≈ 9.966
  2. Time: t = n·DT ≈ 9.966·30 ≈ 299.0 min

Real data rarely lands on exact powers of 2 — decimals for n are totally normal.

Frequently Asked Questions

Q: Is doubling time the same as generation time?

In binary fission / ideal exponential growth, yes — “generation time” is essentially the doubling time.

Q: Does it matter if I use ln or log₁₀?

No. In n = log(N/N₀)/log(2), the log base cancels out.

Q: When should I NOT use this model?

If growth isn’t close to exponential (lag phase, stationary phase, death phase), this can be misleading.

Q: What does a fractional number of doublings mean?

It means the population increased by a factor between two powers of 2. For example, n = 2.5 corresponds to a multiplier of 2^2.5 ≈ 5.66.

Q: Can I use this for decreasing populations?

This calculator assumes growth (N > N₀). If your count decreases, that’s decay and you’d typically use a half-life / decay model instead.