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Growth and Decay Models · Tutorial 1400 of 1495

Computing Growth from a Known Doubling Time

Use the doubling-time model to calculate and interpret exponential growth at specified elapsed times, including fractional doublings.

Intermediate 9 min read

What You'll Learn

  • Use a known doubling time and initial amount to calculate a later amount.
  • Convert elapsed time into a possibly fractional number of doublings.
  • Interpret fractional doublings as exponential growth factors.
  • Calculate and report estimates with appropriate units and rounding.
  • Recognize why rounding the number of doublings to a whole number gives an incorrect prediction.

Using a Known Doubling Time

A doubling time describes how long an exponentially growing quantity takes to become twice as large. If the doubling time is known, we can use it to predict the quantity after a specified elapsed time, even when that time is not a whole number of doubling periods.

The key is to use the fractional number of doubling times that have elapsed. Exponential growth is continuous in the model: an amount does not need to wait for a complete doubling period before changing.

Definition: The doubling time \(T_d\) is the positive length of time required for a quantity following an exponential growth model to double. If the initial amount is \(P_0\), the amount after elapsed time \(T_d\) is \(2P_0\).

We assume throughout that the quantity follows a constant-rate exponential growth model, that \(P_0>0\), and that \(T_d>0\). The elapsed time \(t\) and the doubling time must be expressed in the same units.

The Doubling-Time Prediction Formula

For a known doubling time, the amount after elapsed time \(t\) is found from

$$ P(t)=P_0\,2^{t/T_d}. $$

Here \(t/T_d\) is the number of doubling periods represented by the elapsed time. It need not be an integer. For example, \(t/T_d=2.5\) means that two and a half doubling times have elapsed, so the growth factor is \(2^{2.5}\), not \(2^2\) or \(2^3\).

This formula is consistent with the exponential solution \(P(t)=P_0e^{kt}\). A doubling time means \(P(T_d)=2P_0\), so \(P_0e^{kT_d}=2P_0\), and therefore \(k=\ln(2)/T_d\). Substituting this value into the exponential solution gives

$$ P(t)=P_0e^{(\ln 2)t/T_d} =P_0\,2^{t/T_d}. $$

This also explains why fractional doubling periods work. The exponent \(t/T_d\) changes continuously with \(t\), and the exponential function assigns a growth factor to every real exponent—not only to whole numbers.

Useful interpretation: After \(n\) doubling times, the model predicts a factor of \(2^n\). For a fractional value such as \(n=1.5\), the factor is \(2^{1.5}=2\sqrt{2}\). A fractional doubling is a fractional exponent in the model, not a claim that the amount changes in sudden, countable jumps.

A Reliable Calculation Process

To make a prediction, identify the initial amount, the doubling time, and the requested elapsed time. Then check the units, calculate the ratio \(t/T_d\), and use that ratio as the exponent on 2.

1
Record the known quantities.
Write down \(P_0\), \(T_d\), and \(t\), including their units.
2
Match the time units.
Express \(t\) and \(T_d\) in the same units before dividing.
3
Find the number of doubling times.
Calculate \(t/T_d\), retaining a fractional result when needed.
4
Calculate and interpret.
Evaluate \(P_0\,2^{t/T_d}\), include the amount's units, and round only at the final step if an approximation is requested.

Worked Predictions

Worked Example: A culture after two and a half doubling times

A culture begins with \(1000\) cells and has a doubling time of \(5\) hours. Predict the amount after \(12.5\) hours.

Step 1: Identify the values. The initial amount is \(P_0=1000\) cells, the doubling time is \(T_d=5\) hours, and the elapsed time is \(t=12.5\) hours. The time units already match.

Step 2: Find the number of doubling times.

$$\frac{t}{T_d}=\frac{12.5\text{ hours}}{5\text{ hours}}=2.5.$$

The units cancel, leaving a dimensionless exponent.

Step 3: Apply the formula.

$$P(12.5)=1000\cdot 2^{2.5}=1000\cdot 2^2\cdot 2^{0.5}=1000\cdot4\sqrt{2}=4000\sqrt{2}\text{ cells}.$$

Using \(\sqrt{2}\approx1.41421356\),

$$P(12.5)\approx4000(1.41421356)=5656.85424\text{ cells}.$$

Rounded to the nearest whole cell, the model predicts approximately \(5657\) cells.

Check: The exact result is \(4000\sqrt{2}\). One doubling time earlier, at \(7.5\) hours, the formula gives \(1000\cdot2^{1.5}=2000\sqrt{2}\). Doubling that amount gives \(4000\sqrt{2}\), the calculated result after a further \(5\) hours.

Worked Example: A quantity after three complete doubling times

A growing quantity is initially \(320\) milligrams and doubles every \(6\) days. Find the model prediction after \(18\) days.

Step 1: Identify the values. Here \(P_0=320\) milligrams, \(T_d=6\) days, and \(t=18\) days.

Step 2: Calculate the number of doubling times.

$$\frac{t}{T_d}=\frac{18\text{ days}}{6\text{ days}}=3.$$

Step 3: Calculate the predicted amount.

$$P(18)=320\cdot2^3=320\cdot8=2560\text{ milligrams}.$$

Check: Three successive doublings give \(320\to640\to1280\to2560\) milligrams. The result agrees with the formula, and the elapsed time is exactly three intervals of \(6\) days.

Worked Example: A quarter of a doubling time

A population model starts at \(750\) units and has a doubling time of \(12\) days. Estimate the amount after \(3\) days, to one decimal place.

Step 1: Find the fraction of a doubling time.

$$\frac{t}{T_d}=\frac{3\text{ days}}{12\text{ days}}=\frac14.$$

Step 2: Use the fractional exponent.

$$P(3)=750\cdot2^{1/4}=750\sqrt[4]{2}\text{ units}.$$

Since \(\sqrt[4]{2}\approx1.189207115\),

$$P(3)\approx750(1.189207115)=891.90533625\text{ units}.$$

To one decimal place, the prediction is \(891.9\) units.

Check: The calculated growth factor is \(2^{1/4}\). Applying this same factor four times gives \((2^{1/4})^4=2\), so four intervals of \(3\) days make one \(12\)-day doubling time, as required.

Why the Fractional Exponent Matters

Suppose a doubling time is \(T_d\). An elapsed time of \(mT_d\), where \(m\) is a positive integer, gives \(m\) complete doublings and a factor \(2^m\). But elapsed times are not generally exact multiples of \(T_d\). The same model applies when the ratio is a fraction or any other positive real number.

For example, if \(t=T_d/2\), then the predicted factor is \(2^{1/2}=\sqrt{2}\), not 1 and not 2. If \(t=3T_d/2\), the factor is \(2^{3/2}=2\sqrt{2}\). In both cases, the exponential model gives a specific amount between the values at the neighboring whole doubling times.

The units of \(t/T_d\) cancel, so the exponent is a pure number. The predicted amount has the same units as \(P_0\). If the model represents a count of organisms or people, its mathematical output may be noninteger; a reported count estimate is usually rounded to a whole individual, while the unrounded value remains useful in the calculation.

Common Pitfall: Rounding the Number of Doublings

A frequent error is to round \(t/T_d\) to the nearest whole number before using the formula. That replaces the requested elapsed time with a different one.

For the first example, \(12.5/5=2.5\). Rounding \(2.5\) up to \(3\) and then calculating \(1000\cdot2^3=8000\) cells would predict the amount after \(15\) hours, not after \(12.5\) hours. Rounding down to \(2\) would give \(1000\cdot2^2=4000\) cells, the amount after \(10\) hours. Neither is the requested prediction.

Correction: Keep the full value of \(t/T_d\) as the exponent. In this example, use \(1000\cdot2^{2.5}=4000\sqrt{2}\approx5656.9\) cells, and round only the final amount if needed.

Interpreting the Prediction

A doubling-time calculation is a consequence of an exponential model, not a guarantee that a real-world quantity will follow that model indefinitely. The prediction assumes the same constant relative growth pattern continues throughout the stated interval. If conditions change, or if the quantity does not exhibit that pattern, the calculated value may not describe what actually occurs.

It is also important to distinguish the model's continuous progression from a sequence of observed measurements. A doubling time specifies a multiplicative change over a time interval; it does not mean that the quantity remains unchanged between doublings and then suddenly doubles. The exponential formula provides the model's predicted amount at every elapsed time in the interval.

When reporting a result, state the amount, its units, and whether it is exact or approximate. Keep intermediate values unrounded where practical. In particular, retain a fractional exponent through the calculation rather than approximating it prematurely.

Check Your Understanding

Use \(P(t)=P_0\,2^{t/T_d}\) for each prediction, and keep time units consistent.

  1. A quantity starts at \(400\) units and doubles every \(4\) hours. What does the model predict after \(10\) hours? Express the exact growth factor before approximating.
  2. A culture starts at \(900\) cells and has a doubling time of \(3\) days. How many doubling times have elapsed after \(12\) days, and what amount does the model predict?
  3. What growth factor corresponds to \(t/T_d=0.5\)? Explain why this is not the same as rounding to zero or one doubling.
  4. A model has \(P_0=600\) units and \(T_d=8\) days. Find the predicted amount after \(2\) days, rounded to one decimal place.
  5. Why must \(t\) and \(T_d\) be expressed in the same units before calculating \(t/T_d\)?

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