Why Doubling Time and Half-Life Are Useful
A quantity that changes at a constant proportional rate follows an exponential model. In this model, the same elapsed time always produces the same multiplicative change, regardless of the quantity’s starting value. That makes it possible to describe growth by its doubling time, or decay by its half-life.
These times are not additional assumptions about how a quantity changes. They can be derived directly from the model \(y=y_0e^{kt}\). Once the relationship is derived, it can also be reversed: a measured doubling time or half-life determines the rate constant \(k\).
Deriving the Doubling-Time Formula
For a growing quantity, the doubling time \(T_d\) is the positive elapsed time required for the quantity to become twice its value at the starting time. Starting at \(t=0\), this means \(y(T_d)=2y_0\). Substitute the exponential model and solve:
Since \(y_0>0\), divide both sides by \(y_0\), then take the natural logarithm. For growth, \(k>0\), so the resulting time is positive.
The initial amount does not appear in this formula. For example, under the same constant proportional growth rate, a quantity starting at 10 and one starting at 100 have the same doubling time. The time depends on \(k\), not on \(y_0\).
Deriving the Half-Life Formula
For a decaying quantity, the half-life \(T_{1/2}\) is the positive elapsed time required for the quantity to become half its value at the starting time. Thus \(y(T_{1/2})=\frac12y_0\). Substituting into the model gives:
Divide by \(y_0>0\) and take the natural logarithm. Since \(\ln(\frac12)=-\ln 2\), the equation becomes \(kT_{1/2}=-\ln 2\). Decay requires \(k<0\), so solving for time gives a positive result.
The minus sign matters. A half-life is elapsed time, so it cannot be negative. Writing \(\ln 2/k\) when \(k<0\) gives a negative value and therefore does not give the half-life.
Why the Formulas Do Not Depend on the Starting Time
The same time interval doubles or halves the quantity even if measurement begins at a later time \(t\), not only at \(t=0\). To see why, compare the model’s values at \(t+T\) and \(t\):
The starting time cancels. Thus a doubling interval satisfies \(e^{kT}=2\), and a halving interval satisfies \(e^{kT}=\frac12\). These are precisely the equations used to derive the formulas. This also explains why the times are constant: for a fixed \(k\), the multiplicative change over an interval of length \(T\) is independent of the amount present at the interval’s start.
Calculating the Rate Constant from a Time Measurement
The formulas can be rearranged to find the rate constant. If a growing quantity doubles in \(T_d\) units of time, then \(k=\ln 2/T_d\). If a decaying quantity halves in \(T_{1/2}\) units of time, then \(k=-\ln 2/T_{1/2}\). The negative sign in the half-life expression indicates decay.
These expressions require a positive measured time. A doubling time describes forward growth, so it requires \(k>0\). A half-life describes forward decay, so it requires \(k<0\). If \(k=0\), then \(y(t)=y_0\) is constant and has neither a finite doubling time nor a finite half-life.
Worked Examples
Worked Example: Find a Doubling Time from the Rate Constant
A population follows \(y(t)=y_0e^{0.12t}\), where \(t\) is measured in hours. Find its doubling time and check the result.
Here \(k=0.12\ \text{hour}^{-1}\), so \(k>0\) and the doubling-time formula applies:
To three decimal places, \(T_d=5.776\) hours. Check by substituting this rounded time into the model’s growth factor:
The small difference from 2 is due to rounding the time. Using the unrounded value \(T_d=\ln 2/0.12\) gives \(e^{0.12T_d}=e^{\ln 2}=2\) exactly. The calculated time is consistent with doubling.
Worked Example: Find the Rate Constant from a Doubling Time
A quantity doubles every 6 hours. Find \(k\), including its units, and verify that the model doubles over 6 hours.
Since the quantity grows, use \(k=\ln 2/T_d\), with \(T_d=6\) hours:
To six decimal places, \(k=0.115525\ \text{hour}^{-1}\). The unit is reciprocal hours because the numerator is dimensionless and the denominator is measured in hours. Check the growth factor after 6 hours using the unrounded value of \(k\):
Thus the model predicts exactly a doubling over the stated interval.
Worked Example: Find a Half-Life from a Decay Rate
A decaying quantity follows \(y(t)=y_0e^{-0.08t}\), where \(t\) is measured in days. Find the half-life to three decimal places and check it using the rounded time.
Here \(k=-0.08\ \text{day}^{-1}\). Because \(k<0\), use \(T_{1/2}=\ln 2/(-k)\):
To three decimal places, the half-life is \(8.664\) days. The rounded time gives the exponent \((-0.08)(8.664)=-0.69312\). Check the model’s remaining fraction:
This is approximately one-half, as expected; the small difference from \(0.5\) results from rounding the time. With the unrounded half-life, the exponent is \(-\ln 2\), so the remaining fraction is \(e^{-\ln 2}=\frac12\) exactly.
Worked Example: Find a Decay Rate from a Half-Life
A substance has a half-life of 12 days. Find its proportional rate constant and check the model after 12 days.
For a half-life \(T_{1/2}=12\) days, the rate constant is negative:
To six decimal places, \(k=-0.057762\ \text{day}^{-1}\). Using the exact expression for the check avoids any rounding discrepancy:
Therefore the model predicts that half the initial amount remains after 12 days.
Common Pitfall: Losing the Sign in Decay
A frequent error is to use \(T_{1/2}=\ln 2/k\) for a negative decay constant. For example, if \(k=-0.08\ \text{day}^{-1}\), that incorrect calculation gives:
A negative elapsed time cannot be the forward half-life. The equation \(e^{kT_{1/2}}=\frac12\) instead gives \(kT_{1/2}=-\ln 2\), so the correction is:
Always use the sign condition as a check: \(k>0\) for a finite forward doubling time, and \(k<0\) for a finite forward half-life. The resulting elapsed time should be positive.
Summary
For the constant-rate exponential model \(y(t)=y_0e^{kt}\), the doubling and halving formulas come from setting the model’s multiplicative change equal to 2 or \(\frac12\). They can be used in either direction: find the characteristic time from \(k\), or find \(k\) from the characteristic time.
In every calculation, keep the time units consistent, include reciprocal-time units for \(k\), and check the result by substituting it into the exponential model.
Check Your Understanding
Use the exponential model and the sign conditions to answer these questions. Include units where appropriate.
- A quantity follows \(y(t)=y_0e^{0.05t}\), with \(t\) in minutes. What is its doubling time?
- A growing quantity doubles every 10 hours. What is its proportional rate constant, in reciprocal hours?
- A model has \(k=-0.04\ \text{day}^{-1}\). Find its half-life and state the appropriate units.
- A decaying quantity has a half-life of 5 years. Find \(k\), including its sign and units.
- Why does the doubling time depend on \(k\) but not on \(y_0\)?
