Why Time Units Matter in an Exponential Model
An exponential model can describe an amount that grows or decays at a constant relative rate. A common form is
where \(A_0\) is the amount at time zero, \(t\) is elapsed time, and \(k\) is the signed rate constant. A positive \(k\) describes growth; a negative \(k\) describes decay. The numerical value of \(k\) depends on the time unit used for \(t\).
For example, a rate constant stated in \(\text{hour}^{-1}\) must be paired with elapsed time measured in hours. If the time is instead measured in days, the rate constant must also be converted to \(\text{day}^{-1}\). Changing only one of them changes the exponent, and therefore changes the predicted amount.
The Units of the Exponent
The exponential function \(e^x\) takes a dimensionless input. In \(A(t)=A_0e^{kt}\), the exponent \(kt\) therefore cannot have units. If \(t\) is measured in hours, \(k\) must have units of inverse hours. If \(t\) is measured in days, \(k\) must have units of inverse days.
The units cancel in each product. The numerical value of \(kt\), not just its units, must also be preserved when rewriting the same model in another time unit.
How the Rate and Time Convert Together
Suppose the original time is recorded in an old unit, and one new unit is equal to \(c\) old units. For example, if the old unit is an hour and the new unit is a day, then \(c=24\), since one day is 24 hours.
If \(t_{\text{old}}\) is the numerical elapsed time in old units and \(t_{\text{new}}\) is the numerical elapsed time in new units, the same physical interval satisfies
Because one new unit equals \(c\) old units, the numerical time values are related by \(t_{\text{old}}=c\,t_{\text{new}}\). To preserve the exponent, set \(k_{\text{old}}t_{\text{old}}=k_{\text{new}}t_{\text{new}}\). Substituting the time relation gives the rate conversion rule.
Thus, when the new unit is larger than the old unit, its numerical rate constant is larger by the same factor. This may seem counterintuitive at first: a rate “per day” has a larger numerical value than the corresponding rate “per hour,” because a day contains many hours. Meanwhile, the numerical elapsed time in days is smaller than the time in hours by that factor.
Worked Examples
Worked Example: Convert an Hourly Growth Model to Days
A quantity follows \(A(t)=200e^{0.03t}\), where \(t\) is measured in hours and the rate constant is \(0.03\ \text{hour}^{-1}\). Rewrite the model using days, then find the predicted amount after 10 hours in both forms.
Since one day is 24 hours, the time in days is
The rate constant must be multiplied by 24 to express it per day:
The converted model is \(A(t_{\text{day}})=200e^{0.72t_{\text{day}}}\). Ten hours is \(10/24=5/12\) day. In the original model, the exponent is
In the converted model, it is
Both expressions therefore predict \(A=200e^{0.30}\). As a check, converting the elapsed time back gives \((5/12)(24)=10\) hours, and converting the rate back gives \(0.72/24=0.03\ \text{hour}^{-1}\). The original inputs and prediction are recovered.
Worked Example: Convert a Daily Decay Model to Hours
A decaying amount is modeled by \(A(t)=500e^{-0.04t}\), where \(t\) is in days and \(k=-0.04\ \text{day}^{-1}\). Express the model in hours and calculate the amount after 3 days using each time unit.
Since one day contains 24 hours, the rate per hour is the daily rate divided by 24:
The converted model is \(A(t_{\text{hour}})=500e^{-t_{\text{hour}}/600}\). Three days is \(3(24)=72\) hours. Using days, the exponent is
Using hours, it is
Both forms give \(A=500e^{-0.12}\). Check the rate conversion in reverse: \((-1/600)(24)=-0.04\ \text{day}^{-1}\). The negative sign remains negative because changing units does not change whether the model describes decay.
Worked Example: Convert a Daily Rate to Minutes
Suppose \(A(t)=80e^{0.12t}\), with \(t\) measured in days and \(k=0.12\ \text{day}^{-1}\). Rewrite the model using minutes and evaluate it after \(1.5\) days.
One day is 1440 minutes, so the rate per minute is
The time \(1.5\) days is \(1.5(1440)=2160\) minutes. In days, the exponent is
In minutes, it is
The prediction in either unit is \(A=80e^{0.18}\). To check the conversions, \(2160/1440=1.5\) days and \((1/12000)(1440)=0.12\ \text{day}^{-1}\). Both original values are recovered.
A General Justification
The conversion rule follows directly from requiring the same physical elapsed time to produce the same exponent. The following steps make the argument explicit.
If one new unit equals \(c\) old units, then the numerical value in old units is \(c\) times the numerical value in new units: \(t_{\text{old}}=c\,t_{\text{new}}\).
The model’s prediction depends on \(kt\), so preserving the prediction requires \(k_{\text{old}}t_{\text{old}}=k_{\text{new}}t_{\text{new}}\).
Using \(t_{\text{old}}=c\,t_{\text{new}}\), we obtain \(k_{\text{old}}c\,t_{\text{new}}=k_{\text{new}}t_{\text{new}}\). For a nonzero elapsed time, divide by \(t_{\text{new}}\) to get \(k_{\text{new}}=c\,k_{\text{old}}\).
Then \(k_{\text{new}}t_{\text{new}}=(c\,k_{\text{old}})(t_{\text{old}}/c)=k_{\text{old}}t_{\text{old}}\), as required.
The same conclusion can be checked directly in the full model. The converted expression is \(A_0e^{k_{\text{new}}t_{\text{new}}}\). Since the exponent equals \(k_{\text{old}}t_{\text{old}}\), this is \(A_0e^{k_{\text{old}}t_{\text{old}}}\), exactly the original prediction. The initial amount \(A_0\) does not need to be converted because changing the time unit does not change the amount’s unit.
A Common Unit-Conversion Mistake
A frequent error is to convert the elapsed time but leave the rate constant unchanged. For instance, using \(0.03\ \text{hour}^{-1}\) with \(10/24\) days gives the incorrect exponent
This calculation mixes a rate per hour with a time in days. The units do not cancel: the product is not dimensionless. It would produce \(200e^{0.0125}\), which is not the prediction for the original 10-hour interval.
The correction is to convert the rate too: \(0.03\ \text{hour}^{-1}=0.72\ \text{day}^{-1}\). Then
which matches the original exponent \((0.03)(10)=0.30\). A useful check is to write the units beside the numbers before multiplying. If they do not cancel, the time and rate units have not been matched correctly.
A Reliable Conversion Check
When changing the time unit in a model, use a short sequence of checks rather than relying on memory alone:
- Write down the original unit for \(t\) and the reciprocal unit for \(k\).
- Convert the elapsed time to the new unit.
- Convert the rate constant to the reciprocal of that new unit.
- Multiply the converted values and confirm that the exponent is dimensionless.
- Compare the new exponent with the original \(kt\).
If the exponents agree, the exponential factors agree, and the same initial amount gives the same predicted amount. If the exponents differ, recheck the conversion factor and whether it was multiplied or divided in each conversion.
Check Your Understanding
For each question, focus on how the rate and elapsed time must change together.
- A rate is \(0.02\ \text{hour}^{-1}\). What is the equivalent rate in \(\text{day}^{-1}\)?
- A time interval is 36 hours. Express it in days and explain why the corresponding rate must also be converted.
- A model has \(k=-0.06\ \text{day}^{-1}\). What is the equivalent rate in \(\text{hour}^{-1}\)?
- Why must the product \(kt\) be dimensionless in \(A(t)=A_0e^{kt}\)?
- Describe how you could check that a time-unit conversion preserves an exponential model’s prediction.
