Why Time Units Matter in an Exponential Model
An exponential model can describe an amount that grows or decays continuously. A common form is \(A(t)=A_0e^{kt}\), where \(A_0\) is the amount at time \(t=0\), \(k\) is the rate constant, and \(t\) is elapsed time. For growth, \(k\) is positive; for decay, \(k\) is negative.
The numerical value of \(k\) depends on the time unit used. A rate stated per day cannot be multiplied directly by a time stated in hours unless one of them is first converted. Otherwise, the number in the exponent does not represent the intended elapsed time.
This is not a minor formatting issue. Because the product \(kt\) is in an exponent, a unit mismatch can change the predicted amount substantially. The method for repairing it is straightforward: convert the elapsed time to the unit used by \(k\), or convert \(k\) to the unit used by the elapsed time. Either way, the physical duration and model remain unchanged.
What the Units of a Rate Constant Mean
Suppose \(k=0.08\) per year. This means that the numerical rate constant is \(0.08\) when time is measured in years; its units are inverse years. If \(t\) is measured in years, then \(kt\) has units \((1/\text{year})(\text{year})\), so the units cancel. If \(t\) is instead entered as a number of months, the units do not cancel as intended.
The exponential function \(e^x\) takes a dimensionless input \(x\). Thus, a consistent exponential model must use a numerical rate and a numerical time measured in corresponding units. For instance, \(0.08\) per year pairs with \(1.5\) years, while a rate converted to \(0.08/12\) per month pairs with \(18\) months.
These are two ways to express the same calculation, not two different models. The product in the exponent must agree after conversion.
Worked Example: Decay Over 48 Hours with a Rate per Day
Worked Example: Decay Over 48 Hours with a Rate per Day
A substance has an initial amount of \(800\) milligrams and a half-life of \(6\) days. Find the amount remaining after \(48\) hours. First, use the decay rate in per-day units. The half-life formula gives \(k=-\ln(2)/6\) per day.
Convert the elapsed time: \(48\) hours is \(48/24=2\) days. Now both the rate and time use days:
Since \(2^{-1/3}\approx 0.793700526\), the amount is \(800(0.793700526\ldots)=634.960421\ldots\) milligrams, or \(634.96\) milligrams to the nearest hundredth.
Check: The same elapsed time is \(2\) days, so the exponent is \(-(\ln 2)(2/6)=-\ln(2)/3\), as expected for one third of a six-day half-life. The amount is therefore \(800\cdot2^{-1/3}\), consistent with the half-life model.
A unit-mixing error would be to put \(48\) directly into the formula while keeping \(k\) per day. That produces \(800e^{-(\ln 2/6)(48)}=800e^{-8\ln 2}=800/2^8=3.125\) milligrams. This incorrectly treats \(48\) as \(48\) days, rather than \(48\) hours.
Converting the Rate Instead of the Time
Sometimes it is more convenient to retain elapsed time in its original unit. In that case, convert the rate constant. For the half-life in the previous example, \(6\) days is \(144\) hours. The hourly decay constant is therefore \(-\ln(2)/144\) per hour. Pairing it with \(48\) hours gives the same exponent as pairing the daily constant with \(2\) days.
The numerical rate per hour is smaller than the numerical rate per day because one hour is a shorter time interval than one day. The rate constant's units change along with its numerical value.
Why Either Conversion Gives the Same Exponent
Let \(t_{\text{day}}\) be the numerical time in days and \(t_{\text{hour}}\) the numerical time in hours. Since one day is \(24\) hours, \(t_{\text{hour}}=24t_{\text{day}}\). If \(k_{\text{day}}\) is the rate constant per day, then the equivalent rate per hour is \(k_{\text{hour}}=k_{\text{day}}/24\).
This proves that the two unit choices describe the same exponential factor. The factor \(24\) introduced by expressing time in hours is cancelled by dividing the daily rate by \(24\). More generally, changing the time unit changes the numerical rate constant by the reciprocal conversion factor, so their product stays the same.
This gives a useful diagnostic: if a conversion changes the exponent for the same physical duration, a conversion factor has been omitted or applied in the wrong direction.
Worked Example: Growth Over 18 Months with a Rate per Year
Worked Example: Growth Over 18 Months with a Rate per Year
An account has an initial balance of \(250\) dollars and follows the continuous growth model \(A(t)=250e^{0.08t}\), with \(t\) measured in years. Find the modeled balance after \(18\) months.
Convert \(18\) months to years: \(18/12=1.5\) years. Use \(t=1.5\), because the rate \(0.08\) is per year:
Using \(e^{0.12}\approx1.127496852\), the balance is \(250(1.127496852\ldots)=281.874213\ldots\) dollars, or \(\$281.87\) to the nearest cent.
Check by converting the rate: The equivalent rate per month is \(0.08/12=0.006666\ldots\) per month. Using \(18\) months gives exponent \((0.08/12)(18)=0.12\), the same exponent as before. Thus both unit choices give the same modeled balance.
Incorrect calculation: Substituting \(18\) directly into the per-year model gives \(250e^{(0.08)(18)}=250e^{1.44}\approx250(4.220695816)=\$1{,}055.17\). That calculation treats the duration as \(18\) years. It does not represent \(18\) months.
A Practical Repair Process
When a growth or decay calculation seems unexpectedly large or small, check the units before changing the model. Use this sequence:
For example, \(k=0.02\) per hour requires elapsed time in hours if the rate is left unchanged.
Translate phrases such as “three days” into the unit used by the rate, or decide to convert the rate.
Convert either the time or the rate, and keep track of the conversion factor.
Confirm that the rate and time units cancel and that the exponent agrees with an equivalent conversion.
Calculate the modeled amount and round only at the final step, stating the requested precision.
Worked Example: Growth Over Three Days with a Rate per Hour
Worked Example: Growth Over Three Days with a Rate per Hour
A population is modeled by \(P(t)=100e^{0.02t}\), where \(t\) is measured in hours. Find the modeled population after \(3\) days. Convert the duration to hours: \(3(24)=72\) hours. Substitution gives
Using \(e^{1.44}\approx4.220695816\), we obtain \(P\approx100(4.220695816)=422.069582\), or \(422.07\) to the nearest hundredth.
Check by converting the rate: Since one day contains \(24\) hours, the equivalent rate is \(0.02(24)=0.48\) per day. Using \(3\) days gives exponent \((0.48)(3)=1.44\), matching the hourly calculation.
Incorrect calculation: Using \(t=3\) with the rate \(0.02\) per hour gives \(100e^{0.06}\approx100(1.061836547)=106.18\). This treats the three-day duration as three hours. The corrected calculation uses \(72\) hours, or equivalently \(3\) days with a rate of \(0.48\) per day.
Common Pitfalls and a Final Check
The most common error is to notice the number of elapsed units but overlook which unit the rate constant uses. A phrase like “after 48 hours” does not mean the model should automatically receive \(t=48\). If the rate is per day, first convert \(48\) hours to \(2\) days. Similarly, “after 18 months” is not \(t=18\) in a model whose rate is per year.
Another error is to convert in the wrong direction. If time is converted from days to hours, its numerical value is multiplied by \(24\). A rate per day converted to a rate per hour is divided by \(24\), not multiplied by \(24\). The exponent check reveals the difference: the correct conversion leaves \(kt\) unchanged.
Before accepting an answer, ask two questions: “What time unit does this rate use?” and “Do the rate and time produce a dimensionless exponent?” If the units do not cancel, do not evaluate the exponential yet. Repair the mismatch first, then calculate and check using the equivalent unit conversion.
Check Your Understanding
For each question, pay attention to the units in both the rate and the elapsed time.
- A decay constant is \(0.03\) per hour. What elapsed time in hours corresponds to \(2\) days?
- A model uses a growth rate of \(0.06\) per year. Write the exponent for \(9\) months after converting time to years.
- A rate is \(-\ln(2)/10\) per day. Explain why substituting \(48\) for \(t\) to represent \(48\) hours is incorrect.
- If elapsed time is converted from days to hours, how must a rate per day be converted to keep the exponent unchanged?
- A rate is \(0.04\) per hour and elapsed time is \(5\) days. Give one consistent rate-time pair for evaluating the exponent.
