From a Decay Model to a Threshold Time
A decaying quantity may remain above a specified level for some time and then pass below it. Examples include the amount of a substance remaining in a sample or the number of organisms in a population. The exponential model makes it possible to calculate when the quantity reaches a given threshold.
There is an important distinction between reaching a threshold and being below it. At the exact time the quantity reaches the threshold, the two values are equal. The quantity is strictly below the threshold only at later times. This difference matters whenever a question asks when something “falls below” a limit.
The Exponential Decay Model
Let \(A_0\) be a positive initial amount, let \(k\) be a positive decay constant, and let \(t\) denote time measured from the initial observation. The exponential decay model is
The rate constant \(k\) must have units reciprocal to the units of \(t\), so that the exponent \(kt\) is dimensionless. Because \(k>0\), the amount decreases as time increases. It remains positive for every finite \(t\), even though it approaches zero as \(t\) grows.
Let \(H\) be a specified threshold. To find the time at which the amount equals \(H\), set \(A(t)=H\) and solve for \(t\). A positive threshold can be reached at a nonnegative time only if it is no greater than the initial amount.
In the formula, \(A_0/H\) is a ratio of amounts and therefore has no units. The resulting time \(\tau\) has the same units as \(t\), because \(1/k\) has units of time.
Why the Formula Works
The derivation also clarifies exactly what the answer means. Start with the equality condition \(A(t)=H\). Since the amounts are positive, we can divide by \(A_0\) and take natural logarithms. The logarithm is defined for a positive argument and is the inverse of the exponential function.
Using the model, \(A_0e^{-kt}=H\).
Dividing by \(A_0>0\) gives \(e^{-kt}=H/A_0\). This requires \(H>0\).
Then \(-kt=\ln(H/A_0)\), so \(t=\frac{1}{k}\ln(A_0/H)\). Division by \(k\) is valid because \(k>0\).
For \(0<H<A_0\), the ratio \(A_0/H\) is greater than \(1\), so its logarithm is positive and the equality time is after the initial time. To determine when the quantity is strictly below \(H\), solve the inequality directly:
The last inequality reverses direction when dividing by \(-k\), which is negative. Thus, when \(0<H<A_0\), the equality time \(\tau\) is not itself a time when the quantity is strictly below the threshold: \(A(\tau)=H\), and \(A(t)<H\) exactly when \(t>\tau\).
There is no earliest real time at which the quantity is strictly below \(H\) in this case. If \(t>\tau\), then the time halfway between \(\tau\) and \(t\) is also greater than \(\tau\), and is earlier than \(t\). In practical settings, a question may instead specify allowable observation times, such as whole days; then the first allowable time below the threshold can be identified by checking those times.
Worked Examples
Worked Example: Finding a Threshold Time
A sample contains \(500\) mg of a substance initially and decays with rate constant \(k=0.1\) per hour. When does the model predict that \(100\) mg remains?
Here \(A_0=500\) mg and \(H=100\) mg. First check that the threshold is below the initial amount: \(100<500\). Therefore, a positive equality time is possible. Substitute into the formula:
The reported decimal is rounded to four places. Check by substituting the exact time into the model:
The check gives the specified threshold, and the initial amount \(500\) mg is greater than \(100\) mg, as required for a later downward crossing. The amount is strictly below \(100\) mg for times greater than \(16.0944\) hours, not at the exact equality time.
Worked Example: First Whole-Minute Observation Below a Threshold
A quantity begins at \(240\) units and follows \(A(t)=240e^{-0.3t}\), where \(t\) is measured in minutes. At what time is it \(60\) units, and at which whole-minute observation is it first below \(60\) units?
Because \(0<60<240\), the equality time is
At \(t=\tau\), the amount is exactly \(60\) units. To find the first whole-minute observation strictly below the threshold, check the neighboring whole minutes, \(4\) and \(5\). Using the model:
The approximate values are rounded to four decimal places. The first is above \(60\), and the second is below \(60\). Since the model decreases as time increases, the first whole-minute observation below the threshold is at \(5\) minutes. The equality-time calculation is consistent: \(4<\tau<5\), so the amount has not yet reached \(60\) at minute \(4\), but is below it by minute \(5\).
Worked Example: Checking Whether the Quantity Is Already Below the Threshold
A quantity starts at \(80\) mg and decays according to \(A(t)=80e^{-0.25t}\), with \(t\) measured in hours. When does it reach \(100\) mg, and when is it below \(100\) mg?
First compare the initial amount with the threshold: \(A(0)=80\) mg, which is already less than \(100\) mg. For every \(t\ge0\), \(e^{-0.25t}\le1\), so
Therefore the quantity is already strictly below the threshold at \(t=0\), and remains below it. It never reaches \(100\) mg: the model is decreasing from \(80\) mg, so it cannot later equal a larger amount. Substituting \(t=0\) checks the initial value directly: \(A(0)=80e^0=80\) mg. The threshold-time formula does not apply here, because its condition \(H\le A_0\) is not satisfied.
Classifying the Possible Cases
Before using a logarithm formula, compare the threshold with the initial amount. This check prevents an algebraic answer from being mistaken for a physically possible time.
| Threshold \(H\) | What the model says for \(t\ge0\) |
|---|---|
| \(0<H<A_0\) | The quantity equals \(H\) at \(\tau=\frac{1}{k}\ln(A_0/H)>0\), and is strictly below \(H\) for \(t>\tau\). |
| \(H=A_0\) | The quantity equals \(H\) at \(t=0\), and is strictly below \(H\) for every \(t>0\). |
| \(H>A_0\) | The quantity is already strictly below \(H\) at \(t=0\), and never reaches \(H\). |
| \(H\le0\) | The quantity remains positive, so it never equals or falls below \(H\). |
The final row follows from \(A_0>0\) and \(e^{-kt}>0\): their product is positive at every finite time. A positive decay model can approach zero arbitrarily closely, but it does not reach zero at a finite time.
A Common Mistake: Calling the Equality Time “Below”
A frequent error is to calculate the time at which the amount equals the threshold and then report that same time as when it is strictly below. For the first example, one might correctly find \(\tau=10\ln(5)\) hours and then write \(A(\tau)<100\) mg. That last statement is false. Substitution gives \(A(\tau)=100\) mg exactly.
The correction is to state both facts: the amount reaches \(100\) mg at \(\tau\), and it is strictly below \(100\) mg for \(t>\tau\). If observations occur only at specified times, evaluate the model at those times to find the first observation that satisfies the strict inequality. Also check the initial comparison before calculating: if the amount starts below the threshold, it is already below at time zero and there may be no future equality time.
Why the Distinction Matters
Threshold questions often guide decisions: a process may need to be stopped once an amount is below a limit, or an observation may be scheduled after a quantity is expected to pass a specified level. An equality time gives a precise boundary. It does not, on its own, identify a later observation time or an earliest real time satisfying a strict inequality.
A reliable solution therefore has three parts: compare the initial amount with the threshold, solve the equality equation when the threshold can be reached, and interpret the result using the correct inequality. Substitution into the model checks the equality calculation, while the initial comparison checks that the answer fits the direction of decay.
Check Your Understanding
For each question, assume \(A(t)=A_0e^{-kt}\), with \(A_0>0\), \(k>0\), and \(t\ge0\).
- For \(0<H<A_0\), write the time when \(A(t)=H\). At that time, is \(A(t)\) strictly below \(H\)?
- Why does dividing an inequality by \(-k\) reverse its direction when determining when \(A(t)<H\)?
- If \(H>A_0\), is there a future time when the model reaches \(H\)? What is true at \(t=0\)?
- Can the model reach a threshold \(H=0\) at a finite time? Explain using the model formula.
- If the equality time is \(4.3\) minutes and observations are made only at whole minutes, what model values or comparisons should be checked to identify the first whole-minute observation below the threshold?
