Finding the Time to Reach a Growth Target
A growth model is often used to predict an amount at a specified time. In many applications, the question runs in the other direction: when will the amount reach a chosen target? Examples include estimating when a population will reach a specified size or when an account balance will attain a goal under continuous growth.
For exponential growth, the amount changes at a rate proportional to its current size. If the model starts with amount \(P_0\) at time \(t=0\), and its constant relative growth rate is \(k>0\), then its value at elapsed time \(t\) is
Here \(P_0>0\), and the units of \(k\) are the reciprocal of the time units used for \(t\). For example, if \(t\) is measured in years, \(k\) is measured per year. To find the time at which \(P(t)\) reaches a target \(P_{\text{target}}\), we solve this equation for \(t\).
The Target-Time Formula
The target must be compared with the initial amount before using the formula. The ratio inside the logarithm must be positive, and the comparison determines whether the calculated time represents a nonnegative elapsed time. When \(P_{\text{target}}>P_0\), the ratio is greater than \(1\), so its natural logarithm is positive. Dividing by \(k>0\) then gives \(t>0\).
If \(P_{\text{target}}=P_0\), the model is already at the target at the initial time. If \(0<P_{\text{target}}<P_0\), the formula gives a negative value of \(t\), not a future elapsed time: under this growth model, the amount increases from \(P_0\) for \(t\ge0\), so it does not fall to a smaller target. In this case there is no nonnegative time at which the model reaches the target.
Why the Formula Works
Begin with the equation that says the amount at time \(t\) equals the target. Since \(P_0>0\), divide both sides by \(P_0\). Then use the natural logarithm, which is the inverse of the exponential function, to isolate \(t\).
Use \(P_0e^{kt}=P_{\text{target}}\).
Because \(P_0>0\), this gives \(e^{kt}=P_{\text{target}}/P_0\).
Using \(\ln(e^x)=x\), obtain \(kt=\ln(P_{\text{target}}/P_0)\).
Since \(k>0\), division by \(k\) gives \(t=\ln(P_{\text{target}}/P_0)/k\).
This derivation shows that any time satisfying the model equation must be the value in the formula. It also gives a direct check in the other direction: if the formula produces \(t\), then
Thus the formula does reach the target. It is the only such time because \(e^{kt}\) is strictly increasing in \(t\) when \(k>0\). An increasing function cannot take the same target value at two different times.
Worked Examples
Worked Example: A Population Reaches Twice Its Initial Size
A population begins at \(500\) individuals and follows the model \(P(t)=500e^{0.08t}\), with \(t\) in years. Find when it reaches \(1000\) individuals.
First check the target: \(1000>500\), so it is above the initial amount and the model reaches it at a positive elapsed time. Substitute \(P_0=500\), \(P_{\text{target}}=1000\), and \(k=0.08\) per year:
Rounded to three decimal places, the time is \(8.664\) years. Check by substituting the exact formula for \(t\) into the model:
The model therefore reaches the target at the calculated time.
Worked Example: A Quantity Reaches a Specified Amount
A quantity starts at \(1200\) units and grows according to \(P(t)=1200e^{0.06t}\), where \(t\) is measured in months. Find the time at which it reaches \(1800\) units.
The target exceeds the initial value because \(1800>1200\). The ratio is \(1800/1200=1.5\), and \(k=0.06\) per month. Therefore,
Rounded to three decimal places, the time is \(6.758\) months. Verify using the unrounded expression:
The units are consistent: dividing a logarithm, which is dimensionless, by a rate measured per month gives a time measured in months.
Worked Example: Reaching a Target That Is 1.75 Times the Initial Amount
A sample begins with \(80\) milligrams of a substance, and its amount is modeled by \(P(t)=80e^{0.035t}\), where \(t\) is measured in hours. Find when the model predicts \(140\) milligrams.
Since \(140>80\), this is a larger target and a positive time is expected. The target-to-initial ratio is \(140/80=1.75\). Using \(k=0.035\) per hour gives
Rounded to three decimal places, the predicted time is \(15.989\) hours. Substituting the exact expression for the time verifies the target:
The calculation uses the ratio of target to initial amount, not the difference \(140-80\). The exponential model depends on relative growth, so the ratio is what appears inside the logarithm.
Interpreting the Result and Its Conditions
The calculated time answers a specific question: how long after the model's starting time does it take for the model value to equal the target? It is not the calendar time unless the model's starting time is calendar time zero. Here \(t\) is elapsed time from the point at which \(P_0\) is specified.
Before calculating, check all of the following:
- Initial amount: \(P_0\) must be positive so that the ratio \(P_{\text{target}}/P_0\) is defined as required by the model.
- Growth rate: This formula's growth interpretation assumes \(k>0\).
- Target: For a positive future time, require \(P_{\text{target}}>P_0\). If the target equals \(P_0\), the elapsed time is zero.
- Units: The time unit in the answer must match the time unit used with \(k\). A rate per month gives an answer in months.
The strict increase of the model matters in applications. It means a target above \(P_0\) is reached once, rather than repeatedly, and that a computed solution is not merely one possible time among several. The answer remains a prediction from the chosen model: it is conditional on using the stated constant rate over the time interval considered.
Common Mistake: Multiplying by the Rate Constant
A frequent algebra error is to multiply the logarithm by \(k\) instead of dividing by \(k\). For the model \(P(t)=500e^{0.08t}\), the time to reach \(1000\) is not \(0.08\ln 2\). That incorrect calculation gives
Check this proposed time in the model:
This is far below \(1000\), so the proposed time does not solve the problem. The equation after taking logarithms is \(0.08t=\ln 2\); isolating \(t\) requires dividing both sides by \(0.08\), not multiplying. The corrected result is \(t=\ln 2/0.08\approx8.664\) years, and substitution gives \(500e^{\ln 2}=1000\).
Why the Target Check Comes First
The algebraic expression alone does not tell the whole story. For example, if \(P_0=500\), \(k=0.08\) per year, and the target is \(400\), the formula would produce
This negative value does not describe a future elapsed time. It indicates that the model's value was \(400\) at a time before the stated initial time, if the exponential formula is extended backward. For \(t\ge0\), the model starts at \(500\) and increases, so it cannot reach \(400\). Checking whether the target exceeds the initial amount prevents a negative mathematical solution from being mistaken for a future growth time.
Check Your Understanding
Use the exponential growth model and the target-time formula to answer these questions. State whether each target is reachable at a nonnegative time before calculating.
- If \(P_0=300\), \(k=0.05\) per day, and \(P_{\text{target}}=600\), what is the elapsed time to three decimal places?
- If \(P_0=900\) and the target is \(900\), what is the elapsed time?
- For \(P_0=75\), \(k=0.02\) per hour, and \(P_{\text{target}}=60\), does the model reach the target for \(t\ge0\)? Explain what the formula's sign indicates.
- Why must a rate measured per month be paired with time measured in months in \(P(t)=P_0e^{kt}\)?
- How can substitution into \(P(t)=P_0e^{kt}\) check a proposed time to reach a target?
