Predict an amount or a time
Read the curve in two directions
An account holds 2000 dollars at \(t = 0\) and earns interest continuously at 5% per year, so after \(t\) years it holds \(A = 2000e^{0.05t}\) dollars.
Drag the orange point along the curve. You choose a time, and the dashed lines read off the amount. At \(t = 10\) the amount is \(2000e^{0.5} \approx 3297\) dollars.
Press the button to ask for a time instead. A dashed orange line then marks a goal of 5000 dollars, and the goal slider moves it up or down.
Where the curve meets the line, a dashed drop to the time axis shows when the account reaches the goal. For 5000 dollars the time is \(20\ln 2.5 \approx 18.3\) years.
Evaluate or solve
Every point of the curve has a time and an amount, so knowing one of them fixes the other. Press a button to choose which one you know.
If you know the time, you substitute it into the formula and evaluate. The dashed lines from the point read off the amount.
If you know the amount, you set the formula equal to it and solve for the time. A logarithm undoes the exponential, and the dashed drop gives the time.
Notation
Write the model with the unit of \(t\) and the unit of \(y\). To find the amount at a known time \(t_1\), substitute it and evaluate.
\[y(t_1) = y_0e^{kt_1}\]To find a time, set the model equal to the target amount \(\cy{y_1}\).
\[y_0e^{kt} = \cy{y_1}\]Divide by \(y_0\) so that the exponential is alone. Then take the natural logarithm and divide by \(k\).
\[e^{kt} = \frac{y_1}{y_0} \;\Rightarrow\; \cy{t_1} = \frac{1}{k}\ln\frac{y_1}{y_0}\]Write the answer with its unit, and check that the time is not negative.
A negative time means that the target was reached before the starting time, not after it.
When it fails
A growing amount never falls below its starting amount for \(t \ge 0\). Press the button to choose a target below the start. The dashed line does not meet the curve to the right of the vertical axis.
The formula gives \(t_1 = \frac{1}{k}\ln\frac{y_1}{y_0}\), which is negative here because \(\frac{y_1}{y_0} < 1\). This time is before the start, so the amount does not reach this target in the future.
Example 1: The value of a car
A car is worth 24000 dollars when new. After \(t\) years its value is \(V = 24000e^{-0.15t}\) dollars. Find its value after 5 years.
The question gives the time, so substitute \(t = 5\).
\[\begin{aligned} V(5) &= 24000e^{-0.15\cdot5} \\ &= 24000e^{-0.75} \end{aligned}\]Evaluate with a calculator, which gives \(e^{-0.75} \approx 0.4724\).
\[V(5) \approx 11337\]After 5 years the car is worth about 11337 dollars, less than at the start because the constant \(-0.15\) is negative.
Example 2: A drug leaves the body
The amount of a drug in the blood is \(A = 120e^{-0.3t}\) milligrams, where \(t\) is in hours. When does the amount reach 20 milligrams?
Set the model equal to the target. The dashed orange line marks 20 milligrams, and the curve meets it once.
\[120e^{-0.3t} = 20\]A tempting wrong move: Take the natural logarithm of both sides. The logarithm cancels the exponential.
\[\begin{aligned} -0.3t &= \ln 20 \\ t &= -\frac{\ln 20}{0.3} \approx -10 \end{aligned}\]This time is negative, before the dose. The faint dashed curve extends the model to earlier times, and the ring at \(t = -10\) is far below it. The logarithm of \(120e^{-0.3t}\) is \(\ln 120 - 0.3t\), and this step lost \(\ln 120\).
Divide both sides by 120 first, so that the exponential is alone.
\[e^{-0.3t} = \frac{20}{120} = \frac16\]Take the natural logarithm, which gives \(-\ln 6\). Dividing by \(-0.3\) cancels the two minus signs, so the time is positive.
\[\begin{aligned} -0.3t &= \ln\frac16 = -\ln 6 \\ t &= \frac{\ln 6}{0.3} \end{aligned}\]The amount reaches 20 milligrams after about 5.97 hours.
\[t \approx 5.97\ \text{hours}\]
Practice
An invasive plant covers \(A = 3e^{0.4t}\) square meters after \(t\) days. When does it cover 30 square meters?
Move the dashed line to 30 and read the time on the horizontal axis. Then write the equation for that point.
Solution
Set the model equal to 30 and divide by 3.
\[3e^{0.4t} = 30 \;\Rightarrow\; e^{0.4t} = 10\]Take the natural logarithm and divide by 0.4.
\[t = \frac{\ln 10}{0.4} \approx 5.76\ \text{days}\]
Questions
Questions 1 and 2 use the number of birds \(N = 900e^{-0.2t}\) in a population, where \(t\) is in years.
The dashed line marks 300 birds. At about what time does the population reach 300?
- About 5.5 years
- 3 years
- About 7.5 years
- 10 years
Answer
About 5.5 years
About how many birds are there at \(t = 10\)?
- About 122
- 300
- 180
- 450
Answer
About 122
A culture has \(P = 40e^{0.35t}\) cells, where \(t\) is in hours. Which expression gives the time when \(P = 200\)?
- \(\dfrac{\ln 5}{0.35}\)
- \(\dfrac{\ln 200}{0.35}\)
- \(\dfrac{5}{0.35}\)
- \(0.35\ln 5\)
Answer
\(\dfrac{\ln 5}{0.35}\)
Summary
You substitute a known time to find an amount, and you divide and take a logarithm to find a time.
This works because the logarithm undoes the exponential once the exponential is alone.
You use it when a model is known and a problem asks for an amount or a time.
This is one of four skills in topic 7.8 of Which & Why: AP Calculus.