Use doubling time or half-life
Doubling every three hours
A culture holds 100 cells at \(t = 0\), and the number doubles every 3 hours. The orange bars stand 3 hours apart, and the label ×2 beside each bar is the factor from the bar before it.
Move the slider for the first bar. The heights change, but the factor stays 2: the doubling time does not depend on where the 3 hours begin.
The model \(P = 100e^{kt}\) fits the same curve. In it the factor over 3 hours is \(e^{3k}\), so \(e^{3k} = 2\).
The natural logarithm gives \(3k = \ln 2\), so \(k = \frac{\ln 2}{3} \approx 0.231\) per hour.
A factor over one time span
Move the slider for the time between the bars, and the factor beside each bar changes. Move the slider for the first bar. The factor stays the same, because it depends only on the time between the bars.
Press the button for a doubling time. The factor is 2, and the time between the bars is the doubling time \(T\). Press the button for a half-life. The curve falls, the factor is \(\frac12\), and the time between the bars is the half-life \(H\).
In both cases one time span multiplies the amount by \(e^{kT}\), and this factor fixes \(k\).
Notation
A doubling time \(T\) means that the amount is multiplied by 2 over every time span of length \(T\).
\[y(t + T) = 2\,y(t)\]In the model \(y = y_0e^{kt}\), the factor over a span \(T\) is \(e^{kT}\). The start \(t\) cancels, so the factor does not depend on where the span begins.
\[\frac{y(t + T)}{y(t)} = \frac{y_0e^{kt}e^{kT}}{y_0e^{kt}} = e^{kT}\]Set the factor equal to 2, take the natural logarithm of both sides, and divide by \(T\).
\[\begin{aligned} e^{kT} &= 2 \\ kT &= \ln 2 \\ k &= \frac{\ln 2}{T} \end{aligned}\]For a half-life \(H\), the factor is \(\frac12\). Then \(k\) is negative, because \(\ln\frac12 = -\ln 2\).
\[e^{kH} = \frac12 \;\Rightarrow\; k = \frac{\ln\frac12}{H} = -\frac{\ln 2}{H}\]For any factor \(r\) over a span \(T\), the same steps give \(k\). A drop of 20% per year leaves 80%, so \(r = 0.8\).
\[e^{kT} = r \;\Rightarrow\; k = \frac{\ln r}{T}\]To go the other way, divide \(\ln 2\) by the absolute value of \(k\). A positive \(k\) gives a doubling time, and a negative \(k\) gives a half-life.
\[T = \frac{\ln 2}{|k|}\]
When it fails
Suppose the same quantity is added in every time span, so that the bars stand on the red dashed straight line. The bars are \(y_0\), \(2y_0\), \(3y_0\) and \(4y_0\) tall, and the factors between them are 2, 1.5 and 1.33.
The factor changes, so there is no single doubling time. One exists only when the rate is proportional to the amount.
Example 1: A half-life of eight days
Iodine-131 has a half-life of 8 days. Find the constant \(k\) in \(I = I_0e^{kt}\), where \(t\) is in days. Then find the fraction that remains after 20 days.
Over 8 days the amount is multiplied by \(\frac12\), as the orange bars show. In the model the factor over 8 days is \(e^{8k}\).
\[e^{8k} = \frac12\]Take the natural logarithm of both sides and divide by 8. The constant is negative, because the amount decays.
\[\begin{aligned} 8k &= \ln\tfrac12 = -\ln 2 \\ k &= -\frac{\ln 2}{8} \\ &\approx -0.0866\ \text{per day} \end{aligned}\]The fraction that remains after 20 days is \(\frac{I(20)}{I_0} = e^{20k}\), the height of the orange point.
\[e^{20k} = e^{-\frac{20}{8}\ln 2} = 2^{-2.5} \approx 0.177\]About 17.7% remains. Twenty days is \(\frac{20}{8} = 2.5\) half-lives, so this is the factor \(\frac12\) applied 2.5 times.
Example 2: A drop of 20% each year
The amount of a pollutant in a lake drops by 20% each year. Find the constant \(k\) in \(Q = Q_0e^{kt}\), where \(t\) is in years. Then find the half-life.
A drop of 20% leaves 80%, so each year the amount is multiplied by 0.8. In the model the factor over one year is \(e^{k}\).
A tempting wrong move: A drop of 20% means that \(k = -0.2\).
\[k = -0.2\]With \(k = -0.2\) the factor is \(e^{-0.2} \approx 0.819\) per year, a drop of 18.1%, not 20%. After 3 years the red dashed curve is at 0.549, above the orange bar at 0.512.
Set the one-year factor equal to 0.8 and take the natural logarithm.
\[e^{k} = 0.8 \;\Rightarrow\; k = \ln 0.8 \approx -0.223\]For the half-life \(H\), the factor over \(H\) years is \(\frac12\). The curve meets the dashed orange line at height \(\frac12\) when \(t = H\).
\[\begin{aligned} e^{kH} &= \frac12 \\ H &= \frac{\ln\frac12}{\ln 0.8} = \frac{\ln 2}{-\ln 0.8} \\ &\approx 3.1\ \text{years} \end{aligned}\]
Practice
An investment grows continuously at 4% per year, so \(A = A_0e^{0.04t}\). Find the doubling time.
Doubling means that the factor over \(T\) years is 2. Move the slider for \(T\) until the label beside each bar reads ×2.
Solution
The factor over \(T\) years is \(e^{0.04T}\). For doubling, this factor is 2.
\[e^{0.04T} = 2\]Take the natural logarithm of both sides and divide by 0.04.
\[T = \frac{\ln 2}{0.04} \approx 17.3\ \text{years}\]The answer does not contain \(A_0\), so it is the same for any starting amount.
Questions
Questions 1 and 2 use a substance that decays exponentially. The orange bars show its mass in milligrams at equal time steps, with \(t\) in days.
What is the half-life of the substance?
- 5 days
- 10 days
- 32 days
Answer
5 days
By what factor does the mass change from \(t = 5\) to \(t = 15\)?
- \(\frac14\)
- \(\frac12\)
- \(\frac18\)
Answer
\(\frac14\)
A culture doubles every 12 hours. Which value of \(k\), per hour, fits \(y = y_0e^{kt}\)?
- \(\dfrac{\ln 2}{12}\)
- \(\dfrac{12}{\ln 2}\)
- \(\dfrac{2}{12}\)
- \(\dfrac{\ln 12}{2}\)
Answer
\(\dfrac{\ln 2}{12}\)
Summary
You turn a doubling time, half-life or yearly factor into \(k\) with a logarithm, and reverse the steps to find the time from \(k\).
This works because the exponential multiplies the amount by the same factor \(e^{kT}\) over every time span \(T\).
You use it when a problem gives a repeated factor or asks for one.
This is one of four skills in topic 7.8 of Which & Why: AP Calculus.