Why Use a Log-Linear Plot?
Exponential models describe quantities whose rate of change is proportional to their current amount. In the model \(P(t)=P_0e^{kt}\), \(P_0\) is the amount at time \(t=0\), and \(k\) is the constant relative rate. When \(k>0\), the model describes growth; when \(k<0\), it describes decay.
The model is exponential in \(t\), so its graph as \(P\) versus \(t\) is generally curved. Taking the natural logarithm of the amount converts the model into a linear relationship. This is useful when we want to represent exponential data on a straight-line plot and interpret its slope and intercept.
Deriving the Linear Form
Start with the exponential model \(P(t)=P_0e^{kt}\). To take a natural logarithm, the amount must be positive. In the usual growth-and-decay setting, \(P_0>0\), and since \(e^{kt}>0\) for every real \(t\), the model gives \(P(t)>0\). We may therefore apply \(\ln\) to both sides.
The third line uses the product rule for logarithms, \(\ln(ab)=\ln(a)+\ln(b)\) for positive \(a\) and \(b\). The last line uses \(\ln(e^x)=x\). Thus the transformed model has the form of a line, \(y=a+bt\), with \(y=\ln(P(t))\), \(a=\ln(P_0)\), and \(b=k\).
This is not a different growth model. It is the same model written in transformed coordinates. The vertical coordinate changes from \(P(t)\) to \(\ln(P(t))\); time remains \(t\).
What the Slope and Intercept Mean
In the line \(\ln(P(t))=\ln(P_0)+kt\), the slope measures how quickly the natural logarithm of the amount changes per unit of time. It is the exponential rate constant \(k\), not the rate of change of the amount \(P\) itself at a particular time. Its units are inverse time. For example, if \(t\) is measured in years, \(k\) has units of \(\text{year}^{-1}\).
The vertical intercept is the value of \(\ln(P(t))\) at \(t=0\). Substituting \(t=0\) gives \(\ln(P(0))=\ln(P_0)\). It is therefore the logarithm of the initial amount, not the initial amount itself. To recover the initial amount from the intercept, use the exponential function.
In practice, logarithms are taken of the numerical amount values recorded in a fixed unit, such as grams throughout a dataset. Changing the amount unit changes the numerical intercept and the numerical value of \(P_0\) in that unit. It does not change \(k\), provided the time unit stays the same.
Worked Examples
Worked Example: Transforming a Given Exponential Model
Suppose an amount, measured in milligrams, is modeled by \(P(t)=120e^{0.08t}\), where \(t\) is measured in days. Write the log-linear form and identify the slope and intercept.
Here \(P_0=120\) and \(k=0.08\) per day. Apply the logarithm transformation:
The slope of the plot of \(\ln(P(t))\) against \(t\) is \(0.08\ \text{day}^{-1}\), and the intercept is \(\ln(120)\), approximately \(4.7875\). The intercept is not 120; exponentiating it recovers the initial amount:
Check the transformed equation by exponentiating both sides. Since \(e^{\ln(120)+0.08t}=e^{\ln(120)}e^{0.08t}\), this gives \(P(t)=120e^{0.08t}\), the stated model.
Worked Example: Transforming a Small Measurement Dataset
Suppose measurements in milligrams are recorded at the times shown. Transform the amounts by taking natural logarithms, then identify the line’s slope and intercept.
| Time \(t\) (days) | Measured amount \(P(t)\) (mg) | \(\ln(P(t))\) |
|---|---|---|
| 0 | 50 | \(\ln(50)\approx3.9120\) |
| 1 | 100 | \(\ln(100)\approx4.6052\) |
| 2 | 200 | \(\ln(200)\approx5.2983\) |
For each row, the transformed value is the natural logarithm of the amount. For instance, at \(t=1\), \(\ln(P(1))=\ln(100)\approx4.6052\). The exact logarithmic values can also be written in terms of \(\ln(2)\):
These points lie on the line \(\ln(P(t))=\ln(50)+t\ln(2)\). Its slope is \(\ln(2)\approx0.6931\ \text{day}^{-1}\), and its intercept is \(\ln(50)\approx3.9120\). Exponentiating the intercept gives \(e^{\ln(50)}=50\ \text{mg}\), so the corresponding exponential model is \(P(t)=50e^{(\ln 2)t}\).
Check the model at each measured time. At \(t=0\), \(P(0)=50e^0=50\). At \(t=1\), \(P(1)=50e^{\ln2}=50(2)=100\). At \(t=2\), \(P(2)=50e^{2\ln2}=50(4)=200\). Each predicted amount matches its measurement.
Worked Example: Recovering the Initial Amount from a Line
A log-linear plot, with time measured in hours and amounts recorded in grams, has the line \(\ln(P(t))=5.2983-0.3466t\). Interpret its slope and intercept, and write the corresponding exponential model.
The slope is \(-0.3466\ \text{hour}^{-1}\), so the rate constant is \(k=-0.3466\ \text{hour}^{-1}\). The intercept is \(5.2983\), which represents \(\ln(P_0)\). Exponentiate to recover \(P_0\):
Thus the model is approximately \(P(t)=200.0e^{-0.3466t}\), with \(t\) in hours. Check by transforming this model back into a line:
This reproduces the given line, to the displayed precision. The negative slope corresponds to a negative exponential rate constant.
Why the Transformation Works
The transformation works because logarithms turn multiplication into addition and cancel exponentials. The exponential model combines the initial amount and time-dependent factor by multiplication: \(P(t)=P_0e^{kt}\). Taking the logarithm separates these contributions into an additive constant and a term proportional to time.
Write \(P(t)=P_0e^{kt}\), with \(P_0>0\), so the amount is positive.
Apply \(\ln\) to both sides, giving \(\ln(P(t))=\ln(P_0e^{kt})\).
Use \(\ln(ab)=\ln(a)+\ln(b)\) to obtain \(\ln(P(t))=\ln(P_0)+\ln(e^{kt})\).
Since \(\ln(e^{kt})=kt\), the result is the line \(\ln(P(t))=\ln(P_0)+kt\).
The reverse steps also explain how to return from the line to the exponential model. If a line is written as \(y=a+bt\), with \(y=\ln(P(t))\), then \(k=b\), \(\ln(P_0)=a\), and \(P_0=e^a\). Exponentiating gives \(P(t)=e^{a+bt}=e^ae^{bt}=P_0e^{kt}\).
Common Mistakes and Practical Care
A frequent error is to apply the logarithm to a product incorrectly. The logarithm of a product is a sum; it is not the product of the logarithms.
Another mistake is to interpret the intercept as \(P_0\). The intercept is \(\ln(P_0)\); exponentiating it is necessary to obtain \(P_0\). Likewise, the slope is \(k\), not the amount \(P(t)\) and not the derivative \(P'(t)\). The model’s derivative is \(P'(t)=kP(t)\), so the amount’s rate of change depends on its current value, whereas the log-linear slope remains the constant \(k\).
The natural logarithm is defined only for positive inputs. A zero or negative recorded amount cannot be transformed by \(\ln\) as written. Also keep time units consistent: a slope expressed per day cannot be used unchanged when time is entered in hours. These conditions are part of interpreting the transformed line correctly.
Summary
For a positive exponential model \(P(t)=P_0e^{kt}\), taking natural logarithms gives \(\ln(P(t))=\ln(P_0)+kt\). A plot of \(\ln(P(t))\) against time therefore has slope \(k\) and intercept \(\ln(P_0)\). Exponentiating the intercept recovers \(P_0\), and exponentiating the whole line recovers the original exponential model.
Check Your Understanding
Answer each question using the relationship between an exponential model and its log-linear form.
- For \(P(t)=30e^{0.12t}\), what are the slope and vertical intercept of the plot of \(\ln(P(t))\) against \(t\)?
- A log-linear plot has intercept \(2\). What is the model’s initial amount?
- If time is measured in minutes, what are the units of the slope \(k\)?
- Why must \(P(t)\) be positive to use the natural-log transformation?
- Correct the calculation \(\ln(4e^{3t})=\ln(4)\cdot3t\), and state the resulting slope and intercept.
