Why two observations can determine a model
An exponential model describes a quantity that is multiplied by the same positive factor over each equal unit of time. Its standard form is \(A(t)=A_0b^t\), where \(A(t)\) is the modeled quantity at time \(t\), \(A_0\) is the value at time \(0\), and \(b\) is the multiplicative factor for one unit of time.
If measurements are available at two different times, their ratio reveals how many times the factor \(b\) has been applied. The time difference matters: observations three time units apart compare \(b^3\), not just \(b\). Once \(b\) is found, either observation can be used to calculate \(A_0\).
Fitting formulas from the observations
Order the times so that \(t_1<t_2\), and suppose \(A_1>0\) and \(A_2>0\). The model must satisfy both equations below. Dividing the second equation by the first cancels the initial value \(A_0\).
The ratio \(A_2/A_1\) compares the observed quantities, while \(t_2-t_1\) counts the time units separating them. Solving for the positive factor \(b\), and then using the first observation to find \(A_0\), gives the fitting formulas.
The formula for \(A_0\) is sometimes written in a form that uses both observations symmetrically. Since \(b^{t_2-t_1}=A_2/A_1\), either observation gives the same initial value:
Why the formulas work
The derivation also establishes uniqueness: any positive exponential model that fits the data is forced to have the same parameters. In detail, dividing the model equation at \(t_2\) by the one at \(t_1\) forces a particular positive value of \(b\). Substituting that value into the equation at \(t_1\) forces a particular value of \(A_0\).
A model fitting the observations must satisfy \(A_0b^{t_1}=A_1\) and \(A_0b^{t_2}=A_2\).
Because \(A_0>0\), division cancels it and gives \(b^{t_2-t_1}=A_2/A_1\). The time separation is positive, and the ratio is positive, so taking the positive \((t_2-t_1)\)-th root gives exactly one positive \(b\).
The first condition gives \(A_0=A_1/b^{t_1}\). Since \(A_1>0\) and \(b>0\), this determines exactly one positive \(A_0\).
Using the factor equation, \(A_0b^{t_2}=(A_1/b^{t_1})b^{t_2}=A_1b^{t_2-t_1}=A_1(A_2/A_1)=A_2\). Thus the calculated model fits both observations.
Worked examples
Worked Example: Finding an exact model from a three-unit separation
Suppose \(A(2)=18\) and \(A(5)=486\). Find \(A_0\) and \(b\), then check both observations.
The time separation is \(5-2=3\). The ratio of the later value to the earlier value is \(486/18=27\). Therefore, \(b^3=27\), so \(b=3\).
Use the observation at \(t=2\) to find the initial value:
The fitted model is \(A(t)=2\cdot 3^t\). Check the first observation: \(A(2)=2\cdot 3^2=2\cdot 9=18\). Check the second: \(A(5)=2\cdot 3^5=2\cdot 243=486\). Both observations are reproduced.
Worked Example: Finding a model when the later observation is smaller
Suppose \(A(1)=80\) and \(A(4)=10\). Find a model that fits both values.
The separation is \(4-1=3\), and the ratio is \(10/80=1/8\). Thus \(b^3=1/8\), so \(b=1/2\). Using the observation at \(t=1\),
The fitted model is \(A(t)=160(1/2)^t\). Check at \(t=1\): \(160(1/2)=80\). Check at \(t=4\): \(160(1/2)^4=160/16=10\). Each prediction matches its observation exactly.
Worked Example: Using logarithms for a non-integer factor
Suppose \(A(2)=12\) and \(A(5)=60\). Find the model. Here the time separation is \(3\), and the ratio is \(60/12=5\), so \(b^3=5\) and \(b=\sqrt[3]{5}\). This root is exact, but logarithms give a practical decimal value.
Taking natural logarithms of \(b^3=5\) gives \(3\ln b=\ln 5\), so \(\ln b=\ln(5)/3\). Using \(\ln 5\approx1.609438\), we obtain \(\ln b\approx0.536479\), and therefore \(b\approx e^{0.536479}\approx1.710\).
Now use \(A(2)=12\) to find the initial value. The exact result is \(A_0=12/(\sqrt[3]{5})^2=12/5^{2/3}\). For a decimal calculation, \(\ln A_0=\ln 12-(2/3)\ln 5\). Using \(\ln 12\approx2.484907\) and \(\ln 5\approx1.609438\), this gives \(\ln A_0\approx2.484907-1.072959=1.411948\), hence \(A_0\approx e^{1.411948}\approx4.104\). Decimal values here are rounded to three decimal places.
The model is \(A(t)=(12/5^{2/3})5^{t/3}\). Check exactly at \(t=2\): \((12/5^{2/3})5^{2/3}=12\). At \(t=5\), \((12/5^{2/3})5^{5/3}=12\cdot5=60\). The exact model fits both observations; the decimals are approximations to its parameters.
Reading the fitted parameters carefully
The initial value \(A_0\) means the model’s value at time zero, not necessarily the first observed value. If the first measurement occurs at \(t_1\ne0\), calculate \(A_0\) by dividing \(A_1\) by \(b^{t_1}\). In the first example, for instance, the first observation was at \(t=2\), while the fitted initial value was \(A_0=2\).
The value of \(b\) depends on the unit used for time. If \(t\) is measured in days, \(b\) is the model’s factor per day. If the same time axis is instead measured in hours, the numerical factor per unit changes. The model and its predictions remain consistent when the time values and factor are expressed using the same time unit.
Two positive observations at distinct times always determine one positive exponential model. That mathematical fact does not establish that the real quantity follows an exponential pattern at other times. With only two observations, the model is chosen to fit those observations; additional data are needed to assess how well it describes the wider pattern.
A common mistake: using the wrong exponent
A frequent error is to use the later-to-earlier ratio as \(b\) itself, or to take a root whose degree is one of the time values rather than their difference. The correct relationship is \(b^{t_2-t_1}=A_2/A_1\), so the root depends on the time separation.
For the observations \(A(2)=18\) and \(A(5)=486\), an incorrect calculation would be \(b=27^{1/5}\), using the later time \(5\) as the root’s degree. But the measurements are three time units apart, so their ratio must equal \(b^3\). The correction is \(b=27^{1/3}=3\). Checking the incorrect factor against the data shows the problem: its predicted ratio over three units would be \((27^{1/5})^3=27^{3/5}\), which is not \(27\). Using the separation in the exponent makes the ratio agree.
Check Your Understanding
Use the two-observation formulas to answer these questions. Check any model you calculate by substituting both observed times.
- For observations \(A(0)=7\) and \(A(2)=63\), what are \(b\) and \(A_0\)?
- For observations \(A(2)=48\) and \(A(5)=6\), what equation does the ratio give for \(b\)?
- Why is the degree of the root in the formula for \(b\) the time separation rather than the later time?
- Can two positive observations at distinct times determine an exponential model with positive parameters? State the condition on the observations.
- Does an exact fit to two observations prove that the model fits other times? Explain briefly.
