Comparing Exponential Models with Different Initial Amounts

Comparing Models with the Same Rate Constant

Two quantities can follow exponential models at the same relative rate while starting at different amounts. Their predictions then change together: the ratio of the amounts stays fixed, even though the absolute difference between them may grow or shrink.

This distinction matters when comparing populations, concentrations, or other quantities. A fixed percentage relationship does not mean a fixed numerical gap. To find when two predictions are a specified amount apart, first identify their initial gap, then account for how the shared exponential factor changes it.

Setting Up the Comparison

Let the two models be

where \(A_0\) and \(B_0\) are the initial amounts at \(t=0\), \(k\) is the shared rate constant, and \(t\) is elapsed time. The time unit used for \(t\) must match the unit in which \(k\) is measured. For example, if \(k\) is in years\(^{-1}\), then \(t\) must be in years.

In an amount model, we normally take \(A_0,B_0\geq 0\). The sign of \(k\) describes whether the common exponential factor increases, decreases, or remains constant as time passes: \(k>0\) gives growth, \(k<0\) gives decay, and \(k=0\) gives no change.

Factoring out the shared exponential factor gives the key comparison:

Since \(e^{kt}>0\) for every real \(t\), the sign of \(A(t)-B(t)\) is the same as the sign of \(A_0-B_0\). Thus the models never cross: if one begins above the other, it remains above it. Their absolute gap is the initial gap multiplied by the same factor \(e^{kt}\) that scales either model.

Why the Ratio Stays Constant

When \(B_0\neq 0\), the ratio can also be simplified:

The common factor cancels, so the ratio is independent of time. If both initial amounts are positive, this says that one model is always the same multiple of the other. It does not say the difference is constant: the ratio compares amounts proportionally, while the difference measures their separation in the units of the amounts.

These conclusions follow directly from the factored difference and ratio. For the gap, \(|A_0-B_0|\) is a fixed nonnegative number, so its time dependence comes entirely from \(e^{kt}\). If \(k>0\), \(e^{kt}\) increases with \(t\); if \(k<0\), it decreases; and if \(k=0\), it equals \(1\). In particular, when \(A_0=B_0\), the difference formula gives \(A(t)-B(t)=0\) at all times: the models coincide.

Finding When the Gap Reaches a Target

Suppose the initial amounts are different and we want the models to be separated by a specified positive amount \(d\). The equation to solve is

Dividing by the positive initial gap and taking the natural logarithm gives the candidate time

This formula only provides a valid elapsed time if the target is reachable with \(t\geq0\). When \(k>0\), the gap starts at \(|A_0-B_0|\) and grows, so a target \(d\) is reached at a nonnegative time precisely when \(d\geq |A_0-B_0|\). When \(k<0\), the gap shrinks, so it is reached at a finite nonnegative time precisely when \(0<d\leq |A_0-B_0|\). The gap approaches zero during decay but does not equal zero at any finite time if it starts positive. When \(k=0\), the gap is constant: the target is reached at every time if \(d=|A_0-B_0|\), and otherwise it is never reached.

Worked Examples

A Common Mistake: Treating the Gap as Linear

A tempting but incorrect approach is to assume that the gap changes by a fixed amount per unit time. For the growing models \(120e^{0.05t}\) and \(80e^{0.05t}\), this might lead to the wrong calculation

This treats \(0.05\), which is a relative rate constant, as though it were a fixed number of grams added to the gap each year. But the gap is not \(40+0.05t\). Factoring the actual models shows that it is \(40e^{0.05t}\). The correct calculation is therefore

The substitution check gives \(40e^{0.05(20\ln(3/2))}=60\), confirming the corrected answer. The key distinction is that the common rate constant controls proportional change, not a fixed absolute increment.

How to Choose and Check a Method

When comparing two models with the same \(k\), begin with the quantity the question asks about. For a ratio, divide the models and cancel the shared factor. For a gap, subtract and factor out the shared factor. If a target gap is given, check reachability before interpreting a logarithmic solution as elapsed time.

Why the Comparison Is Useful

Many questions about two exponential predictions can be answered without calculating each amount separately. If the question asks whether the ordering changes, the factored signed difference shows it does not. If it asks for a proportional comparison, the ratio gives the answer immediately. If it asks when the numerical separation reaches a target, the gap formula reduces the problem to one logarithmic equation.

These conclusions depend on the models sharing the same rate constant. If their rates differ, the common exponential factor cannot be canceled or factored out in this way. For the shared-rate case, however, the comparison is especially direct: one common multiplier governs both quantities, their ratio is fixed, and their initial gap is multiplied by that same factor.