From a Fixed-Period Factor to a Continuous Rate
A discrete growth or decay model describes change by applying a fixed multiplier once per time interval. A continuous exponential model describes change through a rate constant \(k\). These descriptions use different quantities: a multiplier records the total change over an interval, while \(k\) describes the continuous relative rate.
The conversion between them is logarithmic. It depends on both the multiplier and the length of the interval over which that multiplier applies. The units of the interval must therefore be stated along with the result.
What Is Being Matched?
Let \(P(t)>0\) be a quantity described by the continuous model \(\frac{dP}{dt}=kP\). Its exponential solution is \(P(t)=P(0)e^{kt}\). Over an interval of length \(\Delta t\), the model therefore gives the ratio
To match a specified discrete multiplier \(a\), we require this ratio to equal \(a\). The conversion finds the \(k\) that makes the continuous model reproduce that multiplier across this interval. It does not, by itself, establish how the quantity changed within the interval; that requires adopting the continuous exponential model.
Deriving the Conversion
The derivation uses the exponential solution and the defining property of the natural logarithm. Since \(a>0\), taking its natural logarithm is valid. Since \(\Delta t>0\), division by the interval length is valid.
Over an interval of length \(\Delta t\), the exponential model multiplies the amount by \(e^{k\Delta t}\).
Set the continuous factor equal to the discrete multiplier: \(e^{k\Delta t}=a\).
Using \(\ln(e^x)=x\), obtain \(k\Delta t=\ln(a)\).
Divide by \(\Delta t\) to get \(k=\frac{\ln(a)}{\Delta t}\).
This also proves that the conversion works: substituting the resulting value of \(k\) gives
Thus the continuous model returns exactly the specified factor over the specified interval. The conditions matter: a positive multiplier is necessary because the exponential factor \(e^{k\Delta t}\) is always positive. A zero multiplier cannot be produced by any finite \(k\); a negative multiplier cannot describe the multiplicative change of a positive amount in this model.
Reading a Percentage Change as a Factor
A percentage change must first be expressed as a multiplier. If the amount increases by a proportion \(p\) during the interval, then the factor is \(a=1+p\). If it decreases by a proportion \(r\), then the factor is \(a=1-r\). For example, a 12% increase means multiplication by \(1.12\), while a 12% decrease means multiplication by \(0.88\).
After converting to a factor, use \(k=\frac{\ln(a)}{\Delta t}\). For \(a>1\), \(\ln(a)>0\), so \(k>0\) and the continuous model grows. For \(0<a<1\), \(\ln(a)<0\), so \(k<0\) and the model decays. If \(a=1\), then \(k=0\), corresponding to no change over the interval.
Worked Conversions
Worked Example: A 21% Increase Over Two Years
A quantity increases by 21% over two years. Find the continuous rate constant in inverse years. If the initial amount is 500 units, check the amount predicted after two years.
A 21% increase gives the multiplier
The interval is \(\Delta t=2\) years, so
Since \(1.21=(1.1)^2\), \(\ln(1.21)=2\ln(1.1)\). Therefore
For an initial amount of 500 units, the continuous model predicts after two years
The check reproduces the original factor: \(605/500=1.21\), a 21% increase. The continuous rate constant is approximately \(0.095310\) per year; it is not the same quantity as the 21% two-year increase.
Worked Example: A 4% Monthly Decrease
A quantity decreases by 4% each month. Find the continuous rate constant in inverse months and check it using an initial amount of 250 units.
A 4% decrease leaves 96% of the amount, so the monthly multiplier is
Here \(\Delta t=1\) month. Thus
The negative sign indicates decay. Starting from 250 units, the continuous model gives after one month
The recovered multiplier is \(240/250=0.96\), which verifies the 4% decrease. The rate constant has units of inverse months because the interval was measured in months.
Worked Example: A Factor of 1.5 Over Four Days
A quantity is multiplied by 1.5 over four days. Find the continuous rate constant in inverse days. Check the conversion for an initial amount of 80 units.
The factor is \(a=1.5\), and \(\Delta t=4\) days. Therefore
Using the unrounded expression for \(k\), the amount after four days is
The ratio \(120/80=1.5\) verifies the conversion. The rounded rate is for reporting; the endpoint check uses the exact expression \(\ln(1.5)/4\).
Converting Back to a Discrete Multiplier
The same relationship can be solved for the multiplier. If the continuous rate constant is \(k\) and the interval length is \(\Delta t\), then the continuous model's factor over that interval is
If a percentage change is wanted, subtract 1 from \(a\): the proportion of change is \(a-1\), and the percentage change is \(100(a-1)\%\). A negative result means a decrease.
Worked Example: A Continuous Rate Over Two Weeks
Suppose \(k=-0.03\text{ week}^{-1}\). Find the multiplier and percentage change over two weeks, then check the result by recovering \(k\).
For \(\Delta t=2\) weeks,
The proportion of change is \(a-1\approx 0.941765-1=-0.058235\), so the percentage change is approximately \(-5.8235\%\), or a decrease of about \(5.8235\%\) over two weeks.
To check, use the unrounded factor \(e^{-0.06}\):
This recovers the given rate constant.
A Common Mistake: Dividing the Percentage by Time
A tempting shortcut is to divide the percentage change by the interval length and call the result the continuous rate constant. That is generally incorrect because the exponential model accumulates a factor \(e^{k\Delta t}\), not a factor obtained by adding the same percentage in equal pieces.
For instance, a 21% increase over two years has multiplier \(1.21\). The incorrect calculation \(21\%/2=10.5\%\) per year treats the total percentage as if it were a continuous rate. Applying the correct conversion gives
not \(0.105\text{ year}^{-1}\). The endpoint check exposes the error: \(e^{(0.105)(2)}=e^{0.21}\approx 1.233678\), which is not the required factor \(1.21\). By contrast, \(e^{2k}=e^{\ln(1.21)}=1.21\). Always convert the percentage to a factor, take its natural logarithm, and then divide by the interval length.
Why the Conversion Is Useful
Discrete data are often reported as percentage changes per period, while a differential equation uses a continuous rate constant. The formula puts those descriptions into a common form, provided the continuous exponential model is appropriate. It also makes the time unit explicit: a rate per month and a rate per year are different numerical descriptions unless the time unit is converted consistently.
The conversion matches the factor over the stated interval. It should not be mistaken for a claim that every process with that observed factor truly follows continuous exponential change at all intermediate times. A model is an assumption about the pattern of change, and the calculated \(k\) is the rate constant for the continuous exponential model that matches the given interval factor.
Check Your Understanding
For each question, state the units of the rate constant and show how you would check the conversion.
- A quantity increases by 8% over one month. What multiplier should be used before applying the logarithmic conversion?
- A quantity is multiplied by \(0.75\) over five years. Write the continuous rate constant in terms of a natural logarithm, and determine its sign.
- Why must a multiplier \(a\) be positive when converting it to a finite continuous rate constant?
- If \(k=0.04\text{ day}^{-1}\), what expression gives the multiplier over three days?
- A student uses \((a-1)/\Delta t\) as the continuous rate constant. Explain why this does not generally reproduce the factor \(a\).
