Growth, Rates, and Doubling Time
An exponential growth model describes a quantity that changes by the same factor over each equal time period. This differs from adding the same amount each period: when a quantity grows exponentially, the amount added typically gets larger as the quantity itself gets larger.
A useful question is how long such a quantity takes to become twice its starting value. This duration is its doubling time. The answer depends on the growth factor and the time unit used in the model. For example, a doubling time calculated in years is not a number of months unless it is converted.
The Growth Model and Its Parameters
For a quantity that grows exponentially, use the model \(A(t)=A_0b^t\). Here \(A_0\) is the amount at time \(t=0\), \(b\) is the growth factor for one time period, and \(t\) is elapsed time measured in those periods. We assume \(A_0>0\) and, for growth, \(b>1\).
The growth factor \(b\) and the per-period growth rate \(r\) describe the same change in two different ways. If \(r\) is written as a decimal, then \(b=1+r\). If the rate is \(p\%\) per period, then \(r=p/100\), so \(b=1+p/100\). For example, growth of \(8\%\) per year means multiplying by \(1.08\) each year.
The rate applies to the amount at the start of each period, not just to the original amount. With an \(8\%\) per-year rate, an amount \(A\) becomes \(1.08A\) after one year. After another year it becomes \(1.08(1.08A)\), not \(A+2(0.08A)\).
Deriving the Doubling-Time Formula
To find the doubling time, set the model’s amount equal to twice its initial value. Because \(A_0>0\), divide both sides by \(A_0\), then take logarithms. For \(b>1\), \(\ln b>0\), so the resulting time is positive.
At the doubling time \(T\), the model requires \(A_0b^T=2A_0\).
Divide by \(A_0\) to obtain \(b^T=2\).
Taking natural logarithms gives \(T\ln b=\ln 2\), so \(T=\ln 2/\ln b\).
This argument also shows why the formula gives the unique doubling time: when \(b>1\), \(b^t\) increases as \(t\) increases. Thus the equation \(b^T=2\) has just one solution. If \(b=1\), the amount stays constant and never doubles. If \(0<b<1\), the model describes decay rather than growth, so it does not have a positive future doubling time.
Worked Example: Find the Doubling Time from an Annual Growth Rate
A population is modeled by \(A(t)=500(1.08)^t\), where \(t\) is measured in years. Find its doubling time.
The growth factor is \(b=1.08\), so the doubling time in years is
Rounded to two decimal places, the model gives a doubling time of \(9.01\) years. The non-integer result is allowed: the exponential model can be evaluated at fractional values of \(t\), so it can describe the estimated time between whole-year observations.
Check by substituting into the model’s growth factor: \(1.08^{9.00647}=\exp(9.00647\ln(1.08))\approx\exp(0.693147)\approx2\). The amount at this time is therefore approximately \(500(2)=1000\), twice the starting amount of \(500\). As an additional check, \(1.08^9\approx1.99900\), just below \(2\), while \(1.08^{10}\approx2.15892\), above \(2\); a doubling time just over nine years is consistent.
Finding a Growth Rate from Doubling Time
The formula can also be rearranged. If a quantity doubles in \(T\) periods, its growth factor \(b\) must satisfy \(b^T=2\). Taking the \(T\)-th root gives \(b=2^{1/T}\), and subtracting \(1\) gives the per-period rate. This requires a positive doubling time measured in the same periods used for the rate.
Worked Example: Find the Annual Rate from a Six-Year Doubling Time
An investment is modeled with an annual growth factor \(b\), and it doubles in six years. Find the annual percentage growth rate under this model.
Here \(T=6\) years. First calculate the annual growth factor:
Subtract \(1\) to obtain the decimal growth rate, then multiply by \(100\) to express it as a percentage:
Rounded to two decimal places, the rate is approximately \(12.25\%\) per year. Check the result by raising the factor to six years: \(1.122462^6\approx2.00000\). Equivalently, the exact factor \(2^{1/6}\) satisfies \((2^{1/6})^6=2\), as required.
Estimating Growth from Two Data Points
Sometimes the doubling time is not given, but observations provide the amount at two times. If the observations are positive and follow the same exponential model, the ratio of the later amount to the earlier amount removes the initial amount. The time difference, rather than either time by itself, determines how many periods the growth factor has acted.
This formula assumes \(t_2>t_1\) and that \(t\) is measured in the time units for which the per-period rate is sought. Once \(b\) is found, calculate \(r=b-1\). The resulting rate is an exponential-model estimate; two points alone do not establish that real-world growth will continue at that rate.
Worked Example: Estimate an Annual Growth Rate from Two Measurements
A population is \(1200\) at the start of a five-year interval and \(1800\) five years later. Assume an exponential model with \(t\) in years. Estimate the annual growth rate.
The ratio of the later amount to the starting amount is
Over five years, the model’s growth factor is therefore \(1.5\), so the annual factor \(b\) satisfies \(b^5=1.5\). Taking the fifth root gives
Thus the annual decimal rate is \(r=b-1\approx0.084472\), or approximately \(8.4472\%\) per year. Rounded to two decimal places, the estimated rate is \(8.45\%\) per year.
Check by substituting back: \(b^5=(1.5^{1/5})^5=1.5\). The model therefore predicts \(A(5)=1200(1.5)=1800\), matching the later observation.
Keep the Time Units Attached to the Answer
The exponent \(t\) counts periods, so the unit used for \(t\) determines the interpretation of both the rate and doubling time. If \(t\) is in years, then \(b\) is an annual factor, \(r\) is an annual rate, and the formula returns a doubling time in years. If \(t\) is in months, the factor and rate are monthly, and the doubling time is in months.
For example, a doubling time of \(18\) months is \(18/12=1.5\) years. But the rate associated with that doubling time, if found using \(T=18\), is a per-month rate. To find a rate per year instead, the time input must be \(1.5\) years, with the model’s growth factor interpreted as annual. Do not attach a new unit to the number without changing the model consistently.
A useful check is to state the unit in words alongside every rate and time. “\(0.08\)” is incomplete on its own; “a decimal rate of \(0.08\) per year,” or “\(8\%\) per year,” specifies the period over which the factor applies.
A Common Mistake: Confusing the Rate with the Factor
A growth rate of \(8\%\) per period is not a growth factor of \(0.08\). The rate is \(r=0.08\), while the factor is \(b=1+r=1.08\). Using \(0.08\) as the base would model a quantity that shrinks to \(8\%\) of its previous value each period, not one that grows by \(8\%\).
For instance, the incorrect substitution \(T=\ln 2/\ln(0.08)\) gives a negative number, since \(\ln(0.08)<0\). A negative result cannot represent the future doubling time of a growing quantity. The correction is to convert the rate to a factor first:
Always identify whether the information gives a factor or a rate before using a formula. Then check that the resulting time unit matches the model.
When These Calculations Are Useful
Doubling time provides a compact way to compare exponential growth rates. A larger growth factor produces a shorter doubling time; a smaller growth factor above \(1\) produces a longer one. Conversely, a shorter positive doubling time implies a larger per-period factor, since \(b=2^{1/T}\).
These calculations are useful for interpreting repeated percentage growth in settings such as population estimates, savings, and other quantities modeled by exponential growth. The result describes what the model predicts under its assumptions. It is not a guarantee that the same rate will persist: the data or situation may change, and an exponential model may no longer be appropriate.
To solve a problem, first write down the period unit and decide whether the known information is a growth factor, a rate, a doubling time, or a pair of measurements. Convert a percentage rate to a factor when needed, select the corresponding formula, and finish by checking the result in the original model.
Check Your Understanding
Use the exponential growth model and include units in your answers.
- A quantity grows by \(5\%\) per year. What growth factor should be used in its yearly model?
- Write the formula for the doubling time when the growth factor is \(b>1\). What unit does the answer have?
- A quantity doubles in \(10\) years. Write an expression for its annual decimal growth rate.
- Measurements are \(400\) at \(t=0\) years and \(500\) at \(t=4\) years. Write an expression for the annual growth factor under an exponential model.
- Why is \(0.06\), rather than \(1.06\), not the factor for growth of \(6\%\) per period?
