From a Situation to a Model
An exponential model is useful when a quantity changes by the same percentage over equal time intervals, or when a continuous rate of change is specified. In applications, the main work is not just calculating with a formula: it is deciding which formula matches the situation, identifying what its parameters mean, and checking that the result makes sense.
This workshop brings together several modeling choices. A fixed percentage change per time interval is represented with a power of a growth or decay factor. A continuous rate is represented with an exponential involving \(e\). If the unknown is the time or exponent, logarithms allow us to solve for it. The time unit, amount unit, and interpretation of the parameters must remain consistent throughout.
Choosing the Exponential Representation
Suppose \(Q(t)\) is the amount of a quantity at time \(t\), and \(Q_0\) is its amount at \(t=0\). If the amount changes by a fixed percentage \(r\) during each time interval, the factor for one interval is \(b=1+r\) for growth, or \(b=1-r\) for decay. Here \(r\) is written as a decimal, and the time variable must be measured in the same units as the interval.
In this form, \(t\) counts the number of intervals. The factor \(b\) is greater than \(1\) for growth and between \(0\) and \(1\) for decay. For instance, a \(12\%\) increase per year gives \(b=1+0.12=1.12\), with \(t\) measured in years.
For a continuous model, the standard representation is \(Q(t)=Q_0e^{kt}\), where \(k\) is the continuous rate per unit of time. A positive \(k\) describes growth; a negative \(k\) describes decay. The units of \(k\) are the reciprocal of the time unit, so that \(kt\) is dimensionless.
These forms describe different assumptions. A discrete factor such as \(1.12\) applies once per stated interval. A continuous rate \(k\) applies continuously. Do not switch between them without converting or fitting the model appropriately.
Fitting a Model to Data
If the initial amount and percentage change are known, they directly determine \(Q_0\) and \(b\). When two data points are supplied instead, the values can determine both the starting scale and the factor, provided the measured amounts are positive and the times are distinct.
To see why the discrete fitting rule works, start with \(Q(t)=Q_0b^t\). Dividing the two model equations cancels the unknown initial amount. The resulting equation can be solved for \(b\), and then either data point determines \(Q_0\).
The ratio \(Q_2/Q_1\) is dimensionless, as a growth factor must be. If the later measurement is smaller, the fitted \(b\) is less than \(1\), indicating decay. The continuous formula follows in the same way from \(Q(t)=Q_0e^{kt}\): dividing the measurements gives \(Q_2/Q_1=e^{k(t_2-t_1)}\), so taking natural logarithms gives the stated value of \(k\).
Worked Example: Fit Growth Data and Find a Threshold Time
Worked Example: Fit Growth Data and Find a Threshold Time
A culture contains \(500\) cells at time \(t=0\) hours and \(864\) cells after \(3\) hours. Assume a constant percentage increase per hour. Fit a model, predict the amount after \(2\) hours, and find when the culture reaches \(1000\) cells.
Use \(Q(t)=Q_0b^t\). The initial value gives \(Q_0=500\). The two data points give \(Q(3)/Q(0)=864/500=1.728\), so \(b^3=1.728\). Since \(1.2^3=1.728\), the hourly factor is \(b=1.2\). Thus the fitted model is \(Q(t)=500(1.2)^t\), with \(t\) in hours and \(Q(t)\) in cells.
At \(t=2\), \(Q(2)=500(1.2)^2=500(1.44)=720\) cells. The model predicts \(720\) cells after \(2\) hours.
For the threshold, set the model equal to \(1000\): \(500(1.2)^t=1000\). Divide by \(500\) to obtain \((1.2)^t=2\). Taking natural logarithms gives \(t\ln(1.2)=\ln 2\), and therefore \(t=\ln(2)/\ln(1.2)\approx3.8018\) hours. To check, \(500(1.2)^{3.8018}\approx500(2)=1000\) cells. The units are hours because \(t\) was measured in hours.
Solving for Time or Amount
When an amount is known and time is unknown, isolate the exponential term first. For a discrete model \(Q(t)=Q_0b^t\), the amount ratio \(Q(t)/Q_0\) must be positive. Taking logarithms then turns the exponent into a factor that can be divided out.
The restriction \(b\ne1\) matters: if \(b=1\), the model is constant, so a different amount is never reached. The continuous model has the parallel solution \(t=\ln(Q(t)/Q_0)/k\), provided \(k\ne0\) and the target-to-initial ratio is positive. A computed time should also fit the context: for example, a negative time may indicate that the target was reached before the chosen starting time.
Worked Example: Use Half-Life to Find Amount and Time
Worked Example: Use Half-Life to Find Amount and Time
A sample has an initial mass of \(640\) grams and a half-life of \(8\) days. Find the mass after \(20\) days, and find when the mass reaches \(100\) grams.
Each \(8\)-day interval multiplies the mass by \(1/2\), so the discrete decay model is \(M(t)=640(1/2)^{t/8}\), with \(t\) in days and \(M(t)\) in grams. After \(20\) days, \(t/8=20/8=2.5\). Thus \(M(20)=640(1/2)^{2.5}=640/(4\sqrt{2})\approx113.14\) grams.
For \(100\) grams, solve \(640(1/2)^{t/8}=100\). Dividing by \(640\) gives \((1/2)^{t/8}=100/640=5/32\). Taking natural logarithms gives \((t/8)\ln(1/2)=\ln(5/32)\). Hence \(t=8\ln(5/32)/\ln(1/2)\approx21.42\) days.
Check by substitution: \(640(1/2)^{21.42/8}\approx640(0.15625)\approx100\) grams. Because the target is less than the initial mass, a time greater than zero is consistent with decay.
Worked Example: Model Continuous Growth
Worked Example: Model Continuous Growth
An account begins with \( \$1000\) and earns interest continuously at a rate of \(5\%\) per year. Find its balance after \(6\) years and the time required to reach \( \$1500\), assuming the rate remains constant and no money is added or withdrawn.
The continuous rate is \(k=0.05\) per year, so the model is \(A(t)=1000e^{0.05t}\), where \(t\) is in years and \(A(t)\) is in dollars. After \(6\) years, \(A(6)=1000e^{0.05(6)}=1000e^{0.3}\approx \$1349.86\).
To find when the balance reaches \( \$1500\), solve \(1000e^{0.05t}=1500\). Dividing by \(1000\) gives \(e^{0.05t}=1.5\). Taking natural logarithms gives \(0.05t=\ln(1.5)\), so \(t=\ln(1.5)/0.05\approx8.11\) years.
For a check, \(1000e^{0.05(8.11)}\approx1000e^{0.4055}\approx \$1500\). The rate has units of inverse years, so its product with time in years is dimensionless; the output retains the account’s unit, dollars.
Common Pitfall: Treating Percent Changes as Additive
A repeated percentage change is multiplicative, not a fixed amount added each interval. For the culture in the first example, a common incorrect calculation is to treat a \(20\%\) hourly increase over \(3\) hours as a \(60\%\) total increase: \(500(1+0.20\cdot3)=800\). This assumes the increase is calculated from the original \(500\) cells each hour.
Under the stated repeated-percentage assumption, each hour’s increase is calculated from the current amount. The correct calculation is \(500(1.20)^3=500(1.728)=864\) cells, matching the observed value. The incorrect linear calculation gives \(800\), so it fails the data check. Always ask whether a percentage is applied repeatedly to the updated amount or just once to the original amount.
A Practical Modeling Checklist
State what the model predicts and keep its units attached to the quantities.
Use a repeated factor for a fixed percentage per interval, or a continuous rate when the situation specifies continuous change.
Use the initial value and rate, or fit the factor or rate from positive data values at distinct times.
When time is unknown, isolate the exponential term and take logarithms. Report the result in the model’s time units.
Substitute it into the model and confirm that its size, units, and growth or decay direction fit the situation.
Exponential models are useful because a small number of interpretable parameters can describe repeated change and support predictions. Their predictions are only as appropriate as their assumptions: a constant percentage or continuous rate may be a useful approximation over one interval of time but not indefinitely. Check whether the context supports using the fitted model over the time range in question.
Check Your Understanding
For each question, state the model or equation you would use, and include units where appropriate.
- A quantity starts at \(300\) and increases by \(10\%\) per hour. Write a discrete model for its amount after \(t\) hours.
- A population changes from \(200\) to \(450\) over \(3\) years, with a constant percentage change per year. What equation determines the annual factor \(b\)?
- A sample begins at \(80\) grams and follows \(S(t)=80(0.9)^t\), where \(t\) is in days. Write an equation that can be used to find when it reaches \(40\) grams.
- An account follows \(B(t)=500e^{0.03t}\), where \(t\) is measured in years and \(B(t)\) in dollars. What are the units and meaning of \(0.03\)?
- Why does the equation \(Q_0b^t=Q_{\text{target}}\) require \(b\ne1\) to determine a unique time for a target different from \(Q_0\)?
