Modeling Pandemic Growth: Find k and Write the Exponential Function
There were 10 people sick in a town during a pandemic, then 3 weeks later there are 75 people sick. If the growth of the infected is modeled using exponential growth, find k, the constant of proportionality. Round to 4 decimal places. Write the function that models the number of sick people in the town A(t), t weeks after the original number.
Already Got the Answer?
Setup:A(t) = A₀ekt with A₀ = 10 and A(3) = 75, so e3k = 7.5.
Result:k = ln(7.5)/3 ≈ 0.6716 and A(t) = 10e0.6716t.
Quick check:10·e0.6716·3 ≈ 10·7.499 ≈ 75 ✓
If your numbers match, the sections on pitfalls, the weekly-factor connection, and the four extension problems are where this page goes further.
Skills This Problem Builds
- Translating "t weeks after the original number" into a model. Because time starts at the original count, the starting value is automatically A₀ = 10, so only one unknown (k) remains.
- Using a logarithm as an "undo" button. The unknown k sits in an exponent, and the natural log is the tool that brings it down to ground level.
- Seeing growth as repeated multiplication. The ratio 75 ÷ 10 = 7.5 is the whole story; the starting value drops out.
- Moving between the two common forms
A₀ektandA₀btusingk = ln b. - Managing rounding. Knowing when to keep full precision and when to round to 4 decimals is a real skill in modeling.
Check Your Foundation
Three quick self-checks. If any feels shaky, review it first and the rest of the page will go much more smoothly:
- What does
ln(ex)equal? (It equalsx. The natural log andexare inverse functions.) - What is the difference between
ln(a/b)andln(a)/b? (They are completely different numbers; see the pitfalls below.) - Can you evaluate
lnandexon your calculator? Make sure you pressln, notlog.
Solution: Method 1 — Solve Directly with the Natural Log
The problem tells us to use exponential growth with a "constant of proportionality" k. That is the standard continuous model, in which the growth rate of the sick population is proportional to the current number of sick people:
Step 1 — Identify the starting value
The problem measures t in weeks after the original number. So at t = 0 there are 10 sick people, which means A₀ = 10. Our model so far is A(t) = 10ekt.
Step 2 — Substitute the second data point
Three weeks later there are 75 people sick, so A(3) = 75:
Step 3 — Isolate the exponential
Divide both sides by 10 so the exponential stands alone:
Notice what this says in plain English: over 3 weeks the outbreak multiplied by 7.5.
Step 4 — Take the natural logarithm of both sides
Since ln(ex) = x, applying ln brings the exponent down:
3k = ln(7.5)
3k ≈ 2.014903
Step 5 — Solve for k and round
Step 6 — Write the model
The units of k are "per week": about 67.16% continuous growth each week. (This is not the same as the population rising by 67% each week. The weekly multiplier is about 1.9574, as the next method shows.)
What the model predicts
A small table makes the pattern concrete. Look at the last column: each week the count is multiplied by almost exactly the same number.
| Week t | A(t) = 10e0.6716t | Ratio to previous week |
|---|---|---|
| 0 | 10.0 | — |
| 1 | 19.6 | × 1.957 |
| 2 | 38.3 | × 1.957 |
| 3 | 75.0 | × 1.957 |
| 4 | 146.8 | × 1.957 |
| 5 | 287.4 | × 1.957 |
Solution: Method 2 — The Per-Week Growth Factor
Method 1 jumps straight to e. This second route builds the model from something more intuitive, "what does the sick count get multiplied by each week?", and only converts to k at the end.
Step 1 — Model with a weekly multiplier b
Write A(t) = 10 · bt, where b is the factor by which the count grows each week.
Step 2 — Use the 3-week data to find b
b³ = 7.5
b = 7.51/3 ≈ 1.95743
So the town's sick population roughly doubles each week (a factor just under 2).
Step 3 — Convert the multiplier into k
The two forms describe the same curve when ek = b, that is, k = ln(b):
This agrees with Method 1, which is no coincidence: ln(7.51/3) = (1/3)·ln(7.5), exactly the formula from before.
b before taking the log. If you round b to 1.96, you get ln(1.96) ≈ 0.6729, which is already off in the third decimal place. Round only at the very end.
Solution: Method 3 — Straightening the Curve with Logarithms
Here is a perspective that statisticians and scientists use constantly: take the log of the data, and an exponential curve becomes a straight line whose slope is k.
Step 1 — Take ln of the model
This is the form y = b + mx with y = ln A, intercept ln 10, and slope k.
Step 2 — Convert both data points
| t (weeks) | A | ln A |
|---|---|---|
| 0 | 10 | 2.302585 |
| 3 | 75 | 4.317488 |
Step 3 — Compute the slope
Same answer, now seen as "rise over run" on a log scale. This is exactly why epidemiologists plot case counts on semi-log graphs: exponential growth shows up as a straight line, and the slope is the growth constant.
The growth constant is k ≈ 0.6716 (per week), and the number of sick people after t weeks is modeled by A(t) = 10e0.6716t.
Verification
A model built from two data points has to reproduce both of them. Test each:
At t = 0:A(0) = 10e0 = 10·1 = 10 ✓
At t = 3:A(3) = 10e0.6716·3 = 10e2.0148 ≈ 10 × 7.499 = 74.99 ≈ 75 ✓
The tiny gap (74.99 versus 75) exists only because k was rounded to 4 decimals. With the unrounded k = 0.671634…, the model gives exactly 75. A gap this small is the signature of correct work, not an error.
Does This Seem Reasonable?
Before trusting any exponential model, it helps to know what range of answers is plausible. The count grew from 10 to 75, a factor of 7.5, in 3 weeks.
- If the population doubled every week, three weeks would give
2³ = 8times the start, or 80. Our actual factor, 7.5, is just a little less than 8, so the weekly multiplier should be just under 2. We found 1.957. ✓ - If k were tiny (say 0.05), the town would barely grow. If k were 2, the factor over 3 weeks would be
e⁶ ≈ 403. Our k ≈ 0.67 sits between those extremes, exactly where a 7.5-fold increase belongs. - Rough log check:
e² ≈ 7.39, and 7.5 is a hair more than that, so3kshould be a hair over 2, givingka hair over 0.667. Indeed 0.6716. ✓
What Trips People Up
Forgetting to divide by the starting value. The equation is 10e3k = 75, so the exponential equals 7.5, not 75. Using 75 gives k = ln(75)/3 ≈ 1.4391, more than twice the correct value. The log is only legal once the exponential stands alone.
Dividing by t inside the log. The 3 multiplies k in the exponent, so you divide the result of the log: k = ln(7.5) ÷ 3. Put the division outside the log, never inside.
Treating the situation as linear. That is the average change of 21.67 people per week, a rate for a straight line, not the continuous growth constant of an exponential. Linear growth adds the same amount each week; exponential growth multiplies by the same factor.
Leaving t out of the exponent. The variable must sit up in the exponent with k: A(t) = 10e0.6716t. Misplacing it gives a straight line, not a curve.
The base of the logarithm must match the base of the exponential. With e in the model you need ln. Pressing log gives log(7.5)/3 ≈ 0.2917, which belongs to a model written with base 10, not e.
The Pattern Behind This
Every "find the constant from two data points" problem in exponential growth is this same formula in disguise. If a quantity starts at A₀ and equals A₁ at time t₁:
With the numbers here: k = ln(75/10)/3 = ln(7.5)/3. Notice that only the ratio A₁/A₀ matters. Going from 10 to 75 and going from 200 to 1500 in 3 weeks produce the identical k, because both multiply by 7.5. That is the defining feature of exponential growth: percentage change per unit of time is constant, no matter how big the population is.
The same structure gives the other forms you will meet:
| Form | Parameter | Here |
|---|---|---|
| A₀ ekt | k = continuous rate | k ≈ 0.6716 |
| A₀ bt | b = ek = factor per unit time | b ≈ 1.9574 |
| A₀ · 2t/T | T = ln 2 / k = doubling time | T ≈ 1.032 weeks |
Beyond the Textbook
- Epidemiology: early in an outbreak, public health agencies estimate the growth rate k from case counts, then compute doubling time and the reproduction number from it. Case curves are plotted on log scales so that this straight-line slope (Method 3) is visible.
- Finance: continuous compounding is
A = Pert. The constant r plays exactly the role of k. - Computing and biology: bacterial cultures, viral spread across a network, and early user growth of a platform all follow the same
A₀ektcurve during their unconstrained phase.
A Brief History
The number e first appeared in 1683, when Jacob Bernoulli asked what happens to compound interest if the compounding is done more and more often: the answer approaches e. Euler later named the constant and made ex central to calculus, and in 1798 Thomas Malthus used geometric (exponential) growth to argue about human populations. Modern epidemic modeling inherits that same idea, which is why a pandemic word problem is really a 300-year-old piece of mathematics.
Try These Variations
These four problems build on each other. Try each on paper first, then reveal the full solution to compare.
Using the model A(t) = 10e^(0.6716t), how many days does it take for the number of sick people to double? Round to the nearest tenth of a day.
Doubling means reaching twice the starting amount, A = 20, but the doubling time does not depend on the starting amount, so we can write the equation as 20 = 10e0.6716t.
Divide by 10: 2 = e0.6716t.
ln 2 = 0.6716t, so t = 0.693147 ÷ 0.6716 ≈ 1.0321 weeks.
1.0321 × 7 ≈ 7.22 days. The sick population doubles about every 7.2 days.
0.6716 × 1.0321 ≈ 0.6931 ≈ ln 2, so 10e0.6931 = 10 × 2 = 20 ✓. This also matches the earlier finding that the weekly multiplier is just under 2.
With A(t) = 10e^(0.6716t), after how many weeks will 500 people be sick? Round to 2 decimal places.
This reverses the unknown: the output is given and the time is unknown. 500 = 10e0.6716t.
Divide by 10: e0.6716t = 50.
0.6716t = ln 50 ≈ 3.912023.
t = 3.912023 ÷ 0.6716 ≈ 5.825, so t ≈ 5.82 weeks (about 41 days).
0.6716 × 5.82 ≈ 3.909 and e3.909 ≈ 49.8, so A ≈ 498, which is 500 up to rounding in t ✓.
Suppose another town also starts with 10 sick people, but its growth constant is only k = 0.3 per week. (a) How many people are sick after 3 weeks? (b) How many weeks does this slower outbreak need to reach 75 sick people, and how much longer is that than the 3 weeks in the original problem? Round to 2 decimal places.
B(t) = 10e0.3t.
B(3) = 10e0.9 = 10 × 2.4596 ≈ 24.6. About 25 people, compared with 75 in the original town.
75 = 10e0.3t, so e0.3t = 7.5.
0.3t = ln 7.5 ≈ 2.0149, so t ≈ 2.0149 ÷ 0.3 ≈ 6.72 weeks.
6.72 − 3 = 3.72, so the slower outbreak takes about 3.72 weeks longer. Notice that the time scales inversely with k: the ratio 0.6716/0.3 ≈ 2.24, and 3 × 2.24 ≈ 6.72. The ratio 7.5 is the same, so the time is inversely proportional to k.
10e0.3 × 6.72 = 10e2.016 ≈ 10 × 7.51 ≈ 75 ✓.
Town A is the town from the original problem: A(t) = 10e^(0.6716t). Town B starts with 4 sick people, and 3 weeks later it has 45 sick people. Assume exponential growth in both towns. (a) Find Town B's growth constant k to 4 decimal places. (b) Which town's outbreak is spreading faster, and by how much per week in k? (c) Town B has fewer cases now, but after how many weeks does it overtake Town A? Round to 2 decimal places.
45 = 4e3k, so e3k = 11.25 and k = ln(11.25) ÷ 3 ≈ 2.420368 ÷ 3 ≈ 0.8068. Town B's model is B(t) = 4e0.8068t.
Compare 0.8068 with 0.6716: Town B is spreading faster, by about 0.1352 per week in k. (Town A's absolute numbers are bigger, but only k measures how fast each outbreak is growing relative to its size.)
Solve 10e0.6716t = 4e0.8068t. Collect the exponentials: divide by 4 and by e0.6716t to get 2.5 = e(0.8068 − 0.6716)t = e0.1352t.
0.1352t = ln 2.5 ≈ 0.9163, so t ≈ 0.9163 ÷ 0.1352 ≈ 6.78. Town B overtakes Town A after about 6.78 weeks.
At t = 6.78: Town A has 10e0.6716 × 6.78 = 10e4.554 ≈ 950, and Town B has 4e0.8068 × 6.78 = 4e5.470 ≈ 950 ✓. The two counts agree to within rounding. The lesson: a larger k always wins in the long run, even against a big head start.
Frequently Asked Questions
2026-10-26