The complete guide to Q = Q0 e^(rt): half-life, doubling time, radioactive decay, population growth, Newton's law of cooling, logistic growth, and compound interest — with six fully worked application problems.
An exponential growth model describes a quantity that changes at a rate proportional to its current value. If a population grows by 5% each year, the number of new members each year depends on how large the population already is — which is the defining characteristic of exponential, not linear, growth.
The general exponential model is written Q(t) = Q0 times e^(rt), where Q0 is the initial amount at time t = 0, r is the continuous rate of change (positive for growth, negative for decay), and t is time in whatever unit the problem specifies.
The Exponential Growth/Decay Model
Q(t) = Q0 times e^(rt)
Q0 = initial quantity (at t = 0)
r = continuous growth rate (r greater than 0: growth, r less than 0: decay)
t = time elapsed
e = Euler's number, approximately 2.71828
This single formula governs a remarkable range of real-world phenomena: bacteria colonies, radioactive isotopes, investment accounts, atmospheric pressure, drug concentration in the bloodstream, and population projections. The key is identifying Q0, r, and t from the problem context.
Exponential growth model
Q(t) = Q0 * e^(rt)
r greater than 0 for growth, r less than 0 for decay. Q0 = initial amount.
Doubling time
t_double = ln(2) / r approx 0.693 / r
Time for quantity to double. Rule of 72: years approx 72 divided by annual rate %.
Half-life
t_half = ln(2) / |r| approx 0.693 / |r|
Time for quantity to halve. r here is the magnitude of the decay rate.
Half-life form
Q(t) = Q0 * (1/2)^(t/h)
h = half-life period. Equivalent to e-based form with r = negative ln(2) divided by h.
Newton's Law of Cooling
T(t) = T_env + (T0 - T_env) * e^(kt)
k less than 0 for cooling. T_env = ambient temp. T0 = initial object temp.
Compound interest (discrete)
A = P(1 + r/n)^(nt)
P = principal, r = annual rate (decimal), n = compounds per year, t = years.
Continuous compounding
A = P * e^(rt)
Limit of discrete formula as n approaches infinity. Always gives the highest return.
Logistic growth model
P(t) = M / (1 + A * e^(-kt))
M = carrying capacity. k greater than 0. A = (M minus P0) divided by P0.
Base conversion
a^x = e^(x * ln(a))
Converts any base-a exponential to e-based form. Useful for calculus.
Finding r from two points
r = ln(Q2 / Q1) / (t2 - t1)
Divide the two equations to eliminate Q0, then take the natural log.
The critical distinction between linear and exponential growth is how much the quantity increases each step. Linear growth adds a constant amount. Exponential growth multiplies by a constant factor. This difference becomes dramatic over long time spans.
| Property | Linear Growth | Exponential Growth |
|---|---|---|
| Formula | Q(t) = Q0 + rt | Q(t) = Q0 * e^(rt) |
| Change per step | Constant amount added | Constant factor multiplied |
| Graph shape | Straight line | Curved (upward or downward) |
| Example | Saving $200/month | Account earning 5% annual interest |
| After 10 steps at rate 2 | Q0 + 20 | Q0 * e^20 approx 485 million times Q0 |
| Identifying from table | Constant differences | Constant ratios |
How to identify exponential from a table of values
Compute ratios of consecutive Q values. If Q(1)/Q(0) = Q(2)/Q(1) = Q(3)/Q(2) (all the same constant), the relationship is exponential. That common ratio equals e^r, so r = ln(ratio). If the differences are constant instead, the model is linear.
Finding growth rate and population at a future time
Problem: A bacteria culture starts with 500 bacteria and doubles every 3 hours. How many bacteria are present after 10 hours? What is the continuous growth rate r?
Step 1 — Identify Q0 and the doubling condition
Q0 = 500
Q(3) = 1000 (doubles at t = 3 hours)
Step 2 — Find the growth rate r
1000 = 500 * e^(r * 3)
2 = e^(3r)
ln(2) = 3r
r = ln(2) / 3 approx 0.6931 / 3 approx 0.2310 per hour
Step 3 — Build the complete model
Q(t) = 500 * e^(0.2310t)
Step 4 — Evaluate at t = 10 hours
Q(10) = 500 * e^(0.2310 * 10) = 500 * e^(2.310)
e^(2.310) approx 10.079
Q(10) approx 500 * 10.079 approx 5,040 bacteria
Sanity check: at t = 3, Q = 500 * 2 = 1000 (doubled). At t = 6, Q = 500 * 4 = 2000 (doubled twice). At t = 9, Q = 500 * 8 = 4000. At t = 10, slightly above 4000. Answer of ~5040 is reasonable.
The doubling time formula tells you how long it takes for a quantity to double under continuous exponential growth. It is derived by setting Q(t) = 2 Q0 and solving for t.
2 Q0 = Q0 * e^(rt_double)
2 = e^(rt_double)
ln(2) = r * t_double
t_double = ln(2) / r approx 0.693 / r
| Growth Rate r (per year) | Doubling Time (exact) | Rule of 72 Estimate |
|---|---|---|
| 1% = 0.01 | ln(2) / 0.01 = 69.3 years | 72 / 1 = 72 years |
| 2% = 0.02 | ln(2) / 0.02 = 34.7 years | 72 / 2 = 36 years |
| 3% = 0.03 | ln(2) / 0.03 = 23.1 years | 72 / 3 = 24 years |
| 6% = 0.06 | ln(2) / 0.06 = 11.6 years | 72 / 6 = 12 years |
| 10% = 0.10 | ln(2) / 0.10 = 6.93 years | 72 / 10 = 7.2 years |
| 23% = 0.23 | ln(2) / 0.23 = 3.01 years | 72 / 23 = 3.1 years |
The Rule of 72 is a mental math shortcut: doubling time in years is approximately 72 divided by the annual interest rate as a percentage. It overestimates slightly for very high rates and underestimates slightly for very low ones.
Carbon-14 dating and the half-life formula
Problem: Carbon-14 has a half-life of 5,730 years. An archaeological artifact is found to contain 30% of its original carbon-14. How old is the artifact?
Step 1 — Set up the model
Q(t) = Q0 * (1/2)^(t/5730)
Alternatively: Q(t) = Q0 * e^(kt) where k = negative ln(2) / 5730 approx negative 0.000121
Step 2 — Apply the given condition (30% remaining)
0.30 * Q0 = Q0 * (1/2)^(t/5730)
0.30 = (1/2)^(t/5730)
Step 3 — Take the natural log of both sides
ln(0.30) = (t/5730) * ln(1/2)
negative 1.2040 = (t/5730) * (negative 0.6931)
t/5730 = 1.2040 / 0.6931 = 1.7373
t = 5730 * 1.7373 approx 9,955 years old
Key Insight: Carbon-14 Dating Range
Carbon-14 is useful for dating artifacts up to about 50,000 years old (roughly 8 to 9 half-lives). Beyond that, so little C-14 remains that measurement error dominates. For older materials, scientists use uranium-238 (half-life 4.47 billion years) or potassium-40 (half-life 1.25 billion years).
Half-life (denoted h) is the time required for a decaying quantity to reduce to half its current value. Like doubling time, it is derived from the core exponential model. The two forms of the half-life equation are fully equivalent.
Form 1 — Half-life base
Q(t) = Q0 * (1/2)^(t/h)
Direct: after t/h half-lives have passed, the quantity is Q0 times (1/2)^n where n is the number of half-lives.
Form 2 — e-base with decay constant
Q(t) = Q0 * e^(negative(ln 2 / h)t)
The decay constant k = negative ln(2)/h. Note that k is negative because the quantity is decreasing.
Converting between the two forms
Start with: Q(t) = Q0 * (1/2)^(t/h)
Rewrite (1/2) = e^(ln(1/2)) = e^(negative ln 2)
So: Q(t) = Q0 * [e^(negative ln 2)]^(t/h) = Q0 * e^(negative(ln 2 / h)t)
Therefore k = negative ln(2) / h
| Isotope | Half-Life | Decay Constant k | Application |
|---|---|---|---|
| Carbon-14 | 5,730 years | approx negative 0.000121 /yr | Archaeological dating |
| Uranium-238 | 4.47 billion years | approx negative 1.55 x 10^-10 /yr | Geological age of rocks |
| Iodine-131 | 8.02 days | approx negative 0.0864 /day | Medical thyroid treatment |
| Radon-222 | 3.82 days | approx negative 0.181 /day | Radiation safety |
| Polonium-210 | 138.4 days | approx negative 0.00501 /day | Research applications |
Compound interest: discrete vs. continuous compounding
Problem: You invest $8,000 at 4.5% annual interest. Compare (a) quarterly compounding and (b) continuous compounding after 20 years. How long does continuous compounding take to double your money?
Part (a) — Quarterly compounding
A = P(1 + r/n)^(nt)
A = 8000(1 + 0.045/4)^(4 * 20)
A = 8000(1.01125)^80
A = 8000 * 2.4380 approx
A approx $19,504 after 20 years
Part (b) — Continuous compounding
A = Pe^(rt) = 8000 * e^(0.045 * 20)
A = 8000 * e^(0.9) = 8000 * 2.4596
A approx $19,677 after 20 years
Continuous earns $173 more than quarterly compounding over 20 years.
Part (c) — Doubling time at continuous 4.5%
t_double = ln(2) / r = 0.6931 / 0.045
t_double approx 15.4 years
Rule of 72: 72 / 4.5 = 16 years (close to the exact answer).
| Compounding | n | Formula | Balance after 20 years |
|---|---|---|---|
| Annually | 1 | 8000(1.045)^20 | $19,316 |
| Semi-annually | 2 | 8000(1.0225)^40 | $19,453 |
| Quarterly | 4 | 8000(1.01125)^80 | $19,504 |
| Monthly | 12 | 8000(1.00375)^240 | $19,557 |
| Daily | 365 | 8000(1+0.045/365)^(7300) | $19,670 |
| Continuously | infinity | 8000 * e^(0.9) | $19,677 |
Finding r from two data points, then projecting forward
Problem: A city had a population of 240,000 in 2005 and 310,000 in 2015. Assuming exponential growth, find the growth rate and predict the population in 2035. When will the population reach 500,000?
Step 1 — Set t = 0 at year 2005, establish model
P(t) = 240000 * e^(rt)
At t = 10 (year 2015): P(10) = 310000
Step 2 — Solve for r
310000 = 240000 * e^(10r)
310000 / 240000 = e^(10r)
1.2917 = e^(10r)
10r = ln(1.2917) = 0.2557
r = 0.02557 approx 2.56% per year (continuous)
Step 3 — Complete model and predict 2035 (t = 30)
P(t) = 240000 * e^(0.02557t)
P(30) = 240000 * e^(0.02557 * 30) = 240000 * e^(0.7671)
e^(0.7671) approx 2.154
P(30) approx 517,000 people in 2035
Step 4 — When does population reach 500,000?
500000 = 240000 * e^(0.02557t)
500000 / 240000 = e^(0.02557t)
ln(2.0833) = 0.02557t
0.7340 = 0.02557t
t = 0.7340 / 0.02557 approx 28.7 years after 2005
Population reaches 500,000 around year 2033 to 2034.
Newton's Law of Cooling states that the rate at which an object cools is proportional to the difference between its temperature and the ambient (surrounding) temperature. This leads to an exponential model for the object's temperature over time.
T(t) = T_env + (T0 minus T_env) * e^(kt) where k is less than 0
T(t) = temperature of the object at time t
T_env = constant ambient temperature
T0 = initial object temperature at t = 0
k = cooling constant (negative, determined from data)
Why k must be negative
If the object is hotter than its surroundings (T0 greater than T_env), then (T0 minus T_env) is positive. For T(t) to decrease toward T_env as t increases, the exponential term e^(kt) must approach 0, which requires k to be negative. The temperature difference decays exponentially.
Newton's Law of Cooling with two stages
Problem: A cup of coffee is brewed at 200°F and placed in a 72°F room. After 8 minutes it has cooled to 170°F. (a) Find the cooling constant k. (b) What is the temperature after 25 minutes? (c) How long until the coffee reaches a comfortable 140°F?
Setup
T(t) = 72 + (200 minus 72) * e^(kt) = 72 + 128 * e^(kt)
Part (a) — Find k using T(8) = 170
170 = 72 + 128 * e^(8k)
98 = 128 * e^(8k)
e^(8k) = 98/128 = 0.7656
8k = ln(0.7656) = negative 0.2674
k = negative 0.03343 per minute
Part (b) — Temperature at t = 25 minutes
T(25) = 72 + 128 * e^(negative 0.03343 * 25)
T(25) = 72 + 128 * e^(negative 0.8358)
T(25) = 72 + 128 * 0.4336
T(25) approx 72 + 55.5 approx 127.5°F
Part (c) — Time to reach 140°F
140 = 72 + 128 * e^(negative 0.03343 t)
68 = 128 * e^(negative 0.03343 t)
e^(negative 0.03343 t) = 68/128 = 0.53125
negative 0.03343 t = ln(0.53125) = negative 0.6325
t = 0.6325 / 0.03343
t approx 18.9 minutes to reach a comfortable 140°F
Pure exponential growth is unrealistic for any real population because resources are limited. The logistic growth model adds a carrying capacity M — the maximum sustainable population — that acts as an upper bound. Growth is fastest when the population is at M/2, then slows as it approaches M.
P(t) = M / (1 + A * e^(negative kt))
M = carrying capacity (maximum population)
k = growth rate constant (positive)
A = (M minus P0) / P0, determined from the initial population P0
As t approaches infinity, P(t) approaches M from below
Behavior of the logistic curve
Applications of logistic growth
Example: A deer population in a wildlife reserve
Carrying capacity M = 1000 deer. Initial population P0 = 100 deer. k = 0.4.
A = (1000 minus 100) / 100 = 900 / 100 = 9
Model: P(t) = 1000 / (1 + 9 * e^(negative 0.4t))
At t = 5: P(5) = 1000 / (1 + 9 * e^(negative 2)) = 1000 / (1 + 9 * 0.1353)
= 1000 / (1 + 1.218) = 1000 / 2.218 approx 451 deer
At t = 10: P(10) = 1000 / (1 + 9 * e^(negative 4)) approx 858 deer
Using radioactive decay with a very long half-life
Problem: Uranium-238 decays to lead-206 with a half-life of 4.47 billion years. A rock sample contains uranium-238 and lead-206 in a 3:1 ratio by mass (meaning 75% of the original U-238 remains as U-238, and 25% has decayed to Pb-206). Estimate the age of the rock.
Step 1 — Interpret the ratio
If 75% of the original U-238 is still present, then Q(t) = 0.75 Q0.
Assumption: all Pb-206 in the sample came from U-238 decay (reasonable for very old rocks with no initial Pb-206).
Step 2 — Set up the decay model
Q(t) = Q0 * (1/2)^(t / 4.47 billion)
0.75 Q0 = Q0 * (1/2)^(t / 4.47)
0.75 = (1/2)^(t / 4.47)
(Using t in billions of years throughout)
Step 3 — Solve for t
ln(0.75) = (t / 4.47) * ln(0.5)
negative 0.2877 = (t / 4.47) * (negative 0.6931)
t / 4.47 = 0.2877 / 0.6931 = 0.4151
t = 4.47 * 0.4151
t approx 1.855 billion years old
For comparison, the Earth's age is approximately 4.54 billion years. Rocks from the Precambrian era (older than 541 million years) commonly show ages in the 1 to 4 billion year range. Uranium-238 decay is one of the most reliable geological clocks because of its extremely long half-life.
Exponential functions can be written using any positive base. The e-based form Q = Q0 e^(rt) is standard for calculus and most science applications. The base-a form Q = Q0 a^t is sometimes more intuitive for interpreting growth factors. Both forms describe the same function.
Converting from e-base to base-a
Start: Q = Q0 * e^(rt)
Write as: Q = Q0 * (e^r)^t
Let a = e^r, so Q = Q0 * a^t
Example: e^(0.05) approx 1.0513, so r = 5% continuous growth equals about 5.13% effective annual growth.
Converting from base-a to e-base
Start: Q = Q0 * a^t
Use: a = e^(ln a)
So: a^t = [e^(ln a)]^t = e^(t ln a)
Q = Q0 * e^(t * ln(a))
Therefore r = ln(a). Example: a = 1.08 (8% annual growth) gives r = ln(1.08) approx 0.0770 continuous rate.
Practical example: Rewriting a bacteria model
Bacteria triples every 4 hours: Q = Q0 * 3^(t/4)
Convert: 3^(t/4) = e^((t/4) * ln 3) = e^(0.2747t)
So Q = Q0 * e^(0.2747t) where r = 0.2747 per hour
A common exam problem gives you two (time, quantity) data points and asks you to build the exponential model. The technique is to divide the two equations to eliminate Q0, then take the natural log to solve for r.
General method
Given: Q(t1) = Q1 and Q(t2) = Q2
Write: Q2 / Q1 = (Q0 e^(r t2)) / (Q0 e^(r t1)) = e^(r(t2 minus t1))
Then: ln(Q2/Q1) = r * (t2 minus t1)
r = ln(Q2/Q1) / (t2 minus t1)
Then Q0 = Q1 * e^(negative r * t1)
Example: Drug concentration in the bloodstream
A drug concentration was 45 mg/L at t = 2 hours and 20 mg/L at t = 8 hours after administration.
r = ln(20/45) / (8 minus 2) = ln(0.4444) / 6 = (negative 0.8109) / 6
r = negative 0.1352 per hour (decay)
Q0 = 45 * e^(negative (negative 0.1352)(2)) = 45 * e^(0.2704) = 45 * 1.3104 approx 59.0 mg/L at t = 0
Model: C(t) = 59.0 * e^(negative 0.1352 t)
Doubling time: t = ln(2) / r
Half-life: h = ln(2) / |k|
k from half-life: k = negative ln(2) / h
Rule of 72: years approx 72 / (rate%)
Base conversion: a^t = e^(t ln a)
Two-point r: r = ln(Q2/Q1) / (t2 minus t1)
The exponential growth model is Q(t) = Q0 times e^(rt), where Q0 is the initial quantity, r is the continuous growth rate (r greater than 0 for growth, r less than 0 for decay), t is time, and e is Euler's number approximately 2.71828. This formula describes any quantity that grows or decays at a rate proportional to its current size.
The half-life formula is t = ln(2) divided by r, where r is the decay rate (as a positive number). Equivalently, if the half-life period is h, then at time t the remaining amount is Q(t) = Q0 times (1/2)^(t/h). To find the age of a sample using carbon-14, set up 0.25 times Q0 = Q0 times (1/2)^(t/5730) and solve for t using logarithms.
The doubling time formula is t = ln(2) divided by r, where r is the continuous growth rate. Since ln(2) is approximately 0.693, the doubling time is approximately 0.693 divided by r. The Rule of 72 gives a quick approximation: doubling time in years is roughly 72 divided by the annual percentage rate. For example, at 6% per year, doubling time is approximately 72 divided by 6 = 12 years.
Newton's Law of Cooling states that an object cools at a rate proportional to the difference between its temperature and the surrounding temperature. The formula is T(t) = T_env plus (T0 minus T_env) times e^(kt), where T_env is the ambient temperature, T0 is the initial object temperature, k is a negative constant, and t is time. To use it, first find k using a known temperature at a specific time, then evaluate T(t) at any desired time.
The logistic growth model is P(t) = M divided by (1 plus A times e^(negative kt)), where M is the carrying capacity (maximum sustainable population), k is the growth rate constant, and A is determined by the initial condition. Unlike pure exponential growth, logistic growth slows as the population approaches M, producing an S-shaped curve. It is used when resources are limited.
Any exponential function of the form y = a^x can be written in e-based form as y = e^(x times ln(a)), because a = e^(ln(a)). Conversely, Q = Q0 times e^(rt) can be written as Q = Q0 times b^t where b = e^r. This is useful because e-based form is standard for calculus, while base-a form is easier to interpret as percentage growth per period.
Given two data points (t1, Q1) and (t2, Q2): write Q2 = Q0 times e^(r times t2) and Q1 = Q0 times e^(r times t1). Divide: Q2 divided by Q1 = e^(r times (t2 minus t1)). Take the natural log of both sides: ln(Q2 divided by Q1) = r times (t2 minus t1). Solve: r = ln(Q2 divided by Q1) divided by (t2 minus t1). Then Q0 = Q1 times e^(negative r times t1).
Graphs, transformations, and inverse relationships between exp and log
Compound interest, half-life, and population models with worked examples
Complete chapter-by-chapter guide to all precalculus topics
Log laws, natural log, change of base, and solving log equations
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Calculus continuation: separable equations and exponential growth derivations
Work through interactive problems on half-life, doubling time, compound interest, and Newton's law of cooling. Step-by-step private tutoring included — free to try.
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