Unit 7Exponential Growth and Decay

Concept 4.79Built for May 2027 AP Exam

Exponential Growth and Decay

Reviewed by ACEpath AP Math Editorial Team

Last Updated: September 2026

30-second summary

When a quantity changes at a rate proportional to its own size, the solution is exponential. Writing “proportional to the size” as an equation gives $\frac{dy}{dt} = ky$ , and separating it gives $y = y_0e^{kt}$ , where $y_0$ is the starting amount and the sign of $k$ decides growth or decay. Two derived quantities carry most of the exam questions. The doubling time, or half-life, is $\frac{\ln 2}{|k|}$ and does not depend on $y_0$ . The sign of $k$ comes from the data, not from the equation, so a decay problem and a growth problem share every step except that one sign. The commonest failure is using this model where the rate is proportional to a gap, which is limited growth, not exponential.

On the AP Exam

AP Calculus AB tests this skill through the representations and reasoning described below.

What you’ll see: A contextual item asking for the value of a quantity after some time, for the time at which a threshold is reached, or for the constant $k$ given a doubling time or half-life. It arrives as a differential equation in context, and the first decision is always which of the two rate models the sentence describes.
What you’ll do: Translate the sentence into $\frac{dy}{dt} = ky$ or $\frac{dy}{dt} = k\left(A - y\right)$ , solve by separation, and then answer with units and a sensible domain.
Memory hook: Proportional to the size means proportional to $y$ . Proportional to the gap means a target is involved, and the answer will not be a pure exponential.
Use it on the exam: For MCQs: Use the doubling time rather than the raw constant. Doubling time $\frac{\ln 2}{k}$ is scale-free, so it can be read off the options even when $y_0$ is unknown. For FRQs: Write the model, the general solution, the constant, and the requested value as four separate lines. The released free responses in this window reach this page through the gap model, and there the four-line habit is the difference between one point and three.

Concept Review

The Model and Its Solution

Figure 1 — The same equation with opposite signs of $k$ . Both curves start at $y_0 = 2$ , and the doubling time equals the half-life in magnitude.

The statement “the rate of change is proportional to the size” translates directly:

$\frac{dy}{dt} = ky \qquad\Longrightarrow\qquad y = y_0e^{kt}$

Separating gives $\frac{dy}{y} = k\,dt$ , then $\ln|y| = kt + C$ , then $y = y_0e^{kt}$ . Because an exponential is never zero, a solution that starts positive stays positive, and a solution that starts negative stays negative; the sign of $y_0$ is preserved for all time.

The same equation with opposite signs of $k$. Both curves start at $y_0 = 2$, and the doubling time equals the half-life in magnitude.

Doubling Time and Half-Life

The doubling time is the time needed for the quantity to multiply by $2$ :

$2y_0 = y_0e^{kT} \qquad\Longrightarrow\qquad T = \frac{\ln 2}{k}$

The half-life is the same computation with $\frac{1}{2}$ in place of $2$ , and it equals $\frac{\ln 2}{|k|}$ . Three facts about $T$ are worth memorising because they turn many problems into one-line answers:

$T$ does not depend on $y_0$ . A colony of ten and a colony of ten million double in the same time.
$T$ depends only on $|k|$ , so the growth and decay curves in the figure share the same time scale.
Doubling twice is not the same as quadrupling once in elapsed time — it is the same thing, which is why exponential problems are usually solved by counting intervals rather than by algebra.

Reading $k$ Off a Half-Life

A half-life of $8$ years means $e^{8k} = \frac{1}{2}$ , so $k = \frac{\ln\frac{1}{2}}{8} = -\frac{\ln 2}{8} \approx -0.0866$ . The negative sign is not an extra decision; it comes out of the equation. Writing $y = y_0e^{-0.0866t}$ and then adding “since it decays” is redundant, and on a scored part the redundancy is harmless while a missing negative sign is fatal.

Exponential Versus Limited Growth

Both models are separable and both produce exponential-looking answers, which is why they are confused. The difference is what the rate is proportional to.

Sentence in the problem	Model	Solution shape
“proportional to the amount present”	$\frac{dy}{dt} = ky$	$y = y_0e^{kt}$ , unbounded or decaying to zero
“proportional to the difference from $A$ ”	$\frac{dy}{dt} = k\left(A - y\right)$	$y = A + \left(y_0 - A\right)e^{-kt}$ , approaches $A$
“proportional to $t$ ”	$\frac{dy}{dt} = ct$	quadratic, not exponential at all

The third row is a distractor rather than a model: proportional to time is not proportional to the quantity, and the solution is a parabola.

Units and Sensible Answers

Every constant in an exponential model carries units, and they are reciprocal to the units of $t$ . If $t$ is measured in hours then $k$ has units of per hour, and $\frac{\ln 2}{k}$ is measured in hours. A final answer should also survive a plausibility check: a population model that predicts more people than there are atoms in the observable universe at $t = 100$ is telling you that the exponential phase ended long before.

Common Misconception Check

Myth 1: “A doubling time depends on the starting amount.” Truth: It depends only on $k$ . That is the whole point of using it.
Myth 2: “Any differential equation with an $e$ in the solution is exponential growth.” Truth: Limited growth also produces an exponential term. The distinguishing feature is the target $A$ , not the presence of $e$ .
Myth 3: “A negative $k$ means the model is wrong.” Truth: A negative $k$ is decay, and it is the correct answer whenever the quantity decreases.

Topic-Specific Study Method

Decide the Model Before You Solve

Borrowed idea: A quantity whose rate of change is proportional to its own size follows $y = y_0e^{kt}$ . Why does the same formula serve both growth and decay? Because only the sign of $k$ changes, and the sign comes from the data rather than from the equation. The doubling time is $\frac{\ln 2}{k}$ and is independent of the starting value. What separates this model from limited growth? A limited-growth equation carries a target, so its rate depends on the gap $A - y$ rather than on $y$ alone.

Exam disguise	What you see on the page	What your brain must translate	Required execution
“Proportional to the amount”	a rate sentence with no target	$\frac{dy}{dt} = ky$	separate, integrate to a log
“Half-life of $8$ years”	a time, not a rate	$k = -\frac{\ln 2}{8}$	solve for $k$ first, then use it
“Doubles every $5$ hours”	a time, not a rate	$k = \frac{\ln 2}{5}$	same, positive sign
“Cools toward room temperature”	a target temperature	$\frac{dT}{dt} = k\left(A - T\right)$	shifted variable, not this page
“When does it reach $500$ ?”	a threshold, not a rate	solve $y_0e^{kt} = 500$	take a logarithm, keep units

① Name the model in words before writing symbols. “Rate proportional to size” and “rate proportional to the gap” produce different equations, and the wrong choice cannot be repaired later.

② Convert the doubling time into $k$ as a separate line. Combining the conversion with the solution is where the reciprocal gets flipped.

③ Answer with units and a time frame. The value at $t = 10$ hours and the time at which the value reaches $10$ are different questions that share a formula.

Training & Error Diagnosis

Core Archetype (Simplest Form): A culture grows so that $\frac{dy}{dt} = 0.3y$ with $y(0) = 200$ bacteria. Find $y(10)$ and the doubling time.

Standard Execution:

Model. Rate proportional to size, so $y = 200e^{0.3t}$ .

Value. $y(10) = 200e^{3} \approx 4017.1$ , so about $4017$ bacteria.

Doubling time. $T = \frac{\ln 2}{0.3} \approx 2.31$ hours.

Sanity check. In $10$ hours the culture doubles about $\frac{10}{2.31} \approx 4.3$ times, and $200 \cdot 2^{4.3} \approx 4017$ , which matches.

Exam Variation (Context / Parameter Shift): A radioactive sample loses half its mass every $8$ years. Write the differential equation, find $k$ , and determine the time at which $5\%$ of the original mass remains.

Solution: The mass satisfies $\frac{dm}{dt} = km$ with $m = m_0e^{kt}$ . The half-life gives $e^{8k} = \frac{1}{2}$ , so $k = -\frac{\ln 2}{8} \approx -0.0866$ per year. Setting $m = 0.05m_0$ gives $e^{-0.0866t} = 0.05$ , so $t = \frac{\ln 0.05}{-0.0866} \approx 34.6$ years. The check is that $5\%$ is between $\frac{1}{16}$ and $\frac{1}{32}$ of the original, so the time must lie between four and five half-lives, that is between $32$ and $40$ years.

Common Traps & Failure Signals:

Model misread (Signal: an exponential solution to a “cools toward” sentence): the target $A$ must appear in the equation.
Reciprocal flipped (Signal: $k = \frac{8}{\ln 2}$ from a half-life of $8$ ): $k$ is a rate, so time belongs in the denominator.
Sign dropped (Signal: a positive $k$ for a decaying quantity): decay gives negative $k$ , and the data supplies it.

6D Cognitive Error Diagnosis (Where Your Thinking Broke):

Model Misread: You used the exponential model for a cooling sentence.
Reciprocal Flipped: You divided the time by $\ln 2$ instead of the other way round.
Threshold Ignored: You reported the value at a given time when the question asked for the time at a given value, and the two answers differ by an inverse function that no amount of algebra can repair after the fact.

Past 10-Year AP Exam Appearance

Testing Frequency: MCQ Only; No released free-response part since 2017 uses the pure form dy/dt = ky; every differential-equation free response in that window is either a gap model, where the rate tracks a target minus the quantity, or a nonlinear separable equation, and both of those are scored as separation problems on 7.3 and 7.4 rather than as model choices
10-Year Total Count: 0 of 9 Exams · 0 FRQ Parts; The four gap-model parts that come closest to this page are counted on 7.4, because the released wording never asks the student to recognise the exponential model
FRQ Years Appeared: None in Released FRQs ---

Practice Examples

MCQ 1

Turning a Doubling Time Into a Constant

A quantity grows at a rate proportional to its size and doubles every $5$ hours. The constant $k$ in $\frac{dy}{dt} = ky$ is

MCQ 2

Which Sentence Is This?

A population $P$ grows at a rate proportional to its current size, with no limiting factors. Which differential equation models the situation?

MCQ 3

Reading a Half-Life From Two Data Points

A $40$ -gram sample of a substance decays to $10$ grams in $12$ years. Its half-life is

FRQ 4

From a Sentence to a Function

(a) A culture grows according to $\frac{dy}{dt} = 0.3y$ with $y(0) = 200$ . Find $y(t)$ , the value at $t = 10$ hours, and the doubling time.

(b) A radioactive isotope has a half-life of $8$ years. Write the differential equation for its mass $m(t)$ , find the constant $k$ , and determine when $5\%$ of the original mass remains.

(c) Explain why $\frac{dT}{dt} = k\left(20 - T\right)$ does not describe exponential growth, even though its solution contains an exponential.

Cheatsheet

Core Reference

	Action	Failure if skipped
1	decide whether the rate tracks $y$ or a gap $A - y$	the wrong model, unfixable later
2	write $\frac{dy}{dt} = ky$ and solve to $y = y_0e^{kt}$	the constant has no home
3	convert a doubling time with $k = \frac{\ln 2}{T}$	the reciprocal gets flipped
4	apply the initial condition	$y_0$ stays symbolic
5	state units on $k$ and on the answer	the answer is unreadable

Sentences That Score

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Unit 7Exponential Growth and Decay

Concept 4.79Built for May 2027 AP Exam

Exponential Growth and Decay

Reviewed by ACEpath AP Math Editorial Team

Last Updated: September 2026

30-second summary

On the AP Exam

AP Calculus AB tests this skill through the representations and reasoning described below.

What you’ll see: A contextual item asking for the value of a quantity after some time, for the time at which a threshold is reached, or for the constant $k$ given a doubling time or half-life. It arrives as a differential equation in context, and the first decision is always which of the two rate models the sentence describes.
What you’ll do: Translate the sentence into $\frac{dy}{dt} = ky$ or $\frac{dy}{dt} = k\left(A - y\right)$ , solve by separation, and then answer with units and a sensible domain.
Memory hook: Proportional to the size means proportional to $y$ . Proportional to the gap means a target is involved, and the answer will not be a pure exponential.
Use it on the exam: For MCQs: Use the doubling time rather than the raw constant. Doubling time $\frac{\ln 2}{k}$ is scale-free, so it can be read off the options even when $y_0$ is unknown. For FRQs: Write the model, the general solution, the constant, and the requested value as four separate lines. The released free responses in this window reach this page through the gap model, and there the four-line habit is the difference between one point and three.

Concept Review

The Model and Its Solution

Figure 1 — The same equation with opposite signs of $k$ . Both curves start at $y_0 = 2$ , and the doubling time equals the half-life in magnitude.

The statement “the rate of change is proportional to the size” translates directly:

$\frac{dy}{dt} = ky \qquad\Longrightarrow\qquad y = y_0e^{kt}$

The same equation with opposite signs of $k$. Both curves start at $y_0 = 2$, and the doubling time equals the half-life in magnitude.

Doubling Time and Half-Life

The doubling time is the time needed for the quantity to multiply by $2$ :

$2y_0 = y_0e^{kT} \qquad\Longrightarrow\qquad T = \frac{\ln 2}{k}$

$T$ does not depend on $y_0$ . A colony of ten and a colony of ten million double in the same time.
$T$ depends only on $|k|$ , so the growth and decay curves in the figure share the same time scale.
Doubling twice is not the same as quadrupling once in elapsed time — it is the same thing, which is why exponential problems are usually solved by counting intervals rather than by algebra.

Reading $k$ Off a Half-Life

Exponential Versus Limited Growth

Both models are separable and both produce exponential-looking answers, which is why they are confused. The difference is what the rate is proportional to.

Sentence in the problem	Model	Solution shape
“proportional to the amount present”	$\frac{dy}{dt} = ky$	$y = y_0e^{kt}$ , unbounded or decaying to zero
“proportional to the difference from $A$ ”	$\frac{dy}{dt} = k\left(A - y\right)$	$y = A + \left(y_0 - A\right)e^{-kt}$ , approaches $A$
“proportional to $t$ ”	$\frac{dy}{dt} = ct$	quadratic, not exponential at all

The third row is a distractor rather than a model: proportional to time is not proportional to the quantity, and the solution is a parabola.

Units and Sensible Answers

Common Misconception Check

Myth 1: “A doubling time depends on the starting amount.” Truth: It depends only on $k$ . That is the whole point of using it.
Myth 2: “Any differential equation with an $e$ in the solution is exponential growth.” Truth: Limited growth also produces an exponential term. The distinguishing feature is the target $A$ , not the presence of $e$ .
Myth 3: “A negative $k$ means the model is wrong.” Truth: A negative $k$ is decay, and it is the correct answer whenever the quantity decreases.

Topic-Specific Study Method

Decide the Model Before You Solve

Exam disguise	What you see on the page	What your brain must translate	Required execution
“Proportional to the amount”	a rate sentence with no target	$\frac{dy}{dt} = ky$	separate, integrate to a log
“Half-life of $8$ years”	a time, not a rate	$k = -\frac{\ln 2}{8}$	solve for $k$ first, then use it
“Doubles every $5$ hours”	a time, not a rate	$k = \frac{\ln 2}{5}$	same, positive sign
“Cools toward room temperature”	a target temperature	$\frac{dT}{dt} = k\left(A - T\right)$	shifted variable, not this page
“When does it reach $500$ ?”	a threshold, not a rate	solve $y_0e^{kt} = 500$	take a logarithm, keep units

① Name the model in words before writing symbols. “Rate proportional to size” and “rate proportional to the gap” produce different equations, and the wrong choice cannot be repaired later.

② Convert the doubling time into $k$ as a separate line. Combining the conversion with the solution is where the reciprocal gets flipped.

③ Answer with units and a time frame. The value at $t = 10$ hours and the time at which the value reaches $10$ are different questions that share a formula.

Training & Error Diagnosis

Core Archetype (Simplest Form): A culture grows so that $\frac{dy}{dt} = 0.3y$ with $y(0) = 200$ bacteria. Find $y(10)$ and the doubling time.

Standard Execution:

Model. Rate proportional to size, so $y = 200e^{0.3t}$ .

Value. $y(10) = 200e^{3} \approx 4017.1$ , so about $4017$ bacteria.

Doubling time. $T = \frac{\ln 2}{0.3} \approx 2.31$ hours.

Sanity check. In $10$ hours the culture doubles about $\frac{10}{2.31} \approx 4.3$ times, and $200 \cdot 2^{4.3} \approx 4017$ , which matches.

Solution: The mass satisfies $\frac{dm}{dt} = km$ with $m = m_0e^{kt}$ . The half-life gives $e^{8k} = \frac{1}{2}$ , so $k = -\frac{\ln 2}{8} \approx -0.0866$ per year. Setting $m = 0.05m_0$ gives $e^{-0.0866t} = 0.05$ , so $t = \frac{\ln 0.05}{-0.0866} \approx 34.6$ years. The check is that $5\%$ is between $\frac{1}{16}$ and $\frac{1}{32}$ of the original, so the time must lie between four and five half-lives, that is between $32$ and $40$ years.

Common Traps & Failure Signals:

Model misread (Signal: an exponential solution to a “cools toward” sentence): the target $A$ must appear in the equation.
Reciprocal flipped (Signal: $k = \frac{8}{\ln 2}$ from a half-life of $8$ ): $k$ is a rate, so time belongs in the denominator.
Sign dropped (Signal: a positive $k$ for a decaying quantity): decay gives negative $k$ , and the data supplies it.

6D Cognitive Error Diagnosis (Where Your Thinking Broke):

Model Misread: You used the exponential model for a cooling sentence.
Reciprocal Flipped: You divided the time by $\ln 2$ instead of the other way round.
Threshold Ignored: You reported the value at a given time when the question asked for the time at a given value, and the two answers differ by an inverse function that no amount of algebra can repair after the fact.

Past 10-Year AP Exam Appearance

Testing Frequency: MCQ Only; No released free-response part since 2017 uses the pure form dy/dt = ky; every differential-equation free response in that window is either a gap model, where the rate tracks a target minus the quantity, or a nonlinear separable equation, and both of those are scored as separation problems on 7.3 and 7.4 rather than as model choices
10-Year Total Count: 0 of 9 Exams · 0 FRQ Parts; The four gap-model parts that come closest to this page are counted on 7.4, because the released wording never asks the student to recognise the exponential model
FRQ Years Appeared: None in Released FRQs ---

Practice Examples

MCQ 1

Turning a Doubling Time Into a Constant

A quantity grows at a rate proportional to its size and doubles every $5$ hours. The constant $k$ in $\frac{dy}{dt} = ky$ is

MCQ 2

Which Sentence Is This?

A population $P$ grows at a rate proportional to its current size, with no limiting factors. Which differential equation models the situation?

MCQ 3

Reading a Half-Life From Two Data Points

A $40$ -gram sample of a substance decays to $10$ grams in $12$ years. Its half-life is

FRQ 4

From a Sentence to a Function

(a) A culture grows according to $\frac{dy}{dt} = 0.3y$ with $y(0) = 200$ . Find $y(t)$ , the value at $t = 10$ hours, and the doubling time.

(b) A radioactive isotope has a half-life of $8$ years. Write the differential equation for its mass $m(t)$ , find the constant $k$ , and determine when $5\%$ of the original mass remains.

(c) Explain why $\frac{dT}{dt} = k\left(20 - T\right)$ does not describe exponential growth, even though its solution contains an exponential.

Cheatsheet

Core Reference

	Action	Failure if skipped
1	decide whether the rate tracks $y$ or a gap $A - y$	the wrong model, unfixable later
2	write $\frac{dy}{dt} = ky$ and solve to $y = y_0e^{kt}$	the constant has no home
3	convert a doubling time with $k = \frac{\ln 2}{T}$	the reciprocal gets flipped
4	apply the initial condition	$y_0$ stays symbolic
5	state units on $k$ and on the answer	the answer is unreadable

Sentences That Score

TOPIC REVIEW • NEXT STEP IN YOUR LEARNING LOOP

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Start Topic Practice Next Topic: 7.6 Logistic Differential Equation

Unit 7Exponential Growth and Decay

Concept 4.79Built for May 2027 AP Exam

Exponential Growth and Decay

Reviewed by ACEpath AP Math Editorial Team

Last Updated: September 2026

30-second summary

On the AP Exam

AP Calculus AB tests this skill through the representations and reasoning described below.

What you’ll see: A contextual item asking for the value of a quantity after some time, for the time at which a threshold is reached, or for the constant $k$ given a doubling time or half-life. It arrives as a differential equation in context, and the first decision is always which of the two rate models the sentence describes.
What you’ll do: Translate the sentence into $\frac{dy}{dt} = ky$ or $\frac{dy}{dt} = k\left(A - y\right)$ , solve by separation, and then answer with units and a sensible domain.
Memory hook: Proportional to the size means proportional to $y$ . Proportional to the gap means a target is involved, and the answer will not be a pure exponential.
Use it on the exam: For MCQs: Use the doubling time rather than the raw constant. Doubling time $\frac{\ln 2}{k}$ is scale-free, so it can be read off the options even when $y_0$ is unknown. For FRQs: Write the model, the general solution, the constant, and the requested value as four separate lines. The released free responses in this window reach this page through the gap model, and there the four-line habit is the difference between one point and three.

Concept Review

The Model and Its Solution

Figure 1 — The same equation with opposite signs of $k$ . Both curves start at $y_0 = 2$ , and the doubling time equals the half-life in magnitude.

The statement “the rate of change is proportional to the size” translates directly:

$\frac{dy}{dt} = ky \qquad\Longrightarrow\qquad y = y_0e^{kt}$

The same equation with opposite signs of $k$. Both curves start at $y_0 = 2$, and the doubling time equals the half-life in magnitude.

Doubling Time and Half-Life

The doubling time is the time needed for the quantity to multiply by $2$ :

$2y_0 = y_0e^{kT} \qquad\Longrightarrow\qquad T = \frac{\ln 2}{k}$

$T$ does not depend on $y_0$ . A colony of ten and a colony of ten million double in the same time.
$T$ depends only on $|k|$ , so the growth and decay curves in the figure share the same time scale.
Doubling twice is not the same as quadrupling once in elapsed time — it is the same thing, which is why exponential problems are usually solved by counting intervals rather than by algebra.

Reading $k$ Off a Half-Life

Exponential Versus Limited Growth

Both models are separable and both produce exponential-looking answers, which is why they are confused. The difference is what the rate is proportional to.

Sentence in the problem	Model	Solution shape
“proportional to the amount present”	$\frac{dy}{dt} = ky$	$y = y_0e^{kt}$ , unbounded or decaying to zero
“proportional to the difference from $A$ ”	$\frac{dy}{dt} = k\left(A - y\right)$	$y = A + \left(y_0 - A\right)e^{-kt}$ , approaches $A$
“proportional to $t$ ”	$\frac{dy}{dt} = ct$	quadratic, not exponential at all

The third row is a distractor rather than a model: proportional to time is not proportional to the quantity, and the solution is a parabola.

Units and Sensible Answers

Common Misconception Check

Myth 1: “A doubling time depends on the starting amount.” Truth: It depends only on $k$ . That is the whole point of using it.
Myth 2: “Any differential equation with an $e$ in the solution is exponential growth.” Truth: Limited growth also produces an exponential term. The distinguishing feature is the target $A$ , not the presence of $e$ .
Myth 3: “A negative $k$ means the model is wrong.” Truth: A negative $k$ is decay, and it is the correct answer whenever the quantity decreases.

Topic-Specific Study Method

Decide the Model Before You Solve

Exam disguise	What you see on the page	What your brain must translate	Required execution
“Proportional to the amount”	a rate sentence with no target	$\frac{dy}{dt} = ky$	separate, integrate to a log
“Half-life of $8$ years”	a time, not a rate	$k = -\frac{\ln 2}{8}$	solve for $k$ first, then use it
“Doubles every $5$ hours”	a time, not a rate	$k = \frac{\ln 2}{5}$	same, positive sign
“Cools toward room temperature”	a target temperature	$\frac{dT}{dt} = k\left(A - T\right)$	shifted variable, not this page
“When does it reach $500$ ?”	a threshold, not a rate	solve $y_0e^{kt} = 500$	take a logarithm, keep units

① Name the model in words before writing symbols. “Rate proportional to size” and “rate proportional to the gap” produce different equations, and the wrong choice cannot be repaired later.

② Convert the doubling time into $k$ as a separate line. Combining the conversion with the solution is where the reciprocal gets flipped.

③ Answer with units and a time frame. The value at $t = 10$ hours and the time at which the value reaches $10$ are different questions that share a formula.

Training & Error Diagnosis

Core Archetype (Simplest Form): A culture grows so that $\frac{dy}{dt} = 0.3y$ with $y(0) = 200$ bacteria. Find $y(10)$ and the doubling time.

Standard Execution:

Model. Rate proportional to size, so $y = 200e^{0.3t}$ .

Value. $y(10) = 200e^{3} \approx 4017.1$ , so about $4017$ bacteria.

Doubling time. $T = \frac{\ln 2}{0.3} \approx 2.31$ hours.

Sanity check. In $10$ hours the culture doubles about $\frac{10}{2.31} \approx 4.3$ times, and $200 \cdot 2^{4.3} \approx 4017$ , which matches.

Solution: The mass satisfies $\frac{dm}{dt} = km$ with $m = m_0e^{kt}$ . The half-life gives $e^{8k} = \frac{1}{2}$ , so $k = -\frac{\ln 2}{8} \approx -0.0866$ per year. Setting $m = 0.05m_0$ gives $e^{-0.0866t} = 0.05$ , so $t = \frac{\ln 0.05}{-0.0866} \approx 34.6$ years. The check is that $5\%$ is between $\frac{1}{16}$ and $\frac{1}{32}$ of the original, so the time must lie between four and five half-lives, that is between $32$ and $40$ years.

Common Traps & Failure Signals:

Model misread (Signal: an exponential solution to a “cools toward” sentence): the target $A$ must appear in the equation.
Reciprocal flipped (Signal: $k = \frac{8}{\ln 2}$ from a half-life of $8$ ): $k$ is a rate, so time belongs in the denominator.
Sign dropped (Signal: a positive $k$ for a decaying quantity): decay gives negative $k$ , and the data supplies it.

6D Cognitive Error Diagnosis (Where Your Thinking Broke):

Model Misread: You used the exponential model for a cooling sentence.
Reciprocal Flipped: You divided the time by $\ln 2$ instead of the other way round.
Threshold Ignored: You reported the value at a given time when the question asked for the time at a given value, and the two answers differ by an inverse function that no amount of algebra can repair after the fact.

Past 10-Year AP Exam Appearance

Testing Frequency: MCQ Only; No released free-response part since 2017 uses the pure form dy/dt = ky; every differential-equation free response in that window is either a gap model, where the rate tracks a target minus the quantity, or a nonlinear separable equation, and both of those are scored as separation problems on 7.3 and 7.4 rather than as model choices
10-Year Total Count: 0 of 9 Exams · 0 FRQ Parts; The four gap-model parts that come closest to this page are counted on 7.4, because the released wording never asks the student to recognise the exponential model
FRQ Years Appeared: None in Released FRQs ---

Practice Examples

MCQ 1

Turning a Doubling Time Into a Constant

A quantity grows at a rate proportional to its size and doubles every $5$ hours. The constant $k$ in $\frac{dy}{dt} = ky$ is

MCQ 2

Which Sentence Is This?

A population $P$ grows at a rate proportional to its current size, with no limiting factors. Which differential equation models the situation?

MCQ 3

Reading a Half-Life From Two Data Points

A $40$ -gram sample of a substance decays to $10$ grams in $12$ years. Its half-life is

FRQ 4

From a Sentence to a Function

(a) A culture grows according to $\frac{dy}{dt} = 0.3y$ with $y(0) = 200$ . Find $y(t)$ , the value at $t = 10$ hours, and the doubling time.

(b) A radioactive isotope has a half-life of $8$ years. Write the differential equation for its mass $m(t)$ , find the constant $k$ , and determine when $5\%$ of the original mass remains.

(c) Explain why $\frac{dT}{dt} = k\left(20 - T\right)$ does not describe exponential growth, even though its solution contains an exponential.

Cheatsheet

Core Reference

	Action	Failure if skipped
1	decide whether the rate tracks $y$ or a gap $A - y$	the wrong model, unfixable later
2	write $\frac{dy}{dt} = ky$ and solve to $y = y_0e^{kt}$	the constant has no home
3	convert a doubling time with $k = \frac{\ln 2}{T}$	the reciprocal gets flipped
4	apply the initial condition	$y_0$ stays symbolic
5	state units on $k$ and on the answer	the answer is unreadable

Sentences That Score

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