The chapter's applications. Builds and uses four model families — unbounded exponential growth, decay described by half-life, Newton's law of cooling with its shifted asymptote, and logistic growth with a carrying capacity — and solves each for a time using §4.6's technique. Chooses between them by asking what the quantity approaches.
Subject: Precalculus · 65 slides · symbolic lesson
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Title
Precalculus · Chapter 4 — Exponential and Logarithmic Functions
§4.7 Exponential and Logarithmic Models, pp. 569-589
Objectives
Five things, each one you can check yourself on paper.
OpenStax, Precalculus, §4.7 Exponential and Logarithmic Models §4.7, pp. 569-589 — the pages these objectives are drawn from
Warm-up
Four model families, and one question separates them.
Discussion prompt
A radioactive sample decays, a cup of coffee cools, and a population grows in a limited habitat. What does each approach in the long run?
Hint: For each, imagine waiting a very long time.
Answer:
The sample approaches zero — the material decays away, never quite vanishing but getting arbitrarily close.
The coffee approaches room temperature, not zero. It stops cooling when it matches its surroundings.
The population approaches a carrying capacity set by food and space, having grown almost exponentially at first.
Three different limits, and each needs a different model. The limiting value is what distinguishes the families, which is why identifying it is the first step in choosing one.
Concept
Every question in this section has the same shape: assemble an exponential model from the data given, then solve it for an unknown time using a logarithm. What varies is the model's form.
\[ A=A_0e^{kt}, \quad A=A_0e^{-kt}, \quad T=T_s+(T_0-T_s)e^{-kt}, \quad P=\frac{L}{1+Ce^{-kt}} \]
The four forms differ in what they approach and in how many constants have to be found, but the solving step is identical throughout: isolate the exponential, take a logarithm, and divide. §4.6 supplied that technique and this section supplies the situations that need it.
Figure (svg): A decision table matching each described situation to the model family it needs, with the distinguishing feature of each noted
OpenStax, Precalculus, §4.7 Exponential and Logarithmic Models §4.7, pp. 569-574
Section
Section 1
Concept
A continuous growth or decay model has an initial amount and a rate constant. The initial amount is usually given directly, and the rate is found from one further data point.
\[ A(t)=A_0e^{kt} \]
The distinction between the continuous rate and the percentage rate is worth being careful about. A continuous rate of 0.05 corresponds to annual growth of about 5.13 percent, because continuous compounding earns slightly more than a single annual application of the same nominal rate.
Figure (svg): A decision table matching each described situation to the model family it needs, with the distinguishing feature of each noted
OpenStax, Precalculus, §4.7 Exponential and Logarithmic Models §4.7, pp. 569-575
Picture it
What the quantity approaches decides which row applies.
Figure (svg): A decision table matching each described situation to the model family it needs, with the distinguishing feature of each noted
The first two rows are this idea. The last two are the shifted and bounded models, which the later ideas take up.
Worked example
One data point beyond the start determines k.
\[ \text{A culture of } 1000 \text{ grows to } 2500 \text{ in } 4 \text{ hours. Model it.} \]
Write the model with the known initial amount
Why: A naught is 1000.
\[ A = 1000 e ^{k t} \]
Substitute the second data point
Why: At 4 hours the amount is 2500.
\[ 2500 = 1000 e ^{4 k} \]
Isolate the exponential
Why: Divide by 1000.
\[ e ^{4 k} = 2.5 \]
Take the natural logarithm and divide
Why: The exponent comes down.
\[ k = \ln(2.5) / 4 \]
Figure (svg): The solution to Worked example find the rate constant shown as a ladder of expressions, one row per legal move
\[ A(t)=1000e^{0.229t} \]
Verify: check the second data point
Why: At t equal to 4 the model gives 1000 times e to the power 0.916, which is about 2500 — matching. Note the base e was used throughout, which is why the rate is a continuous one; using base 2.5 to the power t over 4 would have modelled the same data with a different-looking constant.
OpenStax, Precalculus, §4.7 Exponential and Logarithmic Models §4.7, pp. 570-572
Faded example
A quantity of 500 grows to 1500 in 6 years.
Fill in the blanks
e^3 = 6 \;\Longrightarrow\; k = \frac___}}___}
Why: Dividing by the initial 500 gives 3 on the right, and taking the natural logarithm brings the exponent down. The rate is the natural logarithm of 3 divided by 6, about 0.183. The ratio of the two amounts is what determines the rate; the amounts themselves do not matter separately.
Worked example
This is §4.6's exponential technique applied to a model.
\[ \text{With the same model, when does the culture reach } 10\,000? \]
Set the model equal to the target
Why: The amount is known and the time is not.
\[ 1000 e ^{0.229 t} = 10000 \]
Isolate the exponential
Why: Divide by 1000.
\[ e ^{0.229 t} = 10 \]
Take the natural logarithm
Why: The exponent comes down.
\[ 0.229 t = \ln 10 \]
Divide
Why: By the rate constant.
\[ t = 10.06 \]
Figure (svg): The solution to Worked example solve for a time shown as a ladder of expressions, one row per legal move
\[ t=\frac{\ln 10}{0.229}\approx 10.1 \text{ hours} \]
Verify: sanity-check against the data
Why: The culture went from 1000 to 2500 in 4 hours, so reaching 10000 — four times as much again — should take rather longer than another 4 hours. Just over 10 hours total is consistent. Note that this was exactly §4.6's technique with a model's constants substituted in.
OpenStax, Precalculus, §4.7 Exponential and Logarithmic Models §4.7, pp. 572-575
Trap
\[ k=0.05 \;\Longrightarrow\; \text{the quantity grows exactly } 5\% \text{ per year} \]
Read the rate constant as a percentage
Why: The number 0.05 is interpreted directly as five percent.
The annual growth is reported as exactly 5 percent.
The continuous rate is not the annual percentage. After one year the factor is e to the power 0.05, which is about 1.0513 — an increase of about 5.13 percent.
The difference is small for small rates and grows for larger ones. At a continuous rate of 0.5 the annual increase is about 64.9 percent, not 50.
Compute e to the k to get the annual factor, then subtract 1 for the percentage. The two rates describe the same growth and are not the same number.
Prediction
A quantity is decaying.
Predict first
What is the sign of the rate constant in the model?
Correct: Negative, so the exponent decreases the output.
Why: A negative exponent makes the exponential factor shrink towards zero as time grows, which is what decay means. Some texts write the model with an explicit minus sign and a positive k instead, which is equivalent — but one of the two must carry the negative.
Sorting
Read the sign of the rate constant.
Sort into buckets
Sort each model.
Socratic
A growth model could be written with any base.
Discussion prompt
Why do scientific models almost always use base e rather than base 2 or base 10?
Hint: What does the rate constant mean in each case?
Answer:
With base e the constant k is the instantaneous rate of change per unit amount — the fractional growth per unit time at any instant. That is the quantity physical laws are usually stated in terms of.
With any other base the constant is that rate divided by the logarithm of the base, which is an arbitrary rescaling carrying no independent meaning.
So base e is chosen because it makes the constant mean something. §12.4 makes this precise: the exponential with base e is the one function that is its own derivative, which is why it is the natural base for anything described by a rate of change.
Section
Section 2
Concept
The half-life is the time for a decaying quantity to fall to half its value. It is the same wherever you start measuring, which is what makes it a property of the substance rather than of the sample.
The constancy is a real fact about exponential decay rather than a convention. It follows from the multiplicative structure: going forward by a fixed time multiplies by a fixed factor, so if that factor is one half over some interval, it is one half over every interval of the same length.
Figure (svg): An exponential decay curve with successive half-lives marked, showing the quantity halving over each equal interval regardless of where the interval starts
OpenStax, Precalculus, §4.7 Exponential and Logarithmic Models §4.7, pp. 575-580
Picture it
Each equal interval halves the amount, wherever the interval begins.
Figure (svg): An exponential decay curve with successive half-lives marked, showing the quantity halving over each equal interval regardless of where the interval starts
The intervals are equal in width and each one halves the quantity. That equality is what a half-life asserts and it is not obvious from the formula until it is drawn.
Worked example
The half-life determines the rate constant.
\[ \text{A substance has a half-life of } 30 \text{ years. Write its decay model.} \]
Use the half-life form directly
Why: Base one half, exponent time over half-life.
\[ A = A _{0}(\frac{1}{2}) ^{\frac{t}{30}} \]
Or convert to base e
Why: Set the factor after 30 years to one half.
\[ e ^{30 k} = \frac{1}{2} \]
Take the natural logarithm
Why: The exponent comes down.
\[ 30 k = -\ln 2 \]
Solve for k
Why: Divide.
\[ k = -\frac{\ln 2}{30} \]
Figure (svg): An exponential decay curve with successive half-lives marked, showing the quantity halving over each equal interval regardless of where the interval starts
\[ A=A_0\Bigl(\tfrac{1}{2}\Bigr)^{t/30} = A_0e^{-0.0231t} \]
Verify: check both forms agree at 30 years
Why: The first gives one half directly. The second gives e to the power negative 0.693, which is 0.5 — the same. The two forms are the same model written over different bases, and the half-life form is usually easier to read while the base e form is easier to differentiate.
OpenStax, Precalculus, §4.7 Exponential and Logarithmic Models §4.7, pp. 576-578
Faded example
After 4 half-lives, what fraction of the original is left?
Fill in the blanks
\Bigl(\tfrac416\Bigr)^___} = \tfrac______}
Why: Four half-lives means multiplying by one half four times, giving one sixteenth — about 6.25 percent. The fractions multiply rather than the percentages subtracting, which is why the amount approaches zero without ever reaching it.
Worked example
This is dating: a remaining fraction gives an age.
\[ \text{A sample retains } 20\% \text{ of its original amount. How old is it, with a } 30 \text{ year half-life?} \]
Set the model to the remaining fraction
Why: The initial amount cancels.
\[ (\frac{1}{2}) ^{\frac{t}{30}} = 0.2 \]
Take the natural logarithm of both sides
Why: Bringing the exponent down.
\[ (\frac{t}{30}) \ln(0.5) = \ln(0.2) \]
Solve for the exponent
Why: Divide by the logarithm of one half.
\[ \frac{t}{30} = \ln(0.2) / \ln(0.5) \]
Multiply
Why: By the half-life.
\[ t\text{ about } 69.7\text{ years} \]
Figure (svg): The solution to Worked example solve for an elapsed time shown as a ladder of expressions, one row per legal move
\[ t=30\cdot\frac{\ln 0.2}{\ln 0.5}\approx 69.7 \text{ years} \]
Verify: check against whole half-lives
Why: Two half-lives is 60 years and leaves 25 percent; three is 90 years and leaves 12.5 percent. Twenty percent lies between those, so the age should be between 60 and 90 and closer to 60 — and 69.7 is. Note the initial amount cancelled, which is why dating works without knowing how much was there originally.
OpenStax, Precalculus, §4.7 Exponential and Logarithmic Models §4.7, pp. 578-580
Error analysis
A student computes what remains after three half-lives.
Annotate
On: \( \text{three half-lives} \;\Longrightarrow\; 100\% - 50\% - 50\% - 50\% = -50\% \)
Halving repeatedly never reaches zero, let alone goes negative. Each step multiplies by one half, and repeated multiplication by a positive number stays positive forever.
Prediction
Two samples of the same substance have very different masses.
Predict first
Do they have the same half-life?
Correct: Yes: the half-life does not depend on the amount.
Why: Exponential decay multiplies by a fixed factor over a fixed interval, whatever the current amount, so the time to halve is the same for a gram and for a tonne. This is what makes half-life a property of the substance and why it can be tabulated once for each isotope.
Sorting
Each half-life multiplies by one half.
Sort into buckets
Sort each elapsed time by whether more or less than a quarter remains.
Explain it to yourself
It is the same wherever you start measuring.
Discussion prompt
Explain why, using the multiplicative structure of exponential decay.
Hint: What does advancing time by a fixed amount do to the quantity?
Answer:
Advancing time by a fixed interval multiplies the quantity by a fixed factor, whatever the quantity currently is. That is what exponential decay means.
So if some interval multiplies by one half starting from the beginning, the same interval multiplies by one half starting from anywhere — the factor does not depend on the starting amount.
That is why the half-life is a property of the substance rather than of the sample, and it is what makes dating possible: the remaining fraction determines the elapsed time without any need to know the original quantity, which is never available for an ancient sample.
Section
Section 3
Concept
An object cools towards its surroundings rather than towards zero. What decays exponentially is the difference between the object's temperature and the ambient one.
\[ T(t)=T_s+(T_0-T_s)e^{-kt} \]
The shift is the physical content rather than a mathematical adjustment. Modelling the temperature itself as a pure decay would predict the object freezing, which is why the difference is what the law is stated about — and the same structure describes anything approaching an equilibrium rather than zero.
Figure (svg): A cooling curve approaching a horizontal asymptote at room temperature rather than at zero, with the decaying difference marked as the part that shrinks
OpenStax, Precalculus, §4.7 Exponential and Logarithmic Models §4.7, pp. 580-584
Picture it
The dashed segment is the difference, and it is what shrinks.
Figure (svg): A cooling curve approaching a horizontal asymptote at room temperature rather than at zero, with the decaying difference marked as the part that shrinks
The asymptote sits at room temperature. Subtracting it leaves a pure decay, which is exactly how the model is built and how it is solved.
Worked example
Subtract the ambient temperature to isolate the decaying part.
\[ \text{Coffee at } 90 \text{ degrees cools in a } 20 \text{ degree room, reaching } 60 \text{ after } 10 \text{ minutes.} \]
Identify the ambient temperature
Why: The asymptote.
\[ T _{s} = 20 \]
Find the initial difference
Why: Ninety minus twenty.
\[ 70 \]
Substitute the data point
Why: At 10 minutes the difference is 40.
\[ 70 e ^{-10 k} = 40 \]
Solve for k
Why: Isolate and take a logarithm.
\[ k = -\ln(\frac{4}{7}) / 10\text{ about } 0.0560 \]
Figure (svg): A cooling curve approaching a horizontal asymptote at room temperature rather than at zero, with the decaying difference marked as the part that shrinks
\[ T(t)=20+70e^{-0.056t} \]
Verify: check the data point and the limit
Why: At 10 minutes: 20 plus 70 times e to the negative 0.56, which is 20 plus 40, giving 60 — matching. And as time grows the exponential vanishes and the temperature approaches 20, the room temperature, as it must. Both checks confirm the model rather than just the arithmetic.
OpenStax, Precalculus, §4.7 Exponential and Logarithmic Models §4.7, pp. 581-583
Faded example
An object at 100 degrees cools in a 25 degree room.
Fill in the blanks
T(t) = 25 + 75e^___
Why: The ambient temperature 25 is the asymptote and the initial difference is 100 minus 25, which is 75. The coefficient of the exponential is always the initial difference, not the initial temperature — using 100 there would predict the object approaching 25 from a starting point of 125.
Worked example
Isolate the difference before taking a logarithm.
\[ \text{With that model, when does the coffee reach } 30 \text{ degrees?} \]
Set the model to the target
Why: Thirty degrees.
\[ 20 + 70 e ^{-0.056 t} = 30 \]
Subtract the ambient temperature
Why: Isolating the decaying difference.
\[ 70 e ^{-0.056 t} = 10 \]
Divide and take the logarithm
Why: The exponent comes down.
\[ -0.056 t = \ln(\frac{1}{7}) \]
Divide
Why: By the rate constant.
\[ t\text{ about } 34.7\text{ minutes} \]
Figure (svg): The solution to Worked example solve for a time shown as a ladder of expressions, one row per legal move
\[ t\approx 34.7 \text{ minutes} \]
Verify: check the trend
Why: It took 10 minutes to fall from 90 to 60, and 35 minutes to fall from 90 to 30. The second half of the journey took far longer, which is right: as the difference shrinks the cooling slows. That deceleration is the whole content of the law, and a model predicting a steady fall would be linear rather than exponential.
OpenStax, Precalculus, §4.7 Exponential and Logarithmic Models §4.7, pp. 583-584
Trap
\[ 20+70e^{-0.056t}=30 \;\Longrightarrow\; \ln(20)+\ln(70e^{-0.056t})=\ln 30 \]
Take the logarithm of both sides immediately
Why: The equation is transformed without first isolating the exponential term.
The left side is treated as a sum of logarithms.
The left side is a sum, and there is no property for the logarithm of a sum. The step accomplishes nothing.
Subtract the 20 first, so the exponential term stands alone, and only then take a logarithm.
Isolate before taking any logarithm, exactly as §4.6 required. The ambient temperature is the constant that must come off first, and it is the one piece of the model that is not part of the decay.
Prediction
An object cools according to Newton's law in a room at 18 degrees.
Predict first
What temperature does it approach?
Correct: 18 degrees, the room's temperature.
Why: The exponential term decays to zero, leaving only the ambient temperature. The rate constant decides how quickly it gets there but not where it ends up. This is §4.2's vertical shift moving the asymptote off zero, and the shift is exactly the surrounding temperature.
Discrimination
The limiting value distinguishes them.
Sort into buckets
Sort each situation.
Explain it to yourself
The law is stated about a difference rather than a temperature.
Discussion prompt
Explain why modelling the temperature itself as a pure decay would be wrong.
Hint: What would a pure decay predict in the long run?
Answer:
A pure decay approaches zero, so it would predict the coffee cooling to zero degrees and beyond any sensible limit. That is not what happens; it stops at room temperature.
What actually shrinks is the difference between the object and its surroundings, and that difference genuinely does approach zero — the object stops cooling exactly when it matches the room.
So the model decays the difference and then adds the room temperature back. That addition is §4.2's vertical shift, and it is the mathematical expression of the physical fact that heat flows only while there is a temperature difference to drive it.
Section
Section 4
Concept
A logistic model grows almost exponentially while the quantity is small and levels off as it approaches a carrying capacity built into the formula.
\[ P(t)=\frac{L}{1+Ce^{-kt}} \]
The self-limiting is the point. §2.3 warned that a linear model must eventually break down and left the breakdown outside the model; §4.1's exponential had the same problem. The logistic model puts the limit inside the formula, so it does not need a caveat attached.
Figure (svg): A logistic growth curve, nearly exponential at first and then levelling off towards a horizontal carrying capacity, contrasted with an unbounded exponential
OpenStax, Precalculus, §4.7 Exponential and Logarithmic Models §4.7, pp. 584-589
Picture it
The two curves agree early and diverge completely later.
Figure (svg): A logistic growth curve, nearly exponential at first and then levelling off towards a horizontal carrying capacity, contrasted with an unbounded exponential
The dashed exponential runs off the top of the picture; the logistic curve levels off at its capacity. Early data cannot distinguish them, which is why choosing between them requires knowing whether a limit exists.
Worked example
The capacity is visible in the formula.
\[ \text{For } P(t)=\frac{800}{1+39e^{-0.3t}}, \text{ find the capacity and the initial value.} \]
Identify the capacity
Why: The numerator.
\[ L = 800 \]
Find the initial value
Why: Substitute t equal to zero.
\[ \frac{800}{1 + 39} \]
Compute
Why: Eight hundred over forty.
\[ P(0) = 20 \]
Check the long-run behaviour
Why: The exponential decays to zero.
\[ P\text{ approaches } 800 \]
Figure (svg): A logistic growth curve, nearly exponential at first and then levelling off towards a horizontal carrying capacity, contrasted with an unbounded exponential
\[ L=800, \quad P(0)=20 \]
Verify: check the early behaviour
Why: At t equal to 1 the model gives about 800 over 29.9, which is about 26.8 — an increase of about 34 percent, close to exponential. Early on the denominator is dominated by the large exponential term, so the model behaves almost exponentially, exactly as the picture shows.
OpenStax, Precalculus, §4.7 Exponential and Logarithmic Models §4.7, pp. 585-587
Faded example
For the logistic model with numerator 600 and denominator 1 plus 29 times the decaying exponential.
Fill in the blanks
P(0) = \frac2920}} = ___
Why: At time zero the exponential equals 1, so the denominator is 1 plus 29, which is 30, and 600 over 30 is 20. The constant in the denominator therefore determines how far below the capacity the model starts, and a larger constant means a smaller initial value.
Worked example
Isolate the exponential, then take a logarithm.
\[ \text{With that model, when does the population reach } 400? \]
Set the model to the target
Why: Half the capacity.
\[ \frac{800}{1 + 39 e ^{-0.3 t}} = 400 \]
Clear the denominator
Why: Multiply up and divide.
\[ 1 + 39 e ^{-0.3 t} = 2 \]
Isolate the exponential
Why: Subtract 1 and divide by 39.
\[ e ^{-0.3 t} = \frac{1}{39} \]
Take a logarithm and divide
Why: The exponent comes down.
\[ t\text{ about } 12.2 \]
Figure (svg): The solution to Worked example solve for a time shown as a ladder of expressions, one row per legal move
\[ t=\frac{\ln 39}{0.3}\approx 12.2 \]
Verify: note what is special about half the capacity
Why: Four hundred is exactly half of 800, and that is where a logistic curve grows fastest — the inflection point. Before it the growth accelerates and after it the growth slows, which is visible in the figure as the steepest part of the curve. So this particular time has a meaning beyond being the answer to the question.
OpenStax, Precalculus, §4.7 Exponential and Logarithmic Models §4.7, pp. 587-589
Error analysis
A student identifies the carrying capacity of a logistic model.
Annotate
On: \( P(t)=\frac{500}{1+24e^{-0.2t}}: \quad L=24 \)
The capacity is the numerator, because the denominator approaches 1 in the long run. The constant in the denominator sets where the curve starts rather than where it ends.
Prediction
A logistic curve rises from near zero to its carrying capacity.
Predict first
Where is it growing most quickly?
Correct: At half the carrying capacity.
Why: Early on the quantity is small so the absolute growth is small even though the percentage growth is high. Near the capacity the growth slows to nothing. The maximum absolute rate is exactly halfway, at the curve's inflection point, which is visible as its steepest section.
Sorting
Look for a stated limit.
Sort into buckets
Sort each situation.
Edge cases
A logistic model looks almost exponential when the quantity is small.
Discussion prompt
Can early data distinguish the two models, and what does that imply?
Hint: How different are the two curves in the first few time units?
Answer:
Early data cannot distinguish them. While the quantity is far below the capacity, the logistic curve is nearly indistinguishable from an exponential — the figure shows the two curves lying almost on top of each other at first.
So a fit to early data will match both, and choosing between them requires knowing whether a limit exists, which is a question about the situation rather than about the numbers.
This is §2.4's warning about model choice in a sharper form. Two models agreeing on the data and disagreeing wildly on the extrapolation is exactly the situation where the modeller's knowledge matters more than the fit, and it is why epidemic forecasts differ so much early in an outbreak.
Section
Section 5
Concept
The four families differ in their limiting behaviour. Identifying what the quantity tends towards in the long run selects the model before any algebra begins.
The common structure is worth emphasising because it means only one technique has to be fluent. Build the model, isolate the exponential, take a logarithm, divide. The four families differ in what has to be subtracted or divided out before the isolation, and in nothing else.
Figure (svg): A decision table matching each described situation to the model family it needs, with the distinguishing feature of each noted
OpenStax, Precalculus, §4.7 Exponential and Logarithmic Models §4.7, pp. 569-589
Picture it
Four situations, four models, distinguished by the limiting value.
Figure (svg): A decision table matching each described situation to the model family it needs, with the distinguishing feature of each noted
The right-hand column gives the tell in each case. Reading the situation for its long-run behaviour is faster and more reliable than trying to match the data's shape.
Worked example
Read for the limiting behaviour.
\[ \text{A drug's concentration falls towards zero; a room warms towards } 21 \text{ degrees. Which models?} \]
Read the first limit
Why: Towards zero.
Write its form
Why: No shift needed.
\[ A = A _{0} e ^{-k t} \]
Read the second limit
Why: Towards 21, not zero.
Write its form
Why: Ambient temperature added back.
\[ T = 21 + (T _{0} - 21) e ^{-k t} \]
Figure (svg): A decision table matching each described situation to the model family it needs, with the distinguishing feature of each noted
\[ A_0e^{-kt} \quad \text{and} \quad 21+(T_0-21)e^{-kt} \]
Verify: check each limit
Why: The first approaches zero as the exponential vanishes, matching the description. The second approaches 21, since the exponential vanishes and leaves the constant. Both models' asymptotes agree with what the situations were said to approach, which is the check that the right family was chosen.
OpenStax, Precalculus, §4.7 Exponential and Logarithmic Models §4.7, pp. 580-585
Matching
The limiting value decides.
Match the pairs
Why: Each is identified by what it approaches: zero, room temperature, the school's size, and no limit at all. Reading for the long-run behaviour is faster than examining the data, and it is the only way to distinguish logistic from exponential when only early data is available.
Worked example
Every family reduces to the same three steps.
\[ \text{Show that all four families solve for } t \text{ the same way.} \]
Isolate the exponential term
Why: Subtract constants and divide by coefficients.
\[ e ^{k t} =\text{ some number} \]
Take a logarithm of both sides
Why: Natural is convenient with base e.
Divide by the rate constant
Why: Isolating the time.
\[ t = \ln(...) / k \]
Note what differed
Why: Only the isolation step.
Figure (svg): The solution to Worked example the common solving structure shown as a ladder of expressions, one row per legal move
\[ \text{isolate} \to \text{take } \ln \to \text{divide by } k \]
Verify: check against the worked examples above
Why: The growth example divided by the initial amount; the cooling example subtracted the ambient temperature first; the logistic example cleared a denominator first. All three then took a logarithm and divided by k. The isolation differs and the rest is identical, which is why §4.6 was worth its own section.
OpenStax, Precalculus, §4.7 Exponential and Logarithmic Models §4.7, pp. 572-589
Trap
\[ \text{a rumour in a town of } 5000 \;\Longrightarrow\; N(t)=10e^{0.4t} \]
Fit an exponential to the early spread
Why: The early data grows exponentially, so an exponential model is fitted.
The model is used to predict the number who have heard it after 30 days.
The exponential predicts far more than the town's population. At 30 days it gives over a million, in a town of 5000.
A logistic model is needed, with the capacity set at 5000. It agrees with the early data and levels off at the population size instead of running away.
A stated limit means a logistic model. The early data cannot distinguish the two, so the choice has to come from knowing the situation — which is exactly what §2.4 said about model choice generally.
Prediction
A cooling model is being solved for a time.
Predict first
What must be done before taking a logarithm?
Correct: Subtract the ambient temperature, then divide by the initial difference.
Why: The exponential must stand alone before a logarithm is useful, and in the cooling model it is buried under an added constant and a multiplying coefficient. Removing both in that order isolates it. Taking a logarithm first would produce a logarithm of a sum, which cannot be expanded.
Faded example
For the model 30 plus 50 times e to the negative 0.1t, find when it reaches 40.
Fill in the blanks
50e^10 = -0.1 \;\Longrightarrow\; e^___ = 0.2 \;\Longrightarrow\; t = \frac______}
Why: Subtracting 30 leaves 10 on the right, and dividing by 50 gives 0.2. Taking the natural logarithm and dividing by negative 0.1 gives about 16.1. The two isolation steps — subtract then divide — are what the cooling model's structure requires before the logarithm can be applied.
Real world
Two models can agree on the data and disagree on the forecast.
Discussion prompt
Early in an epidemic, an exponential and a logistic model fit the data equally well. Why does the choice matter so much?
Hint: What do the two predict for the total number infected?
Answer:
They agree on the past and disagree completely on the future. The exponential predicts unbounded growth; the logistic predicts levelling off at a capacity, and the two forecasts differ by orders of magnitude.
The data cannot decide between them, because the logistic curve is nearly exponential while the quantity is far below its capacity. Fit quality is uninformative here, exactly as §2.4 warned.
So the choice rests on knowledge of the situation — whether a limit exists and roughly where. That is why epidemic forecasting is contested early and converges later: the data eventually reaches the region where the two models visibly diverge, and by then the answer matters much less.
Comparison
Fill the blanks from memory. The limiting value is what distinguishes them.
Comparison matrix
| approaches | form | the tell | |
|---|---|---|---|
| exponential growth | no limit | A_0 e^(kt) | a constant percentage rise |
| exponential decay | zero | A_0 e^(-kt) | a constant half-life |
| Newton's cooling | the ambient value | T_s + (T_0 - T_s)e^(-kt) | a nonzero asymptote |
| logistic | the carrying capacity | L / (1 + Ce^(-kt)) | a stated limit or capacity |
The solving technique is identical for all four: isolate the exponential, take a logarithm, divide. Only the isolation differs, and the table's third column says how.
Pattern
Six steps, and the first is the one that chooses everything else.
Step 3's parenthetical is where the cooling model catches people. Its coefficient is the initial DIFFERENCE from the surroundings, not the initial temperature, and using the wrong one shifts the whole curve.
OpenStax Algebra and Trigonometry 2e, §6.7 Exponential and Logarithmic Models §6.7
Check
Halving is multiplicative.
Check your understanding
After 3 half-lives, what fraction of a sample remains?
Answer: A
Why: Each half-life multiplies by one half, so three of them give one half cubed, which is one eighth. The fractions multiply rather than the amounts subtracting, which is why the quantity never reaches zero.
Check
Read the asymptote.
Check your understanding
In the cooling model T equals 22 plus 58 times e to the negative kt, what does the temperature approach?
Answer: A
Why: The exponential term decays to zero, leaving the constant 22, which is the ambient temperature. The 58 is the initial difference and the initial temperature was 80, which is their sum.
Check
The capacity is the numerator.
Check your understanding
For the logistic model with numerator 1200 and denominator 1 plus 19 times a decaying exponential, what is the carrying capacity?
Answer: A
Why: As time grows the exponential decays to zero, the denominator approaches 1, and the model approaches the numerator, 1200. The 19 sets the initial value, which is 1200 over 20, or 60.
Real world
Radiocarbon dating is this section's decay model with the initial amount unknown.
Discussion prompt
Dating a sample requires knowing how much carbon-14 it started with, which nobody measured. How is that possible?
Hint: What cancels when you take the ratio?
Answer:
The initial amount cancels. Setting the model equal to a fraction of the original divides both sides by the initial amount, so only the ratio matters — and the ratio is measurable.
What supplies the ratio is the assumption that the proportion of carbon-14 in living tissue is roughly constant over time, so a fresh sample's proportion stands in for the ancient sample's original one.
So the technique needs a half-life and a present ratio, both measurable, and never needs the original quantity. That cancellation is the mathematical reason dating works at all, and it is why the method's accuracy depends on the constancy assumption rather than on any measurement of the past.
Commit first
State your confidence along with your answer.
Predict first
A population grows in a habitat with limited food. Which model is appropriate?
Correct: Logistic, since there is a carrying capacity.
Why: A stated limit means the growth must level off, which is exactly what the logistic model builds in. An exponential would fit the early data equally well and then predict a population exceeding what the habitat can support — the error §2.3 called model breakdown, avoided here by choosing a model that limits itself.
Explain it
The test of understanding is being able to say why, not just what.
Discussion prompt
Explain to a classmate how to choose between an exponential and a logistic model when both fit the early data.
Hint: What question is the data unable to answer?
Answer:
The data cannot decide, because a logistic curve is nearly exponential while the quantity is far below its capacity. Fitting either to early data will look equally convincing.
The question the data cannot answer is whether a limit exists. That is a fact about the situation — a habitat's food supply, a population's size, a market's total customers — rather than about the numbers.
So the choice comes from understanding what is being modelled. A good explanation stresses that this is not a defect in the mathematics: the two models genuinely agree where the evidence is, and disagree only where there is none, which is exactly where a modeller's judgement has to do the work.
Exit ticket
One honest answer, so the next lesson can start in the right place.
Predict first
Which idea from this lesson would you most want to see again?
Correct: Any of these is a legitimate answer; the useful one is the honest one.
Why: There is no correct choice here. The third catches people out because the coefficient is a difference rather than an amount. The fourth is the least mechanical and the most consequential, since choosing the wrong family produces a model that is wrong in a way no amount of correct algebra can repair.
Connect it up
One page, drawn from memory, is worth more than rereading the section.
Draw it
Sketch all four model families on one set of axes: unbounded growth, decay to zero, decay to a nonzero asymptote, and logistic growth to a capacity. Label each one's limiting value. Beside them, write the single solving procedure that all four share, and note for each family what has to be isolated before the logarithm can be taken.
If your four curves have four visibly different limiting behaviours, and one shared solving procedure written beside them, you have both halves of the section.
Recap
Five things, and the last is the one that decides whether a model is right at all.
| if you remember one thing | it should be this |
|---|---|
| about choosing | the limiting value names the family |
| about half-life | it does not depend on how much is present |
| about cooling | the difference decays, not the temperature |
| about solving | isolate the exponential, take a logarithm, divide - every time |
Section 4.8 closes the chapter by fitting exponential and logarithmic models to data that does not lie exactly on a curve, using §2.4's regression ideas with a logarithmic transformation first.
OpenStax, Precalculus, §4.7 Exponential and Logarithmic Models §4.7, pp. 569-589 — everything on these slides traces back here
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