How do I decide between an exponential and a logarithmic model?

Read the story and the shape. A quantity that changes by a fixed factor or percentage each period — doubling, halving, +5% a year — or that approaches a floor is exponential ($y = ka^x$ or $y = ke^{rx} + c$). A quantity that rises or falls steeply at first and then flattens without a fixed doubling time — a learning curve, a sensory scale — is logarithmic ($y = a\ln x + b$). On a scatter plot, exponential curves have a J or reverse-J shape; logarithmic curves shoot up then bend over and level.

What is the difference between the base-$a$ and base-$e$ forms?

$y = ka^x$ and $y = ke^{rx}$ describe the same behaviour written two ways. Base $a$ uses a per-period factor ($a = 1.05$ for +5%); base $e$ uses a continuous rate $r$. They convert by $a = e^r$, so $r = \ln a$. In AI SL you are normally told which form to use, or you fit whichever the question or the regression menu gives — you are not expected to prefer one on your own.

Why does the $+c$ term matter so much in cooling and medication problems?

Because those quantities do not fade to zero — a coffee stops at room temperature, a drug level may settle at a baseline. The $+c$ sets the horizontal asymptote $y = c$, the limiting value the quantity approaches. Leave it out and you force the curve down to zero, which mis-fits the data and makes every prediction wrong near the floor. Always identify the limit first and set $c$ equal to it.

My GDC errors when I solve a logarithmic equation — what went wrong?

Almost always you are asking it to take the log of zero or a negative number, which is undefined. For $y = a\ln(x - c) + b$ the input must satisfy $x > c$. Check the domain, and give the solver a sensible starting guess or the graph a window that actually contains the solution.

Do I have to take logs by hand, or can I just use the calculator?

Either is accepted for the answer, but AI SL is calculator-active and the GDC method — solver or graph intersection — is the expected, faster route, and it is what our marking engine assumes. If you do take logs, show the line where you apply $\ln$ to both sides so the method mark is on the page; if you use the GDC, still write the equation you solved so the examiner sees the method.

Modelling with exponential and logarithmic…

Revise Modelling with exponential and logarithmic functions for IB Mathematics: Applications and Interpretation (2.4)

Looking for free IB Mathematics: Applications and Interpretation notes on Modelling with exponential and logarithmic functions? This Standard Level lesson covers official syllabus point 2.4 on Mathematics: Applications and Interpretation SL. Use the visual steps, key terms, and quick checks below — then practise with IB criterion marking on Modelling with exponential and logarithmic functions. Built for students searching "IB Maths AI 2.4", "Modelling with exponential and logarithmic functions IB revision", or a free alternative to paid IB note sites.

Modelling with exponential and logarithmic functions

Straight lines describe change at a constant rate, but much of the real world does not oblige. A population grows faster the larger it is; a hot drink cools quickest when it is hottest; a radioactive sample loses the same fraction of itself every half-life; a drug clears from the blood a little more slowly as its concentration falls. These are the natural home of exponential and logarithmic models. This lesson builds the whole 2.4 syllabus: the growth and decay forms $y = ka^x$ and $y = ke^{rx} + c$ , the horizontal asymptote read as a limiting value, the meaning of every parameter, fitting a model on the GDC, logarithmic models, and solving for the time to reach a value. Because AI SL is calculator-active, the marks reward choosing the right model and showing the method the examiner needs — so the worked examples below are laid out exactly the way our engine marks your own practice.

BY THE END, YOU CAN…

1Recognise and use the exponential growth and decay models $y = ka^x$ and $y = ke^{rx} + c$ , and identify the horizontal asymptote $y = c$ as a limiting value.
2Interpret each parameter in context: the initial value, the growth or decay factor $a$ , the continuous rate $r$ , and the limit $c$ .
3Fit an exponential or logarithmic model to data using the GDC, from two points or by regression.
4Solve for the time to reach a given value, either by taking logarithms or by using the GDC solver or graph.
5Apply exponential and logarithmic models to population, cooling, radioactive decay and medication problems, presenting working so method marks and follow-through protect the final answer.

In simple terms

A friendly intro before the formal notes — no formulas yet.

Modelling change that speeds up or levels off

Exponential models describe quantities that change by the same factor in equal steps of time — doubling, halving, growing 5% a year — so they race away or fade towards a floor. Logarithmic models describe the opposite feel: fast at first, then flattening. Pick the shape that matches the story, pin down the parameters, then read the answer off the model.

THINK OF IT LIKE…

Think of a hot coffee left on a desk. It cools quickly at first, when it is far hotter than the room, and more slowly as it nears room temperature — but it never drops below the room. That floor is the limiting value: in the model $T = 68e^{-0.05t} + 20$ the coffee always sits $20$ °C above absolute-nothing-left, because the room is $20$ °C. Exponential decay towards a limit is exactly this shape: a gap that shrinks by the same fraction every minute but never quite closes.

1
Decide the shape from the story: a fixed percentage or factor per period, or an approach to a floor, is exponential; a quantity that shoots up then flattens is logarithmic.
2
Write the general form — $y = ka^x$ , or $y = ke^{rx} + c$ when there is a non-zero limit, or $y = a\ln x + b$ — and note what each letter must mean here.
3
Find the parameters: substitute known points, or use the GDC's exponential or logarithmic regression, keeping full accuracy in the calculator.
4
Use the model: substitute a time to predict a value, or solve for the time to reach a value by taking logs or using the GDC solver or intersection.

Explore the concept

Use the live diagram and synced steps — play it or tap a step card to walk through.

Key formulas

Tap any symbol to reveal exactly what it means and its units.

$y = ka^x$

$y = a\ln x + b$

$y = ke^{rx}$

Full topic notes

Formal explanation with the rigour you need for the exam.

Exponential models: growth, decay and a limiting value

An exponential model changes by a constant factor over equal steps: multiply by the same number each period. That single idea produces runaway growth when the factor exceeds one, and a fade towards a floor when it is between zero and one. Two equivalent forms appear in AI SL. The first uses a per-period factor; the second uses base $e$ and adds a constant so the curve can settle above zero.

Growth / decay factor form: $y = ka^x$ . \n Continuous form with a limit: $y = ke^{rx} + c$ .

Reading the parameters in context is where marks are won. In $P(t) = 8000(1.03)^t$ the initial population is $8000$ and the factor $1.03$ means a $3\%$ rise each year. In $m(t) = 50(0.5)^{t/12}$ the initial mass is $50$ and the base $0.5$ with exponent $t/12$ says the sample halves every $12$ hours. In $T(t) = 68e^{-0.05t} + 20$ the limit is $20$ , the starting temperature is $68 + 20 = 88$ , and the negative rate $-0.05$ marks decay towards that floor.

$k$ — the amplitude / initial size. In $y = ka^x$ the initial value (at $x = 0$ ) is $k$ . In $y = ke^{rx} + c$ the initial value is $k + c$ , because $e^0 = 1$ .
$a$ or $r$ — the growth/decay control. $a > 1$ (equivalently $r > 0$ ) means growth; $0 < a < 1$ (equivalently $r < 0$ ) means decay. They convert by $a = e^r$ .
$c$ — the limiting value. The horizontal asymptote is $y = c$ : as $x \to \infty$ in a decay model the term $ke^{rx} \to 0$ and $y \to c$ . This floor is room temperature, a baseline concentration, a background level.
The gap shrinks by a fixed fraction. In a decay-to-a-limit model the distance $y - c$ is itself a pure exponential $ke^{rx}$ — the quantity closes the same fraction of its remaining gap each period.

Fitting a model to data with the GDC

When a question hands you a table rather than a clean pair of points, use regression. Enter the inputs in one list and the outputs in a second, then choose exponential regression for a $y = ka^x$ shape or logarithmic regression for a $y = a\ln x + b$ shape. The calculator returns the parameters that best fit the data; write them down to full accuracy and quote the model. If the data levels off at a non-zero floor, subtract that floor first (work with $y - c$ ) or use a model form that includes the constant, so the exponential part is fitted cleanly.

Logarithmic models

Logarithmic models fit quantities that change rapidly at first and then flatten — the mirror image of exponential growth. Because they are the inverse of exponentials, their graphs rise (or fall) steeply near the start and bend towards the horizontal. The standard AI SL form uses the natural logarithm.

Logarithmic model: $y = a\ln x + b$ , defined for $x > 0$ .

Here $a$ controls how steeply the curve climbs and its sign sets the direction, while $b$ shifts the whole curve vertically. Because $\ln x$ is undefined for $x \le 0$ , always check the domain before substituting or solving, and make sure any GDC window or solver guess sits in the valid region.

Solving for the time to reach a value

A recurring task is the reverse question: not 'what is the value at this time?' but 'when does the value reach this level?'. Set the model equal to the target and solve for the time. There are two routes, and either earns the marks. Algebraically, isolate the exponential and take natural logs of both sides — logs are the tool that frees a variable trapped in an exponent. On the GDC, use the numerical solver or graph both sides and find the intersection. AI SL is calculator-active, so the GDC route is expected and usually quicker; whichever you use, write down the equation you solved so the method is on the page.

For $y = ke^{rx}$ , taking logs gives $x = \dfrac{1}{r}\ln\!\left(\dfrac{y}{k}\right)$ . \n For $y = ke^{rx} + c$ , first subtract $c$ : $y - c = ke^{rx}$ , then take logs.

Common mistakes examiners penalise

Treating $c$ as zero when the data has a floor — in $y = ke^{rx} + c$ the limiting value is $y = c$ , the horizontal asymptote. Cooling and medication problems level off above zero; identify $c$ first or every parameter is wrong.
Misreading the initial value with a shift — for $y = ke^{rx} + c$ the value at $t = 0$ is $k + c$ , not $k$ , because $e^0 = 1$ . Only in $y = ka^x$ is the initial value simply $k$ .
Confusing growth and decay — $a > 1$ or $r > 0$ is growth; $0 < a < 1$ or $r < 0$ is decay. A negative sign in the exponent means the quantity falls; do not report a decaying model as growing.
Rounding parameters mid-calculation — carry the GDC's full values of $k$ , $a$ , $r$ and $c$ ; round only the final answer to 3 significant figures. A rate rounded early can shift a time answer by whole units.
Not showing the method when solving for time — write the equation you set equal to the target and either the $\ln$ step or the GDC equation solved. An answer with no method risks the method mark even if the number is right.
Ignoring the domain of a logarithmic model — $\ln x$ needs $x > 0$ (and $\ln(x - c)$ needs $x > c$ ). Feeding the GDC a value outside the domain produces an error, not a solution.
Forgetting to subtract the limit before taking logs — to solve $y = ke^{rx} + c$ you must isolate the exponential first: $y - c = ke^{rx}$ , then take logs. Taking logs of $y$ directly is wrong.
Answering the wrong question — 'find the value after $t$ ' means substitute; 'find when it reaches a value' means solve. Substituting when a solve is asked (or vice versa) throws away the whole part.

Model answer — marked the way our engine marks it

In AI SL the marks are analytic: each is tied to a specific line of working — a method mark (M) or an accuracy mark (A) — and an accuracy mark depends on the method mark it follows. Follow-through (FT) means an earlier wrong value need not cost you the marks that depend on it, provided the later step is done correctly on your own figure. The engine also ignores subsequent working once a correct answer appears, and accepts any equivalent form and any correctly-rounded value — and a value found by GDC solver or intersection earns exactly the same marks as one found by taking logs. But that protection only exists if the method is on the page. Study how each mark below is earned by a specific line.

Where this leads

Exponential and logarithmic models are the backbone of AI SL's applied topics. The same growth and decay engine drives compound interest and depreciation in the finance strand, where the per-period factor becomes $1 + \frac{r}{100}$ ; the limiting value you met here as room temperature returns as the carrying capacity in more advanced growth models and as the baseline in regression residuals. The habit built in this lesson — name the model, read the asymptote, interpret each parameter, and show the 'set equal and solve' method whether you take logs or reach for the GDC — is exactly the discipline every calculator-active question rewards. Get the shape and the parameters right, and prediction becomes a single substitution and timing a single solve.

Worked examples

See the formulas applied — reveal one step at a time, like the exam.

EXAMPLE 1

A colony of bacteria is modelled by $N(t) = k a^t$ , where $N$ is the number of bacteria (in thousands) and $t$ is the time in hours. When $t = 0$ there are $5$ thousand bacteria, and when $t = 3$ there are $17$ thousand. \n (a) Find $k$ and $a$ . \n (b) State the growth factor as a percentage increase per hour. \n (c) Predict the number of bacteria after $10$ hours. [6]

1
(a) Find the parameters. At $t = 0$ , $N = k a^0 = k$ , so $k = 5$ . [A1]\nUsing the second point in $N(t) = 5a^t$ : $17 = 5a^3$ , so $a^3 = \dfrac{17}{5} = 3.4$ . [M1: substitute the second point]\n $a = 3.4^{1/3} = 1.5036\ldots = 1.50$ (3 s.f.). [A1]\n(On the GDC, exponential regression on the two points returns the same $k$ and $a$ .)\n\n**(b) Interpret the factor.** $a = 1.50$ means the colony multiplies by about $1.50$ each hour — a growth of about $50\%$ per hour. [A1]\n\n**(c) Predict.** Keep the full stored value of $a$ and evaluate $N(10) = 5 \times (3.4^{1/3})^{10}$ . [M1: substitute $t = 10$ with stored $a$ ]\n $N(10) = 5 \times 3.4^{10/3} = 289.6\ldots$ \nSo about $290$ thousand bacteria after $10$ hours (3 s.f.). [A1]\n\nUsing the rounded $a = 1.50$ gives $\approx 288$ thousand — close, but a reminder to carry the unrounded factor to the final line.

✓ FULLY WORKED

EXAMPLE 2

The time $y$ minutes a trainee takes to assemble a device on their $x$ th attempt is modelled by $y = a\ln x + b$ . On the 1st attempt ( $x = 1$ ) the time is $40$ minutes, and on the 10th attempt it is $22$ minutes. \n (a) Find $a$ and $b$ . \n (b) Estimate the time on the 20th attempt. [5]

1
(a) Find the parameters. At $x = 1$ , $\ln 1 = 0$ , so $y = a(0) + b = b$ ; hence $b = 40$ . [A1]\nAt $x = 10$ : $22 = a\ln 10 + 40$ . [M1: substitute the second point]\n $a\ln 10 = 22 - 40 = -18$ , so $a = \dfrac{-18}{\ln 10} = -7.817\ldots = -7.82$ (3 s.f.). [A1]\nThe model is $y = -7.82\ln x + 40$ (negative $a$ : the time falls as attempts rise).\n\n**(b) Predict.** Using the stored value of $a$ : $y = a\ln 20 + 40 = -7.817\ldots \times \ln 20 + 40$ . [M1: substitute $x = 20$ ]\n $y = -7.817\ldots \times 2.9957\ldots + 40 = 16.5\ldots$ \nSo about $16.6$ minutes on the 20th attempt (3 s.f.). [A1]\n\nNotice the flattening: ten attempts cut the time by $18$ minutes, but the next ten cut it by only about $5$ — the hallmark of a logarithmic model.

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EXAMPLE 3

A radioactive sample has mass modelled by $m(t) = 80e^{-0.03t}$ grams, where $t$ is the time in years. \n (a) Find the mass after $25$ years. \n (b) Find the half-life of the sample (the time for the mass to fall to $40$ g). [5]

1
(a) Evaluate the model. Substitute $t = 25$ : $m(25) = 80e^{-0.03 \times 25} = 80e^{-0.75}$ . [M1: substitute $t = 25$ ]\n $m(25) = 80 \times 0.47236\ldots = 37.789\ldots$ \nSo about $37.8$ g after $25$ years (3 s.f.). [A1]\n\n**(b) Solve for the time.** Set $m(t) = 40$ : $40 = 80e^{-0.03t}$ . [M1: set the model equal to the target]\nBy logs: $e^{-0.03t} = 0.5$ , so $-0.03t = \ln 0.5$ , giving $t = \dfrac{\ln 0.5}{-0.03} = 23.104\ldots$ .\nBy GDC: solve $80e^{-0.03t} = 40$ (solver) or intersect $y = 80e^{-0.03x}$ with $y = 40$ ; the result is the same. [A1]\nThe half-life is about $23.1$ years (3 s.f.). [A1]\n\nBecause $80e^{-0.03t} = 40$ is exactly $e^{-0.03t} = \tfrac{1}{2}$ , the half-life does not depend on the starting mass $80$ — a defining feature of exponential decay.

✓ FULLY WORKED

EXAMPLE 4

A cup of coffee cools according to $T(t) = 68e^{-0.05t} + 20$ (temperature in °C, $t$ in minutes). State the room temperature, find the temperature after $10$ minutes, and find the time for the coffee to reach $40$ °C. [5]

1
Model answer — full working.\n\nRoom temperature. As $t \to \infty$ , the term $68e^{-0.05t} \to 0$ , so $T \to 20$ . The horizontal asymptote $y = 20$ is the limiting value, so the room temperature is $20$ °C.\n\nTemperature after 10 minutes. Substitute $t = 10$ :\n $T(10) = 68e^{-0.05 \times 10} + 20 = 68e^{-0.5} + 20.$ \n $68e^{-0.5} = 68 \times 0.60653\ldots = 41.24\ldots$ , so $T(10) = 41.24\ldots + 20 = 61.24\ldots$ \nThe temperature after $10$ minutes is about $61.2$ °C (3 s.f.).\n\nTime to reach 40 °C. Set $T(t) = 40$ :\n $40 = 68e^{-0.05t} + 20.$ \nSubtract the limit first: $20 = 68e^{-0.05t}$ , so $e^{-0.05t} = \dfrac{20}{68} = 0.2941\ldots$ \nBy logs: $-0.05t = \ln(0.2941\ldots)$ , so $t = \dfrac{\ln(0.2941\ldots)}{-0.05} = 24.47\ldots$ \nBy GDC: solve $68e^{-0.05t} + 20 = 40$ , or intersect $y = 68e^{-0.05x} + 20$ with $y = 40$ — the same value.\nThe coffee reaches $40$ °C after about $24.9$ minutes (3 s.f.).\n\n---\nHow our marking engine awards the 5 marks:\n\n- A1 — room temperature. Awarded for $20$ °C, identified as the $+20$ horizontal asymptote / limiting value. The engine accepts the answer whether justified as 'the asymptote $y = 20$ ' or 'as $t \to \infty$ , $T \to 20$ '.\n- M1 — substitute $t = 10$ . A method mark for correctly substituting into the model, i.e. writing $T(10) = 68e^{-0.05 \times 10} + 20$ (or $68e^{-0.5} + 20$ ). It is the substitution that is rewarded, so it survives an arithmetic slip on the next line.\n- A1 — value at 10 minutes. Awarded for $61.2$ °C (accept $61.24\ldots$ correctly rounded). This accuracy mark depends on the M1 above, and is FT: a candidate who substituted correctly but slipped in arithmetic still earns it if their value follows correctly from their line.\n- M1 — set $T = 40$ and solve. A method mark for forming $68e^{-0.05t} + 20 = 40$ and moving to solve it — either isolating the exponential and taking logs, or setting up the GDC solver / intersection. Both routes earn this single method mark equally.\n- A1 — time. Awarded for $t \approx 24.9$ minutes (accept $24.47\ldots$ / $24.5$ from exact working, or the GDC value, correctly rounded). This A-mark depends on the M1 solve and is FT on the candidate's own earlier figures.\n\n**'Accept equivalent forms and correct rounding.'** The engine accepts the time whether written as $\frac{\ln(20/68)}{-0.05}$ , or $24.5$ , or $24.9$ from a GDC solve, provided the rounding is consistent, and it accepts the value at 10 minutes as $61.2$ or $61.24$ . Once a correct final value appears, later restatements do not lose marks.\n\nBottom line: of the 5 marks, two are method marks that survive an arithmetic slip, and the accuracy marks are shielded by follow-through — but only if the substitution and the 'set equal and solve' lines are on the page. A student who writes just ' $61$ °C' and ' $25$ min' with no working risks the two method marks; a student who shows the substitution, isolates the exponential and states the equation solved keeps the method regardless of a slip in the final number.

✓ FULLY WORKED

How it all connects

The big idea sits in the middle — tap a linked idea to explore the link.

MAIN IDEAModelling with exponential and logarithmic functions

Tap a linked idea to see how it connects back to the main topic — that connection is what examiners reward.

Glossary

Try to recall each definition before you reveal it.

Quick check

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Revision flashcards

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QUESTION · 1 / 12

Exponential model $y = ka^x$

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ANSWER

$k$ is the initial value (the value at $x = 0$ , since $a^0 = 1$ ) and $a$ is the growth or decay factor per unit of $x$ . If $a > 1$ the quantity grows; if $0 < a < 1$ it decays.

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1 / 12 · ← → space

Key takeaways

Review these before you close the topic — retrieval beats re-reading.

✓
$k$ — the amplitude / initial size. In $y = ka^x$ the initial value (at $x = 0$ ) is $k$ . In $y = ke^{rx} + c$ the initial value is $k + c$ , because $e^0 = 1$ .
✓
$a$ or $r$ — the growth/decay control. $a > 1$ (equivalently $r > 0$ ) means growth; $0 < a < 1$ (equivalently $r < 0$ ) means decay. They convert by $a = e^r$ .
✓
$c$ — the limiting value. The horizontal asymptote is $y = c$ : as $x \to \infty$ in a decay model the term $ke^{rx} \to 0$ and $y \to c$ . This floor is room temperature, a baseline concentration, a background level.
✓
The gap shrinks by a fixed fraction. In a decay-to-a-limit model the distance $y - c$ is itself a pure exponential $ke^{rx}$ — the quantity closes the same fraction of its remaining gap each period.

Practice — then mark it

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Get a calculator-active question marked: fit the model, read the limit, and solve for the time with full working

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Modelling with exponential and logarithmic functions

BY THE END, YOU CAN…

1Recognise and use the exponential growth and decay models $y = ka^x$ and $y = ke^{rx} + c$ , and identify the horizontal asymptote $y = c$ as a limiting value.
2Interpret each parameter in context: the initial value, the growth or decay factor $a$ , the continuous rate $r$ , and the limit $c$ .
3Fit an exponential or logarithmic model to data using the GDC, from two points or by regression.
4Solve for the time to reach a given value, either by taking logarithms or by using the GDC solver or graph.
5Apply exponential and logarithmic models to population, cooling, radioactive decay and medication problems, presenting working so method marks and follow-through protect the final answer.

In simple terms

A friendly intro before the formal notes — no formulas yet.

Modelling change that speeds up or levels off

THINK OF IT LIKE…

1
Decide the shape from the story: a fixed percentage or factor per period, or an approach to a floor, is exponential; a quantity that shoots up then flattens is logarithmic.
2
Write the general form — $y = ka^x$ , or $y = ke^{rx} + c$ when there is a non-zero limit, or $y = a\ln x + b$ — and note what each letter must mean here.
3
Find the parameters: substitute known points, or use the GDC's exponential or logarithmic regression, keeping full accuracy in the calculator.
4
Use the model: substitute a time to predict a value, or solve for the time to reach a value by taking logs or using the GDC solver or intersection.

Explore the concept

Use the live diagram and synced steps — play it or tap a step card to walk through.

Key formulas

Tap any symbol to reveal exactly what it means and its units.

$y = ka^x$

$y = a\ln x + b$

$y = ke^{rx}$

Full topic notes

Formal explanation with the rigour you need for the exam.

Exponential models: growth, decay and a limiting value

Growth / decay factor form: $y = ka^x$ . \n Continuous form with a limit: $y = ke^{rx} + c$ .

$k$ — the amplitude / initial size. In $y = ka^x$ the initial value (at $x = 0$ ) is $k$ . In $y = ke^{rx} + c$ the initial value is $k + c$ , because $e^0 = 1$ .
$a$ or $r$ — the growth/decay control. $a > 1$ (equivalently $r > 0$ ) means growth; $0 < a < 1$ (equivalently $r < 0$ ) means decay. They convert by $a = e^r$ .
$c$ — the limiting value. The horizontal asymptote is $y = c$ : as $x \to \infty$ in a decay model the term $ke^{rx} \to 0$ and $y \to c$ . This floor is room temperature, a baseline concentration, a background level.
The gap shrinks by a fixed fraction. In a decay-to-a-limit model the distance $y - c$ is itself a pure exponential $ke^{rx}$ — the quantity closes the same fraction of its remaining gap each period.

Fitting a model to data with the GDC

Logarithmic models

Logarithmic model: $y = a\ln x + b$ , defined for $x > 0$ .

Solving for the time to reach a value

For $y = ke^{rx}$ , taking logs gives $x = \dfrac{1}{r}\ln\!\left(\dfrac{y}{k}\right)$ . \n For $y = ke^{rx} + c$ , first subtract $c$ : $y - c = ke^{rx}$ , then take logs.

Common mistakes examiners penalise

Treating $c$ as zero when the data has a floor — in $y = ke^{rx} + c$ the limiting value is $y = c$ , the horizontal asymptote. Cooling and medication problems level off above zero; identify $c$ first or every parameter is wrong.
Misreading the initial value with a shift — for $y = ke^{rx} + c$ the value at $t = 0$ is $k + c$ , not $k$ , because $e^0 = 1$ . Only in $y = ka^x$ is the initial value simply $k$ .
Confusing growth and decay — $a > 1$ or $r > 0$ is growth; $0 < a < 1$ or $r < 0$ is decay. A negative sign in the exponent means the quantity falls; do not report a decaying model as growing.
Rounding parameters mid-calculation — carry the GDC's full values of $k$ , $a$ , $r$ and $c$ ; round only the final answer to 3 significant figures. A rate rounded early can shift a time answer by whole units.
Not showing the method when solving for time — write the equation you set equal to the target and either the $\ln$ step or the GDC equation solved. An answer with no method risks the method mark even if the number is right.
Ignoring the domain of a logarithmic model — $\ln x$ needs $x > 0$ (and $\ln(x - c)$ needs $x > c$ ). Feeding the GDC a value outside the domain produces an error, not a solution.
Forgetting to subtract the limit before taking logs — to solve $y = ke^{rx} + c$ you must isolate the exponential first: $y - c = ke^{rx}$ , then take logs. Taking logs of $y$ directly is wrong.
Answering the wrong question — 'find the value after $t$ ' means substitute; 'find when it reaches a value' means solve. Substituting when a solve is asked (or vice versa) throws away the whole part.

Model answer — marked the way our engine marks it

Where this leads

Worked examples

See the formulas applied — reveal one step at a time, like the exam.

EXAMPLE 1

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(a) Find the parameters. At $t = 0$ , $N = k a^0 = k$ , so $k = 5$ . [A1]\nUsing the second point in $N(t) = 5a^t$ : $17 = 5a^3$ , so $a^3 = \dfrac{17}{5} = 3.4$ . [M1: substitute the second point]\n $a = 3.4^{1/3} = 1.5036\ldots = 1.50$ (3 s.f.). [A1]\n(On the GDC, exponential regression on the two points returns the same $k$ and $a$ .)\n\n**(b) Interpret the factor.** $a = 1.50$ means the colony multiplies by about $1.50$ each hour — a growth of about $50\%$ per hour. [A1]\n\n**(c) Predict.** Keep the full stored value of $a$ and evaluate $N(10) = 5 \times (3.4^{1/3})^{10}$ . [M1: substitute $t = 10$ with stored $a$ ]\n $N(10) = 5 \times 3.4^{10/3} = 289.6\ldots$ \nSo about $290$ thousand bacteria after $10$ hours (3 s.f.). [A1]\n\nUsing the rounded $a = 1.50$ gives $\approx 288$ thousand — close, but a reminder to carry the unrounded factor to the final line.

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EXAMPLE 2

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(a) Find the parameters. At $x = 1$ , $\ln 1 = 0$ , so $y = a(0) + b = b$ ; hence $b = 40$ . [A1]\nAt $x = 10$ : $22 = a\ln 10 + 40$ . [M1: substitute the second point]\n $a\ln 10 = 22 - 40 = -18$ , so $a = \dfrac{-18}{\ln 10} = -7.817\ldots = -7.82$ (3 s.f.). [A1]\nThe model is $y = -7.82\ln x + 40$ (negative $a$ : the time falls as attempts rise).\n\n**(b) Predict.** Using the stored value of $a$ : $y = a\ln 20 + 40 = -7.817\ldots \times \ln 20 + 40$ . [M1: substitute $x = 20$ ]\n $y = -7.817\ldots \times 2.9957\ldots + 40 = 16.5\ldots$ \nSo about $16.6$ minutes on the 20th attempt (3 s.f.). [A1]\n\nNotice the flattening: ten attempts cut the time by $18$ minutes, but the next ten cut it by only about $5$ — the hallmark of a logarithmic model.

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EXAMPLE 3

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(a) Evaluate the model. Substitute $t = 25$ : $m(25) = 80e^{-0.03 \times 25} = 80e^{-0.75}$ . [M1: substitute $t = 25$ ]\n $m(25) = 80 \times 0.47236\ldots = 37.789\ldots$ \nSo about $37.8$ g after $25$ years (3 s.f.). [A1]\n\n**(b) Solve for the time.** Set $m(t) = 40$ : $40 = 80e^{-0.03t}$ . [M1: set the model equal to the target]\nBy logs: $e^{-0.03t} = 0.5$ , so $-0.03t = \ln 0.5$ , giving $t = \dfrac{\ln 0.5}{-0.03} = 23.104\ldots$ .\nBy GDC: solve $80e^{-0.03t} = 40$ (solver) or intersect $y = 80e^{-0.03x}$ with $y = 40$ ; the result is the same. [A1]\nThe half-life is about $23.1$ years (3 s.f.). [A1]\n\nBecause $80e^{-0.03t} = 40$ is exactly $e^{-0.03t} = \tfrac{1}{2}$ , the half-life does not depend on the starting mass $80$ — a defining feature of exponential decay.

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EXAMPLE 4

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Model answer — full working.\n\nRoom temperature. As $t \to \infty$ , the term $68e^{-0.05t} \to 0$ , so $T \to 20$ . The horizontal asymptote $y = 20$ is the limiting value, so the room temperature is $20$ °C.\n\nTemperature after 10 minutes. Substitute $t = 10$ :\n $T(10) = 68e^{-0.05 \times 10} + 20 = 68e^{-0.5} + 20.$ \n $68e^{-0.5} = 68 \times 0.60653\ldots = 41.24\ldots$ , so $T(10) = 41.24\ldots + 20 = 61.24\ldots$ \nThe temperature after $10$ minutes is about $61.2$ °C (3 s.f.).\n\nTime to reach 40 °C. Set $T(t) = 40$ :\n $40 = 68e^{-0.05t} + 20.$ \nSubtract the limit first: $20 = 68e^{-0.05t}$ , so $e^{-0.05t} = \dfrac{20}{68} = 0.2941\ldots$ \nBy logs: $-0.05t = \ln(0.2941\ldots)$ , so $t = \dfrac{\ln(0.2941\ldots)}{-0.05} = 24.47\ldots$ \nBy GDC: solve $68e^{-0.05t} + 20 = 40$ , or intersect $y = 68e^{-0.05x} + 20$ with $y = 40$ — the same value.\nThe coffee reaches $40$ °C after about $24.9$ minutes (3 s.f.).\n\n---\nHow our marking engine awards the 5 marks:\n\n- A1 — room temperature. Awarded for $20$ °C, identified as the $+20$ horizontal asymptote / limiting value. The engine accepts the answer whether justified as 'the asymptote $y = 20$ ' or 'as $t \to \infty$ , $T \to 20$ '.\n- M1 — substitute $t = 10$ . A method mark for correctly substituting into the model, i.e. writing $T(10) = 68e^{-0.05 \times 10} + 20$ (or $68e^{-0.5} + 20$ ). It is the substitution that is rewarded, so it survives an arithmetic slip on the next line.\n- A1 — value at 10 minutes. Awarded for $61.2$ °C (accept $61.24\ldots$ correctly rounded). This accuracy mark depends on the M1 above, and is FT: a candidate who substituted correctly but slipped in arithmetic still earns it if their value follows correctly from their line.\n- M1 — set $T = 40$ and solve. A method mark for forming $68e^{-0.05t} + 20 = 40$ and moving to solve it — either isolating the exponential and taking logs, or setting up the GDC solver / intersection. Both routes earn this single method mark equally.\n- A1 — time. Awarded for $t \approx 24.9$ minutes (accept $24.47\ldots$ / $24.5$ from exact working, or the GDC value, correctly rounded). This A-mark depends on the M1 solve and is FT on the candidate's own earlier figures.\n\n**'Accept equivalent forms and correct rounding.'** The engine accepts the time whether written as $\frac{\ln(20/68)}{-0.05}$ , or $24.5$ , or $24.9$ from a GDC solve, provided the rounding is consistent, and it accepts the value at 10 minutes as $61.2$ or $61.24$ . Once a correct final value appears, later restatements do not lose marks.\n\nBottom line: of the 5 marks, two are method marks that survive an arithmetic slip, and the accuracy marks are shielded by follow-through — but only if the substitution and the 'set equal and solve' lines are on the page. A student who writes just ' $61$ °C' and ' $25$ min' with no working risks the two method marks; a student who shows the substitution, isolates the exponential and states the equation solved keeps the method regardless of a slip in the final number.

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MAIN IDEAModelling with exponential and logarithmic functions

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QUESTION · 1 / 12

Exponential model $y = ka^x$

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ANSWER

$k$ is the initial value (the value at $x = 0$ , since $a^0 = 1$ ) and $a$ is the growth or decay factor per unit of $x$ . If $a > 1$ the quantity grows; if $0 < a < 1$ it decays.

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Key takeaways

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$k$ — the amplitude / initial size. In $y = ka^x$ the initial value (at $x = 0$ ) is $k$ . In $y = ke^{rx} + c$ the initial value is $k + c$ , because $e^0 = 1$ .
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$a$ or $r$ — the growth/decay control. $a > 1$ (equivalently $r > 0$ ) means growth; $0 < a < 1$ (equivalently $r < 0$ ) means decay. They convert by $a = e^r$ .
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$c$ — the limiting value. The horizontal asymptote is $y = c$ : as $x \to \infty$ in a decay model the term $ke^{rx} \to 0$ and $y \to c$ . This floor is room temperature, a baseline concentration, a background level.
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The gap shrinks by a fixed fraction. In a decay-to-a-limit model the distance $y - c$ is itself a pure exponential $ke^{rx}$ — the quantity closes the same fraction of its remaining gap each period.

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Revise Modelling with exponential and logarithmic functions for IB Mathematics: Applications and Interpretation (2.4)

In simple terms

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Key formulas

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Exponential models: growth, decay and a limiting value

Fitting a model to data with the GDC

Logarithmic models

Solving for the time to reach a value

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Model answer — marked the way our engine marks it

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In simple terms

Explore the concept

Key formulas

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Exponential models: growth, decay and a limiting value

Fitting a model to data with the GDC

Logarithmic models

Solving for the time to reach a value

Common mistakes examiners penalise

Model answer — marked the way our engine marks it

Where this leads

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Discuss Modelling with exponential and logarithmic functions