Exponential Models (AI SL)
An exponential model describes a quantity that grows or shrinks by a constant percentage each time period, rather than by a constant amount - population growth, radioactive decay, compound interest and cooling all follow this pattern. This topic covers setting up and interpreting exponential models, reading off growth or decay rates, and solving for when the model reaches a target value.
What the syllabus says
This topic maps onto the Functions & Modelling unit of the official IB Applications & Interpretation syllabus.
| Code | Syllabus content |
|---|---|
| SL2.5 | One of several modelling types in this section: exponential growth and decay models \(f(x)=ka^x+c\), \(f(x)=ka^{-x}+c\) (for \(a>0\)), \(f(x)=ke^{rx}+c\), with equation of a horizontal asymptote. Contexts include population growth, radioactive decay, cooling of a liquid, spread of a virus, compound interest, depreciation and amortization. (This section also covers linear, quadratic, cubic, direct/inverse variation, and sinusoidal models.) |
| SL2.6 | Modelling skills: given a context, recognise and choose an appropriate model; determine a reasonable domain; find the parameters of a model; use technology to fit and test the model against data. |
Links to compound interest (SL1.4) and geometric sequences (SL1.3), since both describe repeated percentage change.
Key terms
Five words worth knowing cold before you touch the formulas below - each with a worked example showing exactly what it means.
What is an exponential model?
An exponential model is a function like \(f(x)=ka^x+c\) or \(f(x)=ke^{rx}+c\) in which the variable \(x\) appears in the exponent, so the quantity changes by a constant percentage each unit of \(x\), rather than by a constant amount.
e.g. \(N(t)=200(1.15)^t\) models a bacteria culture that grows by 15% every hour.
What is exponential growth vs decay?
Growth means the quantity increases without bound (base \(a>1\) or rate \(r>0\)); decay means it decreases toward the asymptote (base \(0 e.g. \(T(t)=22+58e^{-0.08t}\) decays because the exponent's coefficient \(-0.08\) is negative.
What is a horizontal asymptote?
A horizontal asymptote is the constant value \(y=c\) that an exponential model approaches but never reaches as \(x\) becomes large. It usually represents a stable long-term level in the real-world context.
e.g. For \(T(t)=22+58e^{-0.08t}\), the asymptote is \(T=22\)°C - the room temperature the coffee cools toward.
What is the growth or decay rate?
Writing the base as \(1+r\) (growth) or \(1-r\) (decay) lets you read the percentage change per time period straight off the equation, without any extra calculation.
e.g. In \(N(t)=200(1.15)^t\), the base \(1.15=1+0.15\) means 15% growth per hour.
How do you solve for time in an exponential model?
Isolate the exponential term, then take logarithms of both sides (or use the GDC's intersection or equation solver) to bring the unknown exponent down and solve for \(x\).
e.g. \(200(1.15)^t=1000\Rightarrow(1.15)^t=5\Rightarrow t=\dfrac{\ln5}{\ln1.15}\approx11.5\) hours.
Key formulas
A handful of results cover almost every exponential-modelling question. The tables below summarise them at a glance - the explanations underneath go into more depth on each one.
Formula reference
The general forms of the exponential model are standard results you're expected to know; solving for the exponent relies on the change of base / logarithm relationship, which is on the formula booklet.
| Formula | Used for | Booklet? |
|---|---|---|
| \(f(x)=ka^x+c\) | Growth/decay model, base form | Not in booklet |
| \(f(x)=ke^{rx}+c\) | Growth/decay model, rate form | Not in booklet |
| \(y=c\) | Equation of the horizontal asymptote | Not in booklet |
| \(a^x=b \iff x=\log_a b\) | Solving for an unknown exponent | ✓ Yes |
Base form vs rate form
The same exponential model can be written with a base \(a\) or a continuous rate \(r\) - each is more natural in different contexts.
| Feature | Base form \(ka^x+c\) | Rate form \(ke^{rx}+c\) |
|---|---|---|
| Reads directly as | Percentage change per period | Continuous growth/decay rate |
| Growth condition | \(a>1\) | \(r>0\) |
| Decay condition | \(0| \(r<0\) | |
| Typical context | Population growth, compound interest | Radioactive decay, cooling |
Interpreting the model's parameters
Each letter in an exponential model has a concrete meaning in context - questions often ask you to state these directly.
Initial value \(k\) (or \(k+c\))
Set \(x=0\) to find the starting value of the quantity, since \(a^0=1\) and \(e^0=1\).
\(f(0)=k+c\)
Long-term value \(c\)
As \(x\) grows large, \(a^x\to0\) (for decay) so \(f(x)\to c\) - the horizontal asymptote is the model's long-run limit.
\(y=c\)
Rate of change
The base \(a\) or exponent coefficient \(r\) sets how fast the quantity changes - write \(a\) as \(1\pm r\) to read off a percentage.
Growth: \(a>1\). Decay: \(0
Solving exponential equations
Most questions eventually ask you to find an \(x\)-value given a target \(y\)-value, or vice versa.
Algebraically with logs
Isolate the exponential term, then take \(\ln\) or \(\log\) of both sides to bring the exponent down.
Uses the change-of-base relationship (booklet)Graphically on the GDC
Graph the model and a horizontal line at the target value, then use the intersection tool to read off \(x\) directly.
GDC intersection toolWorked examples
Two full exam-style questions, marked exactly like the real thing. Try each one yourself before checking the worked solution.
A bacteria culture is modelled by \(N(t)=200(1.15)^t\) (\(t\) in hours).
(a) State the initial amount.
(b) Find the amount after 6 hours.
(c) State the hourly percentage growth rate.
(d) Find when \(N\) reaches \(1000.\)
Worked solution
(a) Initial amount. Set \(t=0\): \(N(0)=200(1.15)^0=200\). A1
(b) After 6 hours.
\(N(6)=200(1.15)^6=200(2.31306)\approx463.\) A1
(c) Hourly growth rate. The base \(1.15=1+0.15\), so the culture grows \(15\%\) per hour. A1
(d) \(200(1.15)^t=1000\Rightarrow(1.15)^t=5\Rightarrow t=\dfrac{\ln5}{\ln1.15}\) M1
\(\approx11.5\) hours. A1
A coffee cools as \(T(t)=22+58e^{-0.08t}\,^{\circ}\text{C}\), \(t\) in minutes.
(a) Find the initial temperature.
(b) Find the temperature after 10 minutes, to 1 decimal places.
(c) Find when it reaches \(40\,^{\circ}\text{C}\), to 3 significant figures.
(d) Find the average rate of cooling over the first 10 minutes.
Worked solution
(a) Initial temperature. \(T(0)=22+58e^{0}=22+58=80\)°C. A1
(b) After 10 minutes.
\(T(10)=22+58e^{-0.8}=22+58(0.449329)\approx48.1\text{°C}.\) A1
(c) When it reaches \(40\)°C. \(58e^{-0.08t}=18\Rightarrow e^{-0.08t}=0.310345\): M1
\(t=\frac{\ln 0.310345}{-0.08}\approx14.6\text{ minutes}.\) A1
(d) Average rate \(=\dfrac{48.1-80}{10}\) M1
\(\approx-3.19\)°C/min. A1
Common mistakes
The four slip-ups that account for most of the marks lost on this topic - worth reading before you start practising, not just after you get one wrong.
- Forgetting the constant \(c\) when finding the initial value or asymptote. \(f(0)=k+c\), not just \(k\) - and the asymptote is \(y=c\), not \(y=0\), unless \(c=0\).
- Reading the growth/decay rate straight off the base without subtracting 1. A base of \(1.15\) is 15% growth, not 115% - always write \(a=1+r\) first before quoting a percentage.
- Sign errors when isolating the exponential term before taking logs. Rearranging \(k+c-\text{value}\) incorrectly is a common slip that produces a negative or undefined logarithm.
- Treating the horizontal asymptote as a value the model reaches. The model gets arbitrarily close to \(y=c\) but never actually equals it - useful for explaining why a "time to reach the asymptote" question has no exact answer.
Using your GDC
Every step below is a real button sequence, not a vague "use your calculator" hint - covering the TI-84 Plus, TI-Nspire, and Casio fx-9860/fx-CG50. Pick your model to filter down to just the steps that apply to you.
The heart of AI modelling - find the best-fitting curve for a data set, not just a straight line.
- Enter the data in two lists (x and y).
- STAT → EDIT to enter L1/L2, then STAT → CALC → ExpReg.TI-84
- In a Lists & Spreadsheet page enter the data, then menu → Statistics → Stat Calculations → Exponential Regression.Nspire
- Main menu → Statistics, enter the data in lists, then CALC → REG and pick Exp.Casio
Tip: Turn DiagnosticOn (TI-84: 2nd → 0 → DiagnosticOn) to see \(R^2\). Choose the model with the best \(R^2\) that also makes sense for the context.
After fitting several models (linear, quadratic, exponential...), you need to decide which fits the data best - \(R^2\) is the key tool.
- Fit each candidate model in turn and note the \(R^2\) value each time.
- Turn DiagnosticOn first (2nd → 0, scroll to DiagnosticOn, ENTER) - then \(R^2\) appears after every regression.TI-84
- \(R^2\) is shown automatically after each regression calculation in the Statistics menu.Nspire
- \(R^2\) (displayed as r²) appears in the regression output; run CALC → REG for each model type and compare.Casio
- An exponential model with \(R^2=0.98\) is better than a linear model with \(R^2=0.91\) for the same data, but check the scatter plot too.
Tip: \(R^2\) alone doesn't tell you whether the model is appropriate - a high \(R^2\) on the wrong model type is meaningless.
See the full GDC guide for more calculator models and topics.
Ready to practise properly?
Exponential modelling questions, marked instantly like the real exam.
Quick answers
The questions students on this topic ask most often.
What's the difference between exponential growth and decay?
In \(f(x)=ka^x+c\), growth happens when the base \(a>1\) (the quantity increases over time) and decay happens when \(0
What does the horizontal asymptote mean in context?
The horizontal asymptote \(y=c\) is the value the model approaches but never reaches as \(x\) grows large - it often represents a stable long-term level, such as room temperature for a cooling drink or a maximum sustainable population.
How do I find when an exponential model reaches a target value?
Set the model equal to the target value and solve using logarithms, or graph the model together with a horizontal line at the target value and use your GDC's intersection tool - both are accepted methods.
Can percentage growth or decay rates be read directly from the equation?
Yes. Writing the base as \(1+r\) (growth) or \(1-r\) (decay) lets you read the percentage rate straight off - a base of \(1.15\) means 15% growth per time period, and a base of \(0.92\) means 8% decay per period.
Sub-topics
Exponential Models broken down into its individual skills, each with its own focused page.
Related topics
More Functions & Modelling topics from the same AI SL syllabus unit, in case you want to keep going.