AP®MathsCollege BoardPrecalculusRevision NotesExponential & Logarithmic FunctionsModeling with Exponential & Logarithmic FunctionsApplying Exponential Models

Applying Exponential Models (College Board AP® Precalculus): Revision Note

Written by: Roger B

Reviewed by: Mark Curtis

Updated on 13 April 2026

Applying exponential models

What does the base of an exponential model tell us?

For an exponential model in general form $f (x) = a b^{x}$
- the base $b$ is the growth factor
  - it tells you by what factor the output is multiplied each time the input increases by 1 unit
The growth factor is directly related to the percent change per unit
- If $b > 1$ , the quantity is growing
  - and the percent increase per unit is $(b - 1) \times 100 %$
  - E.g. $b = 1.08$ means an 8% increase per unit input
- If $0 < b < 1$ , the quantity is decaying
  - and the percent decrease per unit is $(1 - b) \times 100 %$
  - E.g. $b = 0.95$ means a 5% decrease per unit input
It is important to remember that the growth factor applies per unit of the input variable
- E.g. if the model is $P (t) = 500 (1.03)^{t}$
- and $t$ is in years
  - then the quantity grows by 3% per year

If you need the growth rate per a different time period, you need to rewrite the model (see below)

How do I handle growth rates over non-unit intervals?

A common exam scenario is when the growth rate is given per one time unit but the model uses a different time unit
- The key is to express the exponent so that it counts the number of growth periods

E.g. a population grows by 5% per month, and $t$ is measured in years
- The monthly growth factor is $1.05$
  - and in $t$ years there are $12 t$ months
- So the model is $P (t) = P_{0} (1.05)^{12 t}$

E.g. a substance decays by 10% per decade, and $t$ is measured in years
- The decay factor per decade is $0.90$
  - and in $t$ years there are $\frac{t}{10}$ decades
- So the model is $A (t) = A_{0} (0.90)^{\frac{t}{10}}$

Examiner Tips and Tricks

Remember that the growth factor for a percent increase of $r %$ is $1 + \frac{r}{100}$ , not $\frac{r}{100}$ .

E.g. a 6.1% increase per quarter means the growth factor is $1.061$ , not $0.061$

Similarly the decay factor for a percent decrease of $r %$ is $1 - \frac{r}{100}$ .

A common error is confusing the exponent direction
- E.g. if the rate is per quarter and $t$ is in years the exponent should be $4 t$ (more growth periods per year), not $t / 4$
  - $t / 4$ would be correct if the rate were per year and $t$ were per quarter

Examiner Tips and Tricks

When a question gives a percent change per some time period and asks for a model in a different time unit, the most common error is getting the exponent backwards (e.g. writing $t / 4$ instead of $4 t$ , or vice versa).

A quick check is to substitute a simple value of $t$ and see if the number of growth periods makes sense.

E.g., if $t = 1$ year and growth is per quarter, there should be 4 growth periods
So the exponent should equal 4

How can equivalent forms of an exponential function reveal different information?

The same exponential function can be written in equivalent forms that highlight different properties
- The general principle is that $b^{x} = (b^{k})^{x / k}$
  - for any nonzero constant $k$
This means you can rewrite the base to show the growth factor over any time interval

E.g. $f (d) = 2^{d}$ (quantity doubles every day, where $d$ is in days)
- Weekly form: $f (d) = (2^{7})^{d / 7} = 128^{d / 7}$
  - the quantity grows by a factor of 128 every 7 days (every week)
- Hourly form: $f (d) = (2^{1 / 24})^{24 d}$
  - the quantity grows by a factor of $2^{1 / 24}$ every hour

E.g. A radioactive substance has a half-life of 8 days
- This means that every 8 days the initial amount is multiplied by $\frac{1}{2} = 0.5$
- The decay model is $h (d) = A_{0} (0.5)^{d / 8}$ , where $d$ is in days
  - Every 8 days, the amount is multiplied by 0.5
- To convert to $t$ measured in hours
  - There are 24 hours in a day
    - so $t = 24 d \Rightarrow d = t / 24$
  - Therefore
    - $k (t) = A_{0} (0.5)^{(t / 24) / 8} = A_{0} (0.5)^{t / 192} = A_{0} {(0 . 5^{1 / 192})}^{t}$
  - This shows the decay factor per hour is $0 . 5^{1 / 192}$
- To convert to $w$ measured in weeks
  - There are 7 days in a week
    - so $d = 7 w$
  - Therefore
    - $h (w) = A_{0} (0.5)^{7 w / 8}$
  - This shows the decay factor per week is $0 . 5^{7 / 8}$

In general, converting between time units involves
- substituting the relationship between the units into the exponent
  - then simplifying

Examiner Tips and Tricks

For unit conversion problems, always verify your answer by checking that the original rate is preserved.

E.g. if the half-life is 8 days, substituting 8 days' worth of the new time unit into your converted model should give a decay factor of 0.5

How do I use an exponential model to make predictions?

Once you have an exponential model, you can predict output values
- by substituting input values into the function
You can also predict input values
- by substituting output values into the function
- and solving to find the corresponding input value
There are a number of things to keep in mind
- Predictions are only meaningful within the contextual constraints of the domain
  - E.g. if $t$ represents time since an event, negative values of $t$ may not make sense
- Models may not remain valid indefinitely
  - Real-world growth and decay often level off or change behavior over time
- The model gives only approximate predictions
  - especially when based on a regression or limited data
- The model may not remain valid beyond the range of data used to construct it
  - Just as with any model

Worked Example

The number of subscribers, in thousands, to a streaming service is modeled by the function $S$ . The number of subscribers is expected to increase by 3.5% each month. At time $t = 0$ years, the service had 200 thousand subscribers. If $t$ is measured in years, which of the following is an expression for $S (t)$ ? (Note: A month is one twelfth of a year.)

(A) $200 (0.035)^{(t / 12)}$

(B) $200 (0.035)^{(12 t)}$

(D) $200 (1.035)^{(12 t)}$

Answer:

The initial value is $S (0) = 200$

This occurs in all the answer options

A 3.5% increase each month means the monthly growth factor is

$1 + 0.035 = 1.035$

Since $t$ is measured in years, and there are 12 months in a year

there are $12 t$ months in $t$ years.

So the model should be

(D) $200 (1.035)^{(12 t)}$

Worked Example

A bacteria colony doubles in size every 6 hours. The number of bacteria can be modeled by the function $p$ given by $p (t) = N_{0} (2)^{t / 6}$ , where $N_{0}$ is the number of bacteria at time $t = 0$ and $t$ is the number of hours since $t = 0$ .

Which of the following functions $q$ models the number of bacteria remaining after $m$ minutes, where $N_{0}$ is the number of bacteria at time $m = 0$ ? (There are 60 minutes in an hour, so $m = 60 t$ .)

(A) $q (m) = N_{0} (2)^{m / 60}$

(B) $q (m) = N_{0} {(2^{1 / 60})}^{6 m}$

(D) $q (m) = N_{0} {(2^{1 / 360})}^{m}$

Answer:

Write $t$ in terms of $m$

$m = 60 t \Rightarrow t = \frac{m}{60}$

Substitute that for $t$ in the original model:

$\begin{array}{rcl} q (m) & = & N_{0} (2)^{(m / 60) / 6} \\ = & N_{0} (2)^{m / 360} \\ = & N_{0} {(2^{1 / 360})}^{m} \end{array}$

That is answer is option (D)

You can check this answer quickly

The colony should double every 6 hours = 360 minutes
Using the model with $m = 360$

$\begin{array}{rcl} q (360) & = & N_{0} {(2^{1 / 360})}^{360} \\ = & N_{0} \cdot 2^{1} \\ = & 2 N_{0} \end{array}$

The initial quantity doubles in that time, as expected

(D) $q (m) = N_{0} {(2^{1 / 360})}^{m}$

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Applying Exponential Models (College Board AP® Precalculus): Revision Note

Applying exponential models

What does the base of an exponential model tell us?

How do I handle growth rates over non-unit intervals?

Examiner Tips and Tricks

Examiner Tips and Tricks

How can equivalent forms of an exponential function reveal different information?

Examiner Tips and Tricks

How do I use an exponential model to make predictions?

Worked Example

Worked Example

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Join the 100,000+ Students that ❤️ Save My Exams

Author: Roger B

Reviewer: Mark Curtis