AP Precalculus Unit 2 Notes: Exponential/Log Models, Semi-Log Graphs, and Compositions

Semi-Log Plots and Linearization

What a semi-log plot is (and why you should care)

A semi-log plot is a graph where one axis is scaled logarithmically and the other axis is scaled normally (linearly). You will most often see:

• Log-linear (semi-log) plot: $x$ axis is linear, $y$ axis is logarithmic.
• Less commonly (but very useful), the $x$ axis is logarithmic and $y$ is linear.

Semi-log plots matter because many real-world relationships are not linear in their raw form, but they become linear after a log-based transformation. When a relationship becomes linear, you can use powerful “line tools”: slope, intercepts, and the idea that “straight line = constant rate (in some sense).” In AP Precalculus, this is especially important for identifying and modeling exponential and logarithmic relationships from data.

A key theme is linearization: transforming variables so that a curve turns into a line.

The core idea of linearization

Linearization means rewriting a model so that it fits the linear form

$Y = mX + b$

where $X$ and $Y$ are transformed versions of your original variables. Once you have a line, you can:

• estimate parameters (like growth rate)
• interpret the slope/intercept in context
• check if a model is reasonable (does the transformed data look roughly linear?)

A common misconception is thinking the original data must look like a line if it’s exponential. Exponential data usually looks curved. The “straightness” shows up after the right transformation.

Linearizing exponential functions (log on the output)

Suppose the original relationship is exponential.

Case A: $y = a b^x$

Here $a > 0$ and $b > 0$, $b \neq 1$.

Take a logarithm of both sides (any base works as long as you’re consistent). Using base 10 log for concreteness:

$\log(y) = \log(a b^x)$

Use log properties:

$\log(y) = \log(a) + \log(b^x)$

$\log(y) = \log(a) + x\log(b)$

Now define the transformed variables:

• $Y = \log(y)$
• $X = x$

Then you have a linear equation:

$Y = (\log(b))X + \log(a)$

So on a plot of $\log(y)$ versus $x$:

• slope $m = \log(b)$
• intercept $b_0 = \log(a)$

To recover the original parameters:

$a = 10^{b_0}$

$b = 10^{m}$

If instead you use natural log, you get the same structure with $\ln$. The slope becomes $\ln(b)$ and the intercept becomes $\ln(a)$.

Case B: $y = a e^{kx}$

This is another common exponential form, especially in science.

Take natural log:

$\ln(y) = \ln(a e^{kx})$

$\ln(y) = \ln(a) + kx$

Let $Y = \ln(y)$ and $X = x$:

$Y = kX + \ln(a)$

So on a plot of $\ln(y)$ versus $x$:

• slope is $k$
• intercept is $\ln(a)$

and you recover $a$ via:

$a = e^{\text{intercept}}$

What “semi-log” means graphically

If you literally use graph paper (or software) with a logarithmic $y$-axis, then equal vertical steps represent equal _multiplicative_ changes in $y$. That matches exponential behavior: exponential growth adds a constant amount to the _log_ of the output per unit $x$.

Linearizing logarithmic functions (log on the input)

A typical logarithmic model is

$y = A + B\ln(x)$

with domain $x > 0$.

This model becomes linear if you treat $\ln(x)$ as the input variable. Let:

• $X = \ln(x)$
• $Y = y$

Then:

$Y = A + BX$

Graphically, this corresponds to “log-scaling the $x$ axis” (or computing $\ln(x)$ values and plotting $y$ against them). A common mistake is trying to take $\ln(y)$ here; that would linearize an exponential model, not a logarithmic one.

Worked example 1: Identify an exponential model using linearization

You measure a bacteria culture at times $x$ (hours) and population $y$:

Step 1: Look for a constant ratio (exponential clue).

Compute successive ratios:

• $260/200 = 1.3$
• $338/260 = 1.3$
• $439.4/338 = 1.3$

Constant ratio suggests $y = a b^x$ with $b = 1.3$.

Step 2: Find $a$ from $x = 0$.

If $y = a b^x$, then at $x = 0$:

$y(0) = a b^0 = a$

So $a = 200$.

Model:

$y = 200(1.3)^x$

Connection to semi-log plotting: If you plotted $\log(y)$ versus $x$, you would get a line with slope $\log(1.3)$.

Worked example 2: Linearize a logarithmic relationship

Suppose a sound intensity model is

$y = 20 + 5\ln(x)$

where $x > 0$.

If you define $X = \ln(x)$ and $Y = y$, then

$Y = 20 + 5X$

So a plot of $y$ versus $\ln(x)$ should look like a straight line with slope 5 and intercept 20.

How to choose the right linearization

A practical approach with data is:

• If equal increases in $x$ multiply $y$ by a roughly constant factor, try an exponential model and plot $\ln(y)$ versus $x$.
• If equal multiplicative changes in $x$ (like doubling $x$) add a roughly constant amount to $y$, try a logarithmic model and plot $y$ versus $\ln(x)$.

Notation reference: common log choices

Different log bases change slopes/intercepts numerically but not the underlying linearity.

Because different classes/texts use $\log$ differently, always check what your calculator/software means by $\log$ (often base 10) and $\ln$ (base $e$).

Exam Focus

• Typical question patterns
  • You’re given a semi-log plot or a line in transformed variables and asked to determine whether the original relationship is exponential and to find the model parameters.
  • You’re asked to interpret slope/intercept on a plot of $\ln(y)$ vs $x$ (or $\log(y)$ vs $x$) in context.
  • You’re asked which transformation (logging $x$ or $y$) would linearize a given type of model.
• Common mistakes
  • Taking $\ln$ of the wrong variable (using $\ln(y)$ when you need $\ln(x)$, or vice versa).
  • Forgetting to transform back (leaving the model as $\ln(y) = mx + b$ instead of solving for $y$).
  • Interpreting slope incorrectly when the log base changes (the line is still linear, but the slope value depends on the chosen base).

Modeling with Exponential and Logarithmic Functions

Modeling as a process (not just “pick a formula”)

A mathematical model is a function that approximates a real relationship between quantities. In this unit, you mainly model situations where change is multiplicative (exponential) or where growth slows down as the input increases (logarithmic).

Modeling matters because it lets you predict, estimate, and interpret. But AP Precalculus expects more than plugging numbers into a formula: you need to justify why a certain model type makes sense and explain what parameters mean.

A strong modeling workflow looks like this:

1. Identify variables and units.
2. Choose a model family (exponential or logarithmic) based on the context and/or data behavior.
3. Determine parameters (using points, ratios, or transformed linear relationships).
4. Interpret parameters in context.
5. Check reasonableness (does the model fit the data and the situation? domain restrictions?).

Exponential modeling

What exponential behavior means

Exponential models are appropriate when a quantity changes by a constant percent rate per equal step in $x$. That “constant percent” idea is the conceptual heart: exponential change is multiplicative.

Two common forms are:

$y = a b^x$

and

$y = a e^{kx}$

Here:

• $a$ is the initial value when $x = 0$.
• $b$ is the growth factor per unit of $x$.
  • If $b > 1$, it’s growth.
  • If $0 < b < 1$, it’s decay.
• $k$ is a growth/decay constant in the base-$e$ form.
  • If $k > 0$, growth.
  • If $k < 0$, decay.

A key connection between the forms is:

$b = e^k$

and equivalently:

$k = \ln(b)$

Finding an exponential model from two points

Suppose you assume $y = a b^x$ and you’re given two points $(x_1, y_1)$ and $(x_2, y_2)$ with $x_1 \neq x_2$.

You have:

$y_1 = a b^{x_1}$

$y_2 = a b^{x_2}$

Divide the equations to eliminate $a$:

$\frac{y_2}{y_1} = b^{x_2 - x_1}$

Solve for $b$:

$b = \left(\frac{y_2}{y_1}\right)^{\frac{1}{x_2 - x_1}}$

Then plug back in to find $a$:

$a = \frac{y_1}{b^{x_1}}$

A common mistake is to treat $b$ as $y_2/y_1$ even when $x_2 - x_1$ is not 1. The exponent $x_2 - x_1$ matters.

Worked example 1: Build an exponential model from two data points

A medication amount in the bloodstream is 80 mg at time 0 hours and 50 mg at time 3 hours. Model as exponential decay.

Let $y = a b^x$ where $x$ is hours.

From $x = 0$:

$80 = a b^0$

So $a = 80$.

Use the point $(3, 50)$:

$50 = 80 b^3$

$b^3 = \frac{50}{80} = 0.625$

$b = (0.625)^{\frac{1}{3}}$

So the model is:

$y = 80(0.625^{\frac{1}{3}})^x$

You could also rewrite with $k$ using $k = \ln(b)$ if needed.

Interpretation: Each hour, the amount is multiplied by $b$, where $b < 1$.

Logarithmic modeling

What logarithmic behavior means

Logarithmic models are useful when growth happens quickly at first and then slows down. Conceptually, logarithms convert multiplication into addition: they measure “how many multiplicative steps” you’ve taken.

A common log model is:

$y = A + B\ln(x)$

with $x > 0$.

• $A$ is the value when $\ln(x) = 0$, which occurs at $x = 1$.
  • So $y(1) = A$.
• $B$ controls how strongly $y$ responds to multiplicative changes in $x$.

A powerful interpretation is: increasing $x$ by a multiplicative factor changes $\ln(x)$ by an additive amount.

For example, multiplying $x$ by $e$ increases $\ln(x)$ by 1, so $y$ increases by $B$.

Finding parameters from two points

If you have two points $(x_1, y_1)$ and $(x_2, y_2)$ with $x_1 > 0$ and $x_2 > 0$, and assume

$y = A + B\ln(x)$

then:

$y_1 = A + B\ln(x_1)$

$y_2 = A + B\ln(x_2)$

Subtract to solve for $B$:

$y_2 - y_1 = B(\ln(x_2) - \ln(x_1))$

$B = \frac{y_2 - y_1}{\ln(x_2) - \ln(x_1)}$

Then find $A$:

$A = y_1 - B\ln(x_1)$

Common mistake: forgetting the domain restriction $x > 0$. If the context allows $x = 0$ (like “time since start”), a pure $\ln(x)$ model may need a horizontal shift, such as $\ln(x + c)$, to make sense.

Worked example 2: Fit a logarithmic model from two points

Suppose a learning score $y$ depends on hours practiced $x$ and data suggest diminishing returns. You are told $y = 60$ at $x = 1$ and $y = 70$ at $x = 4$. Model with $y = A + B\ln(x)$.

Use $x = 1$:

$60 = A + B\ln(1)$

$\ln(1) = 0$, so $A = 60$.

Use $x = 4$:

$70 = 60 + B\ln(4)$

$10 = B\ln(4)$

$B = \frac{10}{\ln(4)}$

Model:

$y = 60 + \frac{10}{\ln(4)}\ln(x)$

Interpretation: The score increases quickly early on, then each additional multiplicative increase in practice time yields smaller gains.

Using semi-log plots as evidence for a model choice

Semi-log plotting isn’t just “a trick to get a line.” It’s also a diagnostic tool:

• If $\ln(y)$ vs $x$ is roughly linear, that supports an exponential model.
• If $y$ vs $\ln(x)$ is roughly linear, that supports a logarithmic model.

In both cases, “roughly linear” is important: real data have noise. AP questions often expect you to recognize the pattern and proceed with a reasonable model, not demand perfect alignment.

Interpreting parameters in context (what AP tends to emphasize)

AP Precalculus often assesses whether you can translate parameters into plain-language meaning.

For $y = a b^x$:

• $a$: starting amount at $x = 0$.
• $b - 1$: percent change per 1 unit increase in $x$, expressed as a decimal.

For $y = a e^{kx}$:

• $k$: continuous growth/decay rate per unit of $x$ (larger magnitude means faster change).

For $y = A + B\ln(x)$:

• $A$: value at $x = 1$.
• $B$: change in $y$ when $x$ is multiplied by $e$.

A frequent error is interpreting $B$ as “change per 1 unit increase in $x$” in a log model. In a log model, the meaningful “step” is multiplicative changes in $x$, not additive ones.

Exam Focus

• Typical question patterns
  • Given a context and a small set of data, decide whether exponential or logarithmic modeling is appropriate and justify using ratios, transformations, or shape.
  • Given a transformed linear relationship (like $\ln(y) = mx + b$), write the original model and interpret $m$ and $b$.
  • Use a model to predict a value or solve for an input (for example, “when will the population reach 10,000?”).
• Common mistakes
  • Mixing up what indicates exponential vs logarithmic behavior (constant ratio vs diminishing returns).
  • Ignoring domain restrictions (especially $x > 0$ for log models).
  • Rounding too early when solving for parameters, which can noticeably change predictions.

Composition of Exponential and Logarithmic Functions

What composition is

Composition of functions means feeding the output of one function into another. If you have functions $f$ and $g$, then

$(f \circ g)(x) = f(g(x))$

Composition matters here because exponential and logarithmic functions are deeply linked: they are inverses (when bases match and domains are respected). Many “simplifications” and many equation-solving steps in this unit are really composition plus inverse-function reasoning.

Exponentials and logarithms as inverses

For base $b > 0$, $b \neq 1$:

• The exponential function is $f(x) = b^x$ (domain all real numbers, range positive real numbers).
• The logarithmic function is $f^{-1}(x) = \log_b(x)$ (domain positive real numbers, range all real numbers).

Being inverses means their compositions return the input (on the appropriate domain):

$b^{\log_b(x)} = x$

for $x > 0$, and

$\log_b(b^x) = x$

for all real $x$.

A very common mistake is forgetting the domain condition on $b^{\log_b(x)} = x$. Since $\log_b(x)$ only exists for $x > 0$, you cannot plug in a negative value of $x$ on that side.

Composition and solving equations

Many exponential/log equations are solved by applying a log or an exponential to both sides. That is essentially composing both sides with an inverse function.

Example 1: Solve an exponential equation using logs

Solve:

$3 \cdot 2^x = 20$

Divide by 3:

$2^x = \frac{20}{3}$

Apply natural log to both sides:

$\ln(2^x) = \ln\left(\frac{20}{3}\right)$

Use the power property:

$x\ln(2) = \ln\left(\frac{20}{3}\right)$

Solve for $x$:

$x = \frac{\ln\left(\frac{20}{3}\right)}{\ln(2)}$

This also connects to the change-of-base idea (next section).

Example 2: Solve a logarithmic equation using exponentials

Solve:

$\log_5(x) = 3$

Rewrite in exponential form:

$x = 5^3$

$x = 125$

The “rewrite” step is inverse-function reasoning: $\log_5(x) = 3$ means “the exponent on 5 that gives $x$ is 3.”

Change of base (logs expressed using other logs)

You often want to compute $\log_b(x)$ using calculator buttons $\ln$ or $\log$. The standard relationship is:

$\log_b(x) = \frac{\ln(x)}{\ln(b)}$

and also:

$\log_b(x) = \frac{\log(x)}{\log(b)}$

This is not just a computational trick: it reflects that logarithms differ only by a constant scale factor. That’s also why semi-log plots using $\ln$ versus $\log$ are both linear for exponentials, but with different slope values.

Common mistake: writing $\log_b(x) = \ln(b)/\ln(x)$ (flipped). The base goes in the denominator.

Composition with transformations (what happens to domain and outputs)

When you compose functions, the domain of the composition depends on where both functions make sense.

Example 3: Compose a log with an exponential

Let

$f(x) = e^{2x}$

$g(x) = \ln(x)$

Compute $(g \circ f)(x)$:

$(g \circ f)(x) = g(f(x)) = \ln(e^{2x})$

Using inverse properties:

$\ln(e^{2x}) = 2x$

So $(g \circ f)(x) = 2x$. This works for all real $x$ because $e^{2x} > 0$, so the input to $\ln$ is always valid.

Example 4: Compose an exponential with a log

Compute $(f \circ g)(x)$:

$(f \circ g)(x) = f(g(x)) = e^{2\ln(x)}$

This simplifies using exponent rules:

$e^{2\ln(x)} = (e^{\ln(x)})^2$

$e^{2\ln(x)} = x^2$

But you must keep the domain restriction from $\ln(x)$: this composition is only defined for $x > 0$.

So the simplified result is:

$(f \circ g)(x) = x^2 \text{ for } x > 0$

A subtle but important misconception: you might be tempted to say $e^{2\ln(x)} = x^2$ for all real $x$, but $\ln(x)$ is not defined for $x \le 0$ in real numbers. The algebraic simplification is correct, but the domain does not expand.

Composition as the engine behind linearization

Linearization itself can be viewed as composition:

• For exponential data, you apply $\ln$ to the output: $Y = \ln(y)$.
• For logarithmic data, you apply $\ln$ to the input: $X = \ln(x)$.

That is you composing your original relationship with a log function in order to create a linear relationship.

Worked example 5: Build a model from a linearized equation

You are told a semi-log linearization produced the line

$\ln(y) = 0.4x + 1.2$

Step 1: Recognize the structure.
This matches $\ln(y) = kx + \ln(a)$, which comes from $y = a e^{kx}$.

So $k = 0.4$ and $\ln(a) = 1.2$.

Step 2: Solve for $a$.

$a = e^{1.2}$

Model:

$y = e^{1.2} e^{0.4x}$

You can combine exponents:

$y = e^{1.2 + 0.4x}$

Interpretation: Each increase of 1 in $x$ multiplies $y$ by $e^{0.4}$.

Exam Focus

• Typical question patterns
  • Evaluate or simplify compositions like $\ln(e^{kx})$ or $b^{\log_b(x)}$ while stating any domain restrictions.
  • Use inverse relationships to solve equations (rewrite between log and exponential forms).
  • Given a linearized model (for example, $\log(y) = mx + b$), write the original exponential function by undoing the log.
• Common mistakes
  • Canceling logs/exponentials with mismatched bases (for example, treating $\ln(10^x)$ as $x$ without accounting for the base difference).
  • Dropping domain restrictions when simplifying (especially compositions involving $\ln(x)$).
  • Misusing properties, such as thinking $\ln(a + b) = \ln(a) + \ln(b)$ (this is false).