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

Semi-log plot

A graph where one axis uses a logarithmic scale and the other uses a linear scale, often used to make exponential or logarithmic relationships appear linear.

Log-linear (semi-log) plot

A semi-log plot with a linear x-axis and a logarithmic y-axis; commonly used to linearize exponential data by plotting log(y) vs x.

Linearization

Transforming variables (e.g., using logs) so a nonlinear relationship becomes a straight-line model of the form Y = mX + b.

Transformed variables (X and Y)

New variables defined from the original ones (such as Y = ln(y) or X = ln(x)) so the relationship fits a linear form.

Exponential model (ab^x form)

A model $y = ab^x$ (a>0, b>0, $b \ne 1$) where y changes by a constant multiplicative factor for each 1-unit increase in x.

Exponential model (ae^{kx} form)

A model $y = ae^{kx}$ where k is a continuous growth/decay constant and a is the value at x = 0.

Constant ratio (exponential clue)

A pattern where successive y-values have roughly the same ratio for equal x-steps, indicating exponential behavior.

Growth factor (b)

In $y = ab^x$, the multiplier applied to y for each 1-unit increase in x; b>1 indicates growth and 0<b<1 indicates decay.

Initial value (a) in an exponential model

In $y = ab^x$ or $y = ae^{kx}$, the value of y when x = 0.

Log-linearization of y = a b^x

Taking logs gives $\log(y) = x \cdot \log(b) + \log(a)$, which is linear in x when plotting log(y) vs x.

Slope on a plot of log(y) vs x (ab^x)

The slope equals log(b) (using the same log base as the plot/transform).

Intercept on a plot of log(y) vs x (ab^x)

The intercept equals log(a), so a can be recovered by exponentiating the intercept in that log base.

Linearization of y = a e^{kx} using ln

Taking natural log gives $\ln(y) = kx + \ln(a)$, a straight line when plotting ln(y) vs x.

Parameter recovery from ln(y) = kx + ln(a)

k is the slope; $a = e^{\text{intercept}}$ where the intercept is ln(a).

Logarithmic model (A + B ln(x))

A model y = A + B ln(x) (x>0) used for growth that slows; it becomes linear by setting X = ln(x) and Y = y.

Linearization of a logarithmic model

Let X = ln(x) and Y = y so y = A + B ln(x) becomes Y = A + B X.

Domain restriction for logarithms

For real-valued logs, the input must be positive (e.g., ln(x) requires x > 0).

Interpretation of A in y = A + B ln(x)

A equals y when x = 1 because ln(1) = 0, so y(1) = A.

Interpretation of B in y = A + B ln(x)

B is the change in y when x is multiplied by e (since ln(ex) = ln(x) + 1).

How semi-log scaling relates to exponential change

On a log-scaled y-axis, equal vertical steps represent equal multiplicative changes in y, matching exponential behavior.

Choosing exponential vs logarithmic from data

Use an exponential model if equal x-steps multiply y by a constant factor; use a logarithmic model if multiplicative changes in x add a roughly constant amount to y.

Exponential model from two points (solving for b)

For $y = ab^x$ with points $(x_1,y_1),(x_2,y_2)$: $b = (y_2/y_1)^{1/(x_2-x_1)}$, then $a = y_1 / b^{x_1}$.

Change-of-base formula

A way to compute logs in a different base: $\log_b(x) = \ln(x)/\ln(b)$ (or log(x)/log(b)).

Exponential and logarithmic inverse relationship

For b>0, $b \ne 1$: $b^{\log_b(x)} = x$ (x>0) and $\log_b(b^x) = x$ (all real x).

Composition domain retention

When simplifying compositions (e.g., $e^{2\ln(x)} = x^2$), the original domain restrictions still apply (here, x > 0).