Exponential and Logarithmic Functions

Unit 2 of AP Precalculus, Exponential and Logarithmic Functions, starts from sequences and ends with logarithms. The thread through it is one idea: linear change adds the same amount each step, exponential change multiplies by the same factor.

Sequences: the discrete version

• Arithmetic: aₙ = a₀ + dn. Constant difference, linear growth.
• Geometric: gₙ = g₀ × rⁿ. Constant ratio, exponential growth.

An exponential function f(x) = a × bˣ is the continuous version of a geometric sequence.

Exponential growth and decay

In f(x) = a × bˣ with a > 0:

• b > 1 is growth, 0 < b < 1 is decay.
• A rate of r percent per step means b = 1 + r for growth and b = 1 − r for decay.
• The horizontal asymptote is y = 0, moved by any vertical shift.

Logarithms are inverses

log_b(x) = y means exactly bʸ = x. The graph of y = log_b(x) is the reflection of y = bˣ in the line y = x, with a vertical asymptote at x = 0.

The rules come straight from the exponent rules:

• log(ab) = log a + log b
• log(a/b) = log a − log b
• log(aᵏ) = k log a

Worked example

A car bought for 20,000 loses 15 percent of its value each year. When is it worth 10,000?

1. Model: V(t) = 20000 × 0.85ᵗ.
2. Set up: 20000 × 0.85ᵗ = 10000, so 0.85ᵗ = 0.5.
3. Take logs: t ln 0.85 = ln 0.5, so t = ln 0.5 / ln 0.85 ≈ 4.27.
4. The car is worth 10,000 after about 4.3 years.

A common mistake is to subtract 15 percent of 20,000 every year, which gives a linear model and an answer of 3.3 years.

The free response approach

• Name the model type and justify it: equal ratios over equal intervals mean exponential.
• Show the log step when solving; a calculator answer alone may not earn the reasoning point.
• Interpret in context, with units: about 4.3 years, not only 4.27.