Topic 2.11 Logarithmic Functions: analyze the parent logarithmic function and its transformations, including its domain, range, vertical asymptote, and increasing or decreasing behavior.

What this topic is asking

The College Board (Topic 2.11) wants you to analyze the logarithmic function $f(x) = \log_b x$ and its transformations. You must state its domain ($x > 0$), its range (all reals), its vertical asymptote ($x = 0$), and whether it increases or decreases (depending on the base). Then you must apply shifts, stretches and reflections and track how each changes these features.

The parent logarithmic function

These two anchor points, $(1, 0)$ and $(b, 1)$, let you sketch any logarithmic graph quickly, and the domain and asymptote come straight from the inverse relationship with the exponential (Topic 2.10).

Increasing or decreasing by the base

The natural logarithm $\ln x$ (base $e$) and the common logarithm $\log x$ (base $10$) both increase, since their bases exceed $1$.

Transformations of logarithmic graphs

Logarithmic graphs transform by the same rules as any function (Topic 1.12). The crucial feature to track is the vertical asymptote, which moves with horizontal shifts and bounds the domain.

Why logarithms grow so slowly

The logarithm is the slowest-growing standard increasing function, and this is a direct consequence of being the exponential's inverse. To increase the output of $\log_b x$ by $1$, the input must be multiplied by $b$, not increased by a fixed amount. So moving from output $1$ to output $2$ requires the input to jump from $b$ to $b^2$, a multiplicative leap. This is why the graph flattens out: enormous increases in $x$ produce only small increases in the output. Recognizing the logarithm as "the inverse of fast exponential growth" makes its slow, ever-rising shape feel inevitable rather than surprising.

A point that recurs in exam questions is that the vertical asymptote, not $x = 0$ by default, is always the boundary of the shifted domain. After a horizontal shift the asymptote and the domain move together, so reading the asymptote off the inside of the logarithm (set the inside greater than zero) gives both at once. Anchoring on "the inside must be positive" produces the domain and the asymptote in one step and avoids the common error of leaving the asymptote at $x = 0$ after a shift.

Try this

Q1. State the domain and vertical asymptote of $f(x) = \log(x - 5)$. [1 point]

• Cue. $x - 5 > 0$ gives domain $x > 5$ and vertical asymptote $x = 5$.

Q2. Through which two points does $f(x) = \log_4 x$ pass? [1 point]

• Cue. $(1, 0)$ since $\log_4 1 = 0$, and $(4, 1)$ since $\log_4 4 = 1$.