Logarithmic functions

The function $f(x) = \log_b x$ answers one question for every input: b to what power gives x? Reading the graph at x = 8 on $y = \log_2 x$ returns 3, because $2^3 = 8$. Every feature of the graph follows from this reading.

Two points anchor every log graph, whatever the base: (1, 0), because $b^0 = 1$ for any base, and (b, 1), because $b^1 = b$. If you can place those two points, you can sketch the curve and identify the base from a graph.

Since exponents can be any real number, the range is all reals. Since $b^{\text{anything}}$ is always positive, only positive inputs have an answer: the domain is $(0, \infty)$. Zero and negative numbers are not slow cases or special cases; they are simply not in the function's world.

For b > 1, $\log_b x$ is increasing everywhere on its domain, and concave down: each additional unit of input buys less output than the one before. That combination fools the eye. The curve flattens, so students conclude it must level off at some ceiling.

It never does. There is no horizontal asymptote. To push $\log_2 x$ past any target, just feed it a big enough power of 2: $\log_2 x$ reaches 100 at $x = 2^{100}$, reaches 1000 at $x = 2^{1000}$, and so on forever. The outputs grow without bound; they just demand exponentially more input for each step. Slow, but unstoppable.

That slowness is real, though, and worth naming: a log function eventually grows more slowly than any positive power of x, even $x^{0.001}$. Log growth is the slowest unbounded growth in this course.

The asymptote a log function does have is vertical: the line x = 0. As inputs shrink toward zero from the right, the outputs of $\log_b x$ (for b > 1) plunge: $\log_2(1/2) = -1$, $\log_2(1/4) = -2$, $\log_2(1/1024) = -10$. In limit language,

$$x \to 0^+ \implies \log_b x \to -\infty \quad (b > 1).$$

Note where this wall is NOT: it is not at x = 1. The point (1, 0) is where the graph crosses the x-axis, an intercept, not an asymptote. The graph passes through it at a perfectly ordinary slope. And the wall is an input-side fact: log outputs are allowed to be negative; log inputs are not allowed to reach 0.

For $0 < b < 1$, the reading still works ($\log_{1/2} 8 = -3$ because $(1/2)^{-3} = 8$), but the graph flips: the function is decreasing, climbing out of $+\infty$ near x = 0 and drifting down forever. Domain, vertical asymptote, and the anchor point (1, 0) are unchanged. The same graph appears if you negate: $\log_{1/2} x = -\log_2 x$.

In the general form $f(x) = a\log_b x$, the coefficient a is a vertical stretch (and a reflection when negative). It rescales outputs; it never moves the domain, the wall at x = 0, or the intercept at (1, 0), because a times 0 is still 0.