Graphing Logarithmic Functions

Idea

A logarithm graph is an exponential graph reflected in the line $y = x$ . Everything about its shape follows from that one fact.

The exponential had a horizontal floor it never touched. Reflecting turns that floor into a vertical wall.

Rule

$y = \log_a x$ has domain $x > 0$ and range every real $y$

the line $x = 0$ is a vertical asymptote

the curve passes through $(1, 0)$ for every base

How it is used

1. Two shapes, and the base picks one

Base	Shape	Example
$a > 1$	rises, slowly and forever	$y = \log_2 x$
$0 < a < 1$	falls, slowly and forever	$y = \log_{1/2} x$

The two are mirror images in the $x$ -axis, because $\log_{1/a} x = -\log_a x$ .

2. Three points fix the curve

Every logarithm graph passes through the same three landmarks, written in terms of its base.

\left(\frac{1}{a}, -1\right),

(1, 0),

(a, 1)

For $y = \log_2 x$ that is $\left(\frac{1}{2}, -1\right)$ , $(1, 0)$ and $(2, 1)$ . Plot those, follow the shape, and the sketch is done.

Exponential and logarithm

Try this. Compare bases 2 and 0.5: growth becomes decay. Move the input to 1; the logarithm is zero for either base.

BasePositive input x0.5

Base 3 · Blue: y = 3ˣ · Green: y = log_b(x) · Dashed: y = x. At x = 0.5, y = -0.631.

3. The asymptote is a wall, not a floor

As $x$ shrinks towards $0$ the curve dives without limit, but never crosses to the left.

\log_2 \frac{1}{1024} = -10,

\log_2 \frac{1}{2^{100}} = -100

There is no bottom. The values are unbounded below, which is what "range is every real $y$ " means.

Going the other way the curve keeps climbing, just very slowly: $\log_2 x$ needs $x$ to double to gain a single unit of height.

4. It is the exponential, reflected

	$y = a^x$	$y = \log_a x$
domain	every real $x$	$x > 0$
range	$y > 0$	every real $y$
asymptote	horizontal, $y = 0$	vertical, $x = 0$
passes through	$(0, 1)$	$(1, 0)$

Each row is the previous one with the coordinates swapped. That is what reflecting in $y = x$ does.

5. Shifts move the wall

The asymptote sits wherever the argument is zero.

y = \log_a(x - h) \;\Rightarrow\; \text{asymptote } x = h, \quad \text{domain } x > h

A constant added outside, as in $\log_a x + k$ , lifts the curve but leaves the wall where it was.

Worked example

Describe the graph of $f(x) = \log_2(x - 3) + 1$ .

Remember

the asymptote sits where the argument is zero $\log_a 1 = 0$ and $\log_a a = 1$ a constant outside shifts up, not sideways

Find the wall. The argument is zero at $x = 3$ , so that is the vertical asymptote and the domain is $x > 3$ .

x - 3 > 0 \;\Rightarrow\; x > 3

Use $\log_2 1 = 0$ to get the first point. The argument is $1$ when $x = 4$ .

f(4) = \log_2 1 + 1 = 1

Use $\log_2 2 = 1$ for the second. The argument is $2$ when $x = 5$ .

f(5) = \log_2 2 + 1 = 2

The base is above $1$ , so the curve rises. It climbs from the wall at $x = 3$ through $(4, 1)$ and $(5, 2)$ , gaining one unit each time $x - 3$ doubles.

Explore

Both curves are drawn here. Drag the probe and watch the pair move together.

The green logarithm and the blue exponential are the same curve seen from two sides of the dashed line $y = x$ . Whenever the probe sits at $(p, q)$ on one, its partner sits at $(q, p)$ on the other. The exponential flattens onto its horizontal floor on the left; the logarithm dives down its vertical wall near $x = 0$ . Those are the same behaviour with the axes exchanged.

Now change the base. Above $1$ both curves climb; below $1$ both fall. The reflection holds either way, because it comes from the definition rather than from the shape.