How to graph logarithmic functions: Points, shifts, and reflections

If you can plot a few powers of 3, you already have the points you need to graph a logarithm. We’ll swap the coordinates, then see what happens when we shift and reflect the graph. The video uses a different shift and adds a vertical compression.

Start with the exponential inverse

For a base $b > 0$ with $b \neq 1$ , a logarithm tells us the exponent on that base. The same relationship can be written in two ways:

y = \log_{b} (x) \Leftrightarrow x = b^{y}

The inverse of $y = 3^{x}$ is $y = \log_{3} (x)$ . Swapping the coordinates of each point reflects the exponential curve across $y = x$ . For example, $(1, 3)$ on the exponential becomes $(3, 1)$ on the logarithm. The same idea applies to $\ln (x)$ , which uses base $e$ .

The point (1, 3) on the exponential becomes (3, 1) on the logarithm.

Plot a few points and mark the asymptote

It’s easiest to choose small integer values for $y$ and use $x = 3^{y}$ to find the matching inputs. For instance, $y = - 1$ gives $x = \frac{1}{3}$ .

Points on the graph of log base 3
Input x	Output y
$\frac{1}{9}$	$- 2$
$\frac{1}{3}$	$- 1$
$1$	$0$
$3$	$1$
$9$	$2$

Powers of 3 are positive, so the logarithm’s domain is $x > 0$ . As $y$ decreases without bound in $x = 3^{y}$ , $x$ approaches zero from the right. That gives the vertical asymptote $x = 0$ . The logarithm can produce any real output, so its range is all real numbers.

The parent logarithm increases when $b > 1$ and decreases when $0 < b < 1$ . Either way, its x-intercept is $(1, 0)$ because $b^{0} = 1$ .

Example: Shift right, then reflect across the x-axis

\begin{array}{l} f (x) = \log_{3} (x) \\ g (x) = \log_{3} (x - 1) \\ h (x) = - \log_{3} (x - 1) \end{array}

Replacing $x$ with $x - 1$ moves each point one unit to the right. The parent’s point $(3, 1)$ moves to $(4, 1)$ , and the asymptote moves from $x = 0$ to $x = 1$ .

Multiplying the output by $- 1$ keeps $x$ the same and changes the sign of $y$ , so the reflection is across the x-axis. The point $(4, 1)$ becomes $(4, - 1)$ . The domain remains $x > 1$ , and $(2, 0)$ stays on both graphs because its y-coordinate is zero.

The reflection changes the y-values. Both graphs have vertical asymptote x = 1 and x-intercept (2, 0).

Find the new domain and asymptote from the input

For $f (x) = - 2 \ln (2 x + 6) - 4$ , the input to the logarithm has to be positive. Set $2 x + 6 > 0$ to find the domain, $x > - 3$ . Its boundary is the vertical asymptote $x = - 3$ . The changes outside the logarithm change the outputs, so they don’t move that boundary.

How to graph logarithmic functions: Points, shifts, and reflections

Start with the exponential inverse

Plot a few points and mark the asymptote

Example: Shift right, then reflect across the x-axis

Find the new domain and asymptote from the input

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