Graphing Logarithmic Functions: Examples

Graphing logarithmic functions starts with the parent graph $y = \log_b x$ . It passes through $(1, 0)$ and $(b, 1)$ , gets close to the vertical asymptote $x = 0$ without touching it, and is the mirror image of $y = b^x$ across the line $y = x$ . To graph any log function, shift and stretch this parent graph.

Before this lesson: You should know Logarithms, Exponential Functions, Inverse Functions and Transformations of Functions.

Why a Log Graph Looks the Way It Does

A logarithm undoes an exponent, so $y = \log_2 x$ is the inverse of $y = 2^x$ . To graph an inverse, swap the x- and y-values of each point.

$y = 2^x$	$y = \log_2 x$
$(-2, \frac{1}{4})$	$(\frac{1}{4}, -2)$
$(-1, \frac{1}{2})$	$(\frac{1}{2}, -1)$
$(0, 1)$	$(1, 0)$
$(1, 2)$	$(2, 1)$
$(2, 4)$	$(4, 2)$
$(3, 8)$	$(8, 3)$

The exponential graph hugs the x-axis on the left. So the log graph hugs the y-axis near the bottom. The exponential graph shoots up fast, so the log graph rises more and more slowly.

The blue solid curve y = 2 to the x passes through (0, 1), (1, 2) and (2, 4). The orange dashed curve y = log base 2 of x passes through (1, 0), (2, 1) and (4, 2). A gray dotted line y = x sits between them, and each curve is the reflection of the other across that line. — Swapping the coordinates of each point on y = 2ˣ gives a point on y = log₂ x. The two graphs are mirror images across y = x.

Key Features of the Parent Graph

Here are the key features of the parent function $y = \log_b x$ with base $b \gt 1$ , including its domain and range:

Feature	Value
Domain	$x \gt 0$ (you can only take the log of a positive number)
Range	All real numbers
x-intercept	$(1, 0)$ , because $\log_b 1 = 0$
Another key point	$(b, 1)$ , because $\log_b b = 1$
Vertical asymptote	$x = 0$ (the y-axis)
Direction	Rises from left to right, but more and more slowly

A vertical asymptote is a vertical line the graph gets closer and closer to but never touches. You saw these in graphing rational functions. The log graph has no y-intercept, because $x = 0$ is not in the domain. For a base between 0 and 1, the graph falls from left to right instead.

The Transformed Log Function

y = a \log_b (x - h) + k

Part	What it does
$h$	Shifts the graph right $h$ (left if $h$ is negative). The asymptote moves to $x = h$ .
$k$	Shifts the graph up $k$ (down if $k$ is negative).
$a$	Stretches the graph by a factor of $a$ . A negative $a$ flips it over the x-axis.

Steps to graph a log function:

Find the vertical asymptote. Set the inside of the log equal to 0 and solve. For $\log_b (x - h)$ , the asymptote is $x = h$ .
Start with three parent points: $(\frac{1}{b}, -1)$ , $(1, 0)$ and $(b, 1)$ .
Move each point. Add $h$ to the x-value. Multiply the y-value by $a$ , then add $k$ .
Plot the new points and draw a smooth curve. Make it bend toward the asymptote without touching it.
State the domain ( $x \gt h$ ) and the range (all real numbers).

Worked Examples

Example 1: Graph a parent log function

Graph $y = \log_3 x$ . State the domain, range and asymptote.

Choose x-values that are powers of 3: $\frac{1}{3}$ , 1, 3 and 9.

3^{-1} = \frac{1}{3}

Find $y$ at $x = \frac{1}{3}$ .

\log_3 \frac{1}{3} = -1

Write it as a log.

3^0 = 1

Find $y$ at $x = 1$ .

\log_3 1 = 0

Write it as a log.

3^1 = 3

Find $y$ at $x = 3$ .

\log_3 3 = 1

Write it as a log.

3^2 = 9

Find $y$ at $x = 9$ .

\log_3 9 = 2

Write it as a log.

Plot $(\frac{1}{3}, -1)$ , $(1, 0)$ , $(3, 1)$ and $(9, 2)$ . Draw a curve that drops toward the y-axis on the left.

Answer: Domain $x \gt 0$ , range all real numbers, vertical asymptote $x = 0$ . The graph rises slowly to the right.

Example 2: Shift right and up

Graph $f(x) = \log_2 (x - 3) + 1$ .

x - 3 = 0

Set the inside equal to 0.

x = 3

Add 3 to both sides. This is the vertical asymptote.

Read the shifts: $h = 3$ (right 3) and $k = 1$ (up 1).

(\frac{1}{2} + 3, -1 + 1) = (3\frac{1}{2}, 0)

Move $(\frac{1}{2}, -1)$ .

(1 + 3, 0 + 1) = (4, 1)

Move $(1, 0)$ .

(2 + 3, 1 + 1) = (5, 2)

Move $(2, 1)$ .

(4 + 3, 2 + 1) = (7, 3)

Move $(4, 2)$ .

Check $(7, 3)$ . Use $\log_2 4 = 2$ , then add 1: $f(7) = \log_2 4 + 1 = 2 + 1 = 3$

Plot the points and draw the curve to the right of $x = 3$ .

Answer: Asymptote $x = 3$ , domain $x \gt 3$ , range all real numbers. The x-intercept is $(3\frac{1}{2}, 0)$ .

A grid with x from −1 to 10 and y from −3 to 5. The gray dashed parent curve y = log base 2 of x passes through (1, 0), (2, 1) and (4, 2). The blue solid curve f(x) = log base 2 of (x − 3) + 1 is the same shape moved right 3 and up 1, through (4, 1), (5, 2) and (7, 3). An orange dashed vertical line marks the asymptote x = 3. — Every point of y = log₂ x moves right 3 and up 1. The asymptote moves from x = 0 to x = 3.

Example 3: Shift left and reflect

Graph $g(x) = -\log_2 (x + 4)$ .

x + 4 = 0

Set the inside equal to 0.

x = -4

Subtract 4 from both sides. This is the vertical asymptote.

Read the parts: $h = -4$ (left 4) and $a = -1$ (flip over the x-axis).

(\frac{1}{2} - 4, -1 \cdot (-1)) = (-3\frac{1}{2}, 1)

Move $(\frac{1}{2}, -1)$ .

(1 - 4, -1 \cdot 0) = (-3, 0)

Move $(1, 0)$ .

(2 - 4, -1 \cdot 1) = (-2, -1)

Move $(2, 1)$ .

(4 - 4, -1 \cdot 2) = (0, -2)

Move $(4, 2)$ .

Plot the points. The curve falls from left to right.

Answer: Asymptote $x = -4$ , domain $x \gt -4$ , range all real numbers. The graph falls from left to right and crosses the y-axis at $(0, -2)$ .

Example 4: Find the domain, asymptote and x-intercept

For $h(x) = \log (2x - 6)$ , find the domain, the vertical asymptote and the x-intercept.

2x - 6 \gt 0

The inside must be positive.

2x \gt 6

Add 6 to both sides.

x \gt 3

Divide both sides by 2 (see multi-step inequalities)

x = 3

The asymptote is where the inside equals 0.

\log (2x - 6) = 0

For the x-intercept, set $h(x) = 0$ .

2x - 6 = 10^0

Rewrite in exponential form (see logarithms)

2x - 6 = 1

Simplify the power.

2x = 7

Add 6 to both sides.

x = 3\frac{1}{2}

Divide both sides by 2.

Answer: Domain $x \gt 3$ , vertical asymptote $x = 3$ , x-intercept $(3\frac{1}{2}, 0)$ .

Example 5: Word problem with a growth model

A seedling’s height in inches $t$ weeks after it sprouts is modeled by $H(t) = 5 \log_2 (t + 1) + 2$ . Find its height at 0, 1, 3 and 7 weeks. Then find when it reaches 22 inches.

5 \log_2 1 + 2 = 5 \cdot 0 + 2 = 2

Find $H(0)$ using $\log_2 1 = 0$ .

5 \log_2 2 + 2 = 5 \cdot 1 + 2 = 7

Find $H(1)$ using $\log_2 2 = 1$ .

5 \log_2 4 + 2 = 5 \cdot 2 + 2 = 12

Find $H(3)$ using $\log_2 4 = 2$ .

5 \log_2 8 + 2 = 5 \cdot 3 + 2 = 17

Find $H(7)$ using $\log_2 8 = 3$ .

5 \log_2 (t + 1) + 2 = 22

Set the height to 22.

5 \log_2 (t + 1) = 20

Subtract 2 from both sides.

\log_2 (t + 1) = 4

Divide both sides by 5.

t + 1 = 2^4

Rewrite in exponential form.

t + 1 = 16

Simplify the power.

t = 15

Subtract 1 from both sides.

Answer: The heights are 2, 7, 12 and 17 inches. It reaches 22 inches at 15 weeks. The graph rises fast at first, then slows down. Each time $t + 1$ doubles, the height grows only 5 more inches.

Common Mistakes

Mistake	Example of the error	Fix
Using the wrong sign for the asymptote	$y = \log (x - 3)$ has asymptote $x = -3$	Set $x - 3 = 0$ . The asymptote is $x = 3$ .
Shifting the wrong way	$\log_2 (x + 4)$ moved right 4	A plus sign inside moves the graph left: left 4.
Including the asymptote in the domain	Domain of $\log x$ is $x \ge 0$	The log of 0 is undefined, so the domain is $x \gt 0$ .
Mixing up log and exponential key points	$y = \log_2 x$ passes through $(0, 1)$	That point is on $y = 2^x$ . The log graph passes through $(1, 0)$ .
Letting the curve touch the asymptote	Drawing the graph crossing $x = 0$	The curve gets close to the asymptote but never reaches it.

Related Methods

Graph from the exponential function

If you can graph $y = b^x$ , you can graph $y = \log_b x$ . Swap the coordinates of each point and reflect the graph across $y = x$ . The horizontal asymptote $y = 0$ of the exponential graph becomes the vertical asymptote $x = 0$ of the log graph.

The natural log graph

The graph of $y = \ln x$ has base the number e, which is about 2.718. It passes through $(1, 0)$ and $(e, 1) \approx (2.72, 1)$ , with asymptote $x = 0$ . It sits between the graphs of $\log_2 x$ and $\log_3 x$ .

Bases between 0 and 1

When $0 \lt b \lt 1$ , the graph falls from left to right. For example, $y = \log_{\frac{1}{2}} x$ passes through $(\frac{1}{2}, 1)$ , $(1, 0)$ and $(2, -1)$ . It is the same as $y = -\log_2 x$ .

Practice Problems

List four points on $y = \log_4 x$ using x-values $\frac{1}{4}$ , 1, 4 and 16.
Find the vertical asymptote and domain of $y = \log_5 (x - 2)$ .
Find the vertical asymptote, domain and range of $y = \ln (x + 6) - 1$ .
Graph $y = \log_2 x + 3$ by moving the points $(\frac{1}{2}, -1)$ , $(1, 0)$ , $(2, 1)$ and $(4, 2)$ .
Graph $y = \log_3 (x - 1)$ by moving the points $(\frac{1}{3}, -1)$ , $(1, 0)$ , $(3, 1)$ and $(9, 2)$ .
Graph $y = -\log_3 x$ . Does it rise or fall from left to right?
Graph $y = 2 \log_2 (x + 1)$ . Give the asymptote and the y-intercept.
Find the domain of $y = \log (3x + 12)$ .
Find the x-intercept of $y = \log_2 (x - 5)$ .
A store models its daily visitors after spending $d$ hundred dollars on ads as $V(d) = 200 \log_2 (d + 2)$ . Find $V(0)$ and $V(6)$ . Then find how much it must spend to get 800 visitors.

Show answers

$4^{-1} = \frac{1}{4}$ , so $\log_4 \frac{1}{4} = -1$
$4^0 = 1$ , so $\log_4 1 = 0$
$4^1 = 4$ , so $\log_4 4 = 1$
$4^2 = 16$ , so $\log_4 16 = 2$
Answer: $(\frac{1}{4}, -1)$ , $(1, 0)$ , $(4, 1)$ , $(16, 2)$

x - 2 = 0

Set the inside equal to 0.

x = 2

Add 2 to both sides.

Answer: asymptote $x = 2$ , domain $x \gt 2$

x + 6 = 0

Set the inside equal to 0.

x = -6

Subtract 6 from both sides.

The shift down 1 does not change the range.
Answer: asymptote $x = -6$ , domain $x \gt -6$ , range all real numbers

Shift up 3, so add 3 to each y-value

(\frac{1}{2}, -1 + 3) = (\frac{1}{2}, 2)

Shift up 3: add 3 to the y-value.

(1, 0 + 3) = (1, 3)

Shift up 3: add 3 to the y-value.

(2, 1 + 3) = (2, 4)

Shift up 3: add 3 to the y-value.

(4, 2 + 3) = (4, 5)

Shift up 3: add 3 to the y-value.

Answer: plot these points; the asymptote stays $x = 0$

Shift right 1, so add 1 to each x-value

(\frac{1}{3} + 1, -1) = (1\frac{1}{3}, -1)

Shift right 1: add 1 to the x-value.

(1 + 1, 0) = (2, 0)

Shift right 1: add 1 to the x-value.

(3 + 1, 1) = (4, 1)

Shift right 1: add 1 to the x-value.

(9 + 1, 2) = (10, 2)

Shift right 1: add 1 to the x-value.

Answer: plot these points; the asymptote is $x = 1$

Multiply each y-value of $y = \log_3 x$ by −1
$(\frac{1}{3}, -1)$ becomes $(\frac{1}{3}, 1)$
$(1, 0)$ stays $(1, 0)$
$(3, 1)$ becomes $(3, -1)$
$(9, 2)$ becomes $(9, -2)$
Answer: it falls from left to right

x + 1 = 0

Find the asymptote: set the inside equal to 0.

x = -1

Subtract 1 from both sides.

(\frac{1}{2} - 1, 2 \cdot (-1)) = (-\frac{1}{2}, -2)

Move $(\frac{1}{2}, -1)$ .

(1 - 1, 2 \cdot 0) = (0, 0)

Move $(1, 0)$ .

(2 - 1, 2 \cdot 1) = (1, 2)

Move $(2, 1)$ .

(4 - 1, 2 \cdot 2) = (3, 4)

Move $(4, 2)$ .

Answer: asymptote $x = -1$ , y-intercept $(0, 0)$

3x + 12 \gt 0

The inside must be positive.

3x \gt -12

Subtract 12 from both sides.

x \gt -4

Divide both sides by 3.

Answer: $x \gt -4$

\log_2 (x - 5) = 0

Set $y = 0$ .

x - 5 = 2^0

Rewrite in exponential form.

x - 5 = 1

Simplify the power.

x = 6

Add 5 to both sides.

Answer: $(6, 0)$

10.

V(0) = 200 \log_2 2 = 200 \cdot 1 = 200

Find $V(0)$ using $\log_2 2 = 1$ .

V(6) = 200 \log_2 8 = 200 \cdot 3 = 600

Find $V(6)$ using $\log_2 8 = 3$ .

200 \log_2 (d + 2) = 800

Set $V(d) = 800$ .

\log_2 (d + 2) = 4

Divide both sides by 200.

d + 2 = 2^4

Rewrite in exponential form.

d + 2 = 16

Simplify the power.

d = 14

Subtract 2 from both sides.

Answer: 200 and 600 visitors; it must spend 14 hundred dollars, or $1,400

FAQ

What does the graph of a logarithmic function look like?

The parent graph $y = \log_b x$ with $b \gt 1$ starts low next to the y-axis, crosses the x-axis at $(1, 0)$ and rises more and more slowly to the right. It never touches the y-axis.

How do you find the vertical asymptote of a log function?

Set the inside of the log equal to 0 and solve for $x$ . For $y = \log (x - 7)$ , the asymptote is $x = 7$ .

What is the domain of a logarithmic function?

The inside of the log must be positive. For $y = \log_b (x - h)$ , the domain is $x \gt h$ . The range is always all real numbers.

How are log graphs and exponential graphs related?

They are inverses, so each one is the reflection of the other across the line $y = x$ . The point $(0, 1)$ on $y = b^x$ matches the point $(1, 0)$ on $y = \log_b x$ .

Summary

$y = \log_b x$ passes through $(\frac{1}{b}, -1)$ , $(1, 0)$ and $(b, 1)$ , with vertical asymptote $x = 0$ .
The log graph is the reflection of $y = b^x$ across $y = x$ .
For $y = a \log_b (x - h) + k$ , the asymptote is $x = h$ and the domain is $x \gt h$ .
Shift each point right $h$ , multiply its y-value by $a$ , then shift it up $k$ .
The range of every log function is all real numbers.

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