Lesson 3 of 4Exponential & Logarithmic Functions

Unit 3 · Lesson 3 of 4

Graphs of Logarithmic Functions

$y = lo g_{b} x$ is the inverse of $y = b^{x}$ , so it is that graph reflected in the line $y = x$ — turning the horizontal asymptote into a vertical one at $x = 0$ .

4 min read11 practice questionsInteractiveAsk AI about this →

Key things to remember

1 $y = lo g_{b} x$ is the inverse of $y = b^{x}$ ; its graph is the reflection of $y = b^{x}$ in the line $y = x$ .
2Parent function: domain $x > 0$ , range $y \in R$ , vertical asymptote $x = 0$ , $x$ -intercept $(1, 0)$ , and it passes through $(b, 1)$ .
3Transformed form: $y = a lo g_{b} (k (x - d)) + c$ — same parameter roles as every other function family.
4The vertical asymptote moves to $x = d$ ; the vertical shift $c$ does not move it.
5Domain of a transformed logarithm: solve $k (x - d) > 0$ . It is $x > d$ when $k > 0$ and $x < d$ when $k < 0$ .
6For $b > 1$ , $y = lo g_{b} x$ has no maximum and no horizontal asymptote — it rises forever, just very slowly.

You do not need to plot a logarithmic function point by point, because you already know what it looks like. A logarithm undoes an exponential, which makes $y = lo g_{b} x$ the inverse function of $y = b^{x}$ — and the graph of any inverse is the original reflected in the line $y = x$ . Every feature of the logarithmic graph is a feature of the exponential graph with its coordinates swapped.

Fig 1Reflecting the exponential curve in the line y = x produces the logarithmic curve; the asymptote turns from horizontal to vertical.

Swapping the columns

To get points on $y = lo g_{2} x$ , take points on $y = 2^{x}$ and swap the coordinates. This is not a trick — it is exactly what taking an inverse means: the input and output trade places.

Point on $y = 2^{x}$	Point on $y = lo g_{2} x$	Reason
$(- 1, \frac{1}{2})$	$(\frac{1}{2}, - 1)$	$2^{- 1} = \frac{1}{2}$
$(0, 1)$	$(1, 0)$	$2^{0} = 1$
$(1, 2)$	$(2, 1)$	$2^{1} = 2$
$(2, 4)$	$(4, 2)$	$2^{2} = 4$
$(3, 8)$	$(8, 3)$	$2^{3} = 8$

Swapping coordinates turns the y-intercept (0, 1) into the x-intercept (1, 0).

Fig 2The anatomy of y = log b of x: a vertical asymptote at x = 0, an x-intercept at (1, 0), and slow unbounded growth to the right.

Key features of the parent function

Domain: $x > 0$ . The argument of a logarithm must be positive, so the graph lives entirely to the right of the $y$ -axis.
Range: all real numbers, $y \in R$ . This is the exponential's range $y > 0$ and domain "all reals" swapping places.
Vertical asymptote at $x = 0$ — the reflection of the exponential's horizontal asymptote $y = 0$ . The curve dives downward forever as $x$ approaches 0 from the right.
$x$ -intercept at $(1, 0)$ for every base, because $lo g_{b} 1 = 0$ always. There is no $y$ -intercept, since $x = 0$ is not in the domain.
Passes through $(b, 1)$ , because $lo g_{b} b = 1$ — this is how you read the base off a graph.
For $b > 1$ the function increases forever, but more and more slowly. It has no upper bound and no horizontal asymptote — it just takes a very long time to get anywhere.

AExponential y = bˣ

Domain: all real numbers
Range: y > 0
Horizontal asymptote y = 0
y-intercept (0, 1)
Grows explosively fast

BLogarithmic y = log_b x

Domain: x > 0
Range: all real numbers
Vertical asymptote x = 0
x-intercept (1, 0)
Grows unboundedly but very slowly

Transformations

Logarithmic functions transform using the same framework as every other function family you have met. Nothing new has to be memorized — but the asymptote is vertical now, so it is $d$ , not $c$ , that moves it.

y = a lo g_{b} (k (x - d)) + c

Parameter	Effect	Moves the asymptote?
a	Vertical stretch or compression by \|a\|; reflection in the x-axis if a < 0.	No
k	Horizontal stretch or compression by 1/\|k\|; reflection in the y-axis if k < 0.	No
d	Horizontal translation: d units right if d > 0, left if d < 0.	Yes — to x = d
c	Vertical translation: c units up if c > 0, down if c < 0.	No

Loading interactive…

Drag the parameters and watch the vertical asymptote follow d while the vertical shift c slides the curve without moving it.

How a transformation moves the asymptote

For an exponential, the vertical shift moved the horizontal asymptote. For a logarithm it is the opposite: the horizontal shift moves the vertical asymptote, from

x = 0

x = d

. A vertical shift slides the curve up or down but leaves the asymptote exactly where it was, because the curve still plunges at the same

x

-value. The fastest reliable way to find the domain is then to ignore the transformation vocabulary entirely and just demand that the argument be positive: solve

k (x - d) > 0

. If

k > 0

that gives

x > d

and the curve sits to the right of its asymptote; if

k < 0

it gives

x < d

and the curve has been flipped to the left of it.

Factor before you read off the parameters

An argument like

8 - 2 x

hides its parameters. Factor out the coefficient of

x

first:

8 - 2 x = - 2 (x - 4)

, which reveals

k = - 2

and

d = 4

. Reading

d

straight off an unfactored expression is a guaranteed way to get the asymptote wrong.

VideoGraphs a transformed logarithm two ways: by applying the shifts and reflections to the parent graph, and by rewriting it in exponential form and plotting points, identifying the vertical asymptote along the way.Watch on YouTube

Worked examples

Reveal one step at a time and try to beat it to the next line.

Example 1

State the domain, range, asymptote, and two points on

y = lo g_{5} x

Example 2

For

y = 3 lo g_{2} (x - 4) + 1

, describe the transformations, state the asymptote and domain, and find the

x

-value where

y = 4

Example 3

Find the asymptote and domain of

y = lo g_{5} (8 - 2 x)

, and state which side of the asymptote the graph lies on.

Example 4

The graph of

y = lo g_{3} x

is stretched vertically by a factor of 2, then translated 6 units left and 5 units down. Write the equation and state the new domain.

Where marks get lost

Giving the asymptote of $y = lo g_{2} (x - 5)$ as $y = 5$ or $y = 0$ . The asymptote of a logarithmic graph is vertical, so it must be written as $x = 5$ .
Writing the domain as $x > 0$ for every logarithm, even after a translation. Always solve "argument $> 0$ " for the actual function in front of you.
Assuming the graph always sits to the right of its asymptote. When $k < 0$ the graph is reflected in the $y$ -axis and sits to the left, as in $y = lo g (3 - x)$ .
Reading $d$ off an unfactored argument: in $y = lo g (2 x + 6)$ the asymptote is not $x = - 6$ . Factor first — $2 x + 6 = 2 (x + 3)$ — so the asymptote is $x = - 3$ .
Thinking the curve levels off or has a maximum because it looks flat on a calculator screen. It keeps climbing forever; $lo g_{10} x$ reaches 6 only when $x$ is a million.

Flashcards 8

Relationship between

y = b^{x}

and

y = lo g_{b} x

Domain of

y = lo g_{b} x

Range of

y = lo g_{b} x

Asymptote of

y = lo g_{b} x

+ 4 more

Practice

State the equation of the asymptote of

y = lo g_{2} (x - 5)

What is the domain of

y = lo g_{3} (2 x - 6)

Find the value of

y

on the graph of

y = lo g_{2} x

when

x = 32

What is the range of

y = lo g_{b} x

(with

b > 1

State the domain of

y = lo g (4 - x)

The graph of

y = lo g_{7} x

passes through the point

(x, 1)

. Find

x

For

y = lo g_{2} (x - 3) + 4

, find

y

when

x = 11

Which transformation moves the vertical asymptote of

y = a lo g_{b} (k (x - d)) + c

State the asymptote of

y = lo g_{5} (3 x + 12)

Match each feature of

y = b^{x}

to the corresponding feature of

y = lo g_{b} x

Horizontal asymptote y = 0

y-intercept (0, 1)

Domain: all real numbers

Range: y > 0

Passes through (1, b)

Order the steps to find the domain and asymptote of

y = - 2 lo g_{3} (6 - 3 x) + 1

1Read off the parameters: k = −3 and d = 2
2Solve 6 − 3x > 0 to get the domain x < 2
3State the asymptote as x = d, so x = 2
4Factor the coefficient of x out of the argument: 6 − 3x = −3(x − 2)

Spot a mistake on this page? Let us know