Transformations of Logarithmic Functions

You’ve already transformed parabolas, exponential functions, and sine curves with the same four parameters. Logarithmic functions work exactly the same way. The one thing to keep an eye on is the vertical asymptote: it moves left and right with the graph, and it decides the domain.

Key ideas

The transformed logarithmic function

$$y = a\log_b\big(k(x - d)\big) + c$$

With no base written, $y = a\log\big(k(x - d)\big) + c$ is a transformation of the common logarithm $y = \log x$. The parameters do the same jobs as for any function:

Parameter	Effect
$a$	vertical stretch or compression by $\lvert a \rvert$; reflection in the $x$-axis if $a \lt 0$
$k$	horizontal stretch or compression by $\dfrac{1}{\lvert k \rvert}$; reflection in the $y$-axis if $k \lt 0$
$d$	horizontal translation
$c$	vertical translation

The mapping rule is $(x, y) \to \left(\dfrac{x}{k} + d,\ ay + c\right)$.

The asymptote moves with d

The parent $y = \log_b x$ has the vertical asymptote $x = 0$. A vertical line is only moved by horizontal translations, so after the transformation:

• the vertical asymptote is $x = d$
• the domain is $x \gt d$ if $k \gt 0$, or $x \lt d$ if $k \lt 0$ (a reflection in the $y$-axis flips the graph to the other side of the asymptote)
• the range is still all real numbers

Stretches and vertical shifts don’t move the asymptote at all. (Compare exponential functions, where the asymptote is horizontal and moves with $c$.)

Key points to map

Parent	Key points
$y = \log x$	$(0.1, -1)$, $(1, 0)$, $(10, 1)$
$y = \log_2 x$	$\left(\tfrac{1}{2}, -1\right)$, $(1, 0)$, $(2, 1)$, $(4, 2)$, $(8, 3)$
$y = \log_b x$	$\left(\tfrac{1}{b}, -1\right)$, $(1, 0)$, $(b, 1)$

Map the key points and the asymptote, then draw a smooth curve that approaches the new asymptote.

Two descriptions of the same graph

Some transformations of a log graph can be described in two ways. For example, $y = \log(10x)$ is a horizontal compression by a factor of $\tfrac{1}{10}$, but it’s also the same graph as $y = \log x + 1$, a translation $1$ unit up. Example 4 shows why. You saw the same thing with exponential functions, where $3^{x + 2} = 9\left(3^x\right)$ is both a horizontal translation and a vertical stretch.

Worked examples

Example 1: Describing transformations

Describe how $y = -2\log(x - 3) + 1$ relates to $y = \log x$. State the asymptote, domain, and range.

Solution. Here $a = -2$, $k = 1$, $d = 3$, and $c = 1$.

• A vertical stretch by a factor of $2$ and a reflection in the $x$-axis.
• A translation $3$ units right and $1$ unit up.

The asymptote moves right $3$: $x = 3$. Since $k \gt 0$, the domain is $\{x \in \mathbb{R} \mid x \gt 3\}$. The range is $\{y \in \mathbb{R}\}$.

Example 2: Sketching with mapping notation

Sketch $y = 2\log_2(x + 3) - 1$. State the asymptote, domain, range, and intercepts.

Solution. $a = 2$, $d = -3$, and $c = -1$, so the mapping rule is $(x, y) \to (x - 3,\ 2y - 1)$.

$y = \log_2 x$	$\left(\tfrac{1}{2}, -1\right)$	$(1, 0)$	$(2, 1)$	$(4, 2)$	$(8, 3)$
$y = 2\log_2(x + 3) - 1$	$\left(-\tfrac{5}{2}, -3\right)$	$(-2, -1)$	$(-1, 1)$	$(1, 3)$	$(5, 5)$

The asymptote $x = 0$ moves left $3$, to $x = -3$.

The asymptote moves left with the graph, from $x = 0$ to $x = -3$.

Domain $\{x \in \mathbb{R} \mid x \gt -3\}$; range $\{y \in \mathbb{R}\}$.

$y$-intercept: $y = 2\log_2 3 - 1 \approx 2.17$, which matches the graph crossing the $y$-axis between $(-1, 1)$ and $(1, 3)$.

$x$-intercept: set $y = 0$:

$$2\log_2(x + 3) - 1 = 0 \quad\Rightarrow\quad \log_2(x + 3) = \tfrac{1}{2} \quad\Rightarrow\quad x + 3 = 2^{\frac{1}{2}} \quad\Rightarrow\quad x = \sqrt{2} - 3 \approx -1.59$$

Example 3: Factoring out k

Describe the transformations in $y = \log(2x - 8)$, map the points $(0.1, -1)$, $(1, 0)$, and $(10, 1)$, and state the asymptote and domain.

Solution. First factor out the coefficient of $x$:

$$y = \log(2x - 8) = \log\big(2(x - 4)\big)$$

So $k = 2$ and $d = 4$: a horizontal compression by a factor of $\tfrac{1}{2}$, then a translation $4$ units right. The mapping rule is $(x, y) \to \left(\tfrac{x}{2} + 4,\ y\right)$:

$$(0.1, -1) \to (4.05, -1), \qquad (1, 0) \to (4.5, 0), \qquad (10, 1) \to (9, 1)$$

The asymptote is $x = 4$ and the domain is $\{x \in \mathbb{R} \mid x \gt 4\}$.

Check: at $x = 9$, $y = \log(18 - 8) = \log 10 = 1$. ✓ Also, the argument $2x - 8$ is positive exactly when $x \gt 4$, which confirms the domain.

Example 4: The same graph, two ways

Show that $y = \log(1000x)$ and $y = \log x + 3$ have the same graph. Describe each as a transformation of $y = \log x$.

Solution. Pick any $x \gt 0$ and write it as a power of $10$: $x = 10^m$, so $\log x = m$. Then

$$\log(1000x) = \log\left(10^3 \cdot 10^m\right) = \log\left(10^{m + 3}\right) = m + 3 = \log x + 3$$

So the two equations give the same $y$ for every $x$, and the graphs are identical.

• $y = \log(1000x)$: a horizontal compression by a factor of $\tfrac{1}{1000}$.
• $y = \log x + 3$: a translation $3$ units up.

Check with a point: $(10, 1)$ on $y = \log x$ maps to $(0.01, 1)$ under the compression and to $(10, 4)$ under the translation. Both are on $y = \log(1000x)$: $\log 10 = 1$ ✓ and $\log 10\,000 = 4$ ✓.

Common mistakes

Moving the asymptote with $c$. A log graph’s asymptote is vertical, so only $d$ moves it. $y = \log(x + 2) + 5$ has asymptote $x = -2$, not $y = 5$.

Reading the horizontal shift with the wrong sign. $\log(x + 3)$ moves the graph left $3$, and the domain becomes $x \gt -3$.

Not factoring out $k$. $y = \log(2x - 8)$ is $\log\big(2(x - 4)\big)$: the shift is $4$ right, not $8$. A quick check: the asymptote is where the argument equals $0$, and $2x - 8 = 0$ at $x = 4$.

Getting the domain backwards after a reflection in the $y$-axis. If $k \lt 0$, the graph is on the left of the asymptote. For $y = \log(-x)$ the domain is $x \lt 0$.

Using the exponential’s key points. The log graph doesn’t pass through $(0, 1)$. Start from $(1, 0)$ and $(b, 1)$ (for base $10$, $(1, 0)$ and $(10, 1)$).

Practice

1. (Warm-up) State the asymptote and domain of $y = \log(x + 5) - 2$.

Solution

Asymptote $x = -5$; domain $\{x \in \mathbb{R} \mid x \gt -5\}$.

2. (Warm-up) Describe the transformations that take $y = \log x$ to $y = 3\log x + 4$.

Solution

A vertical stretch by a factor of $3$, then a translation $4$ units up. The asymptote stays at $x = 0$.

3. (Warm-up) Describe the transformation that takes $y = \log x$ to $y = \log(-x)$, and state the domain.

Solution

A reflection in the $y$-axis. The domain is $\{x \in \mathbb{R} \mid x \lt 0\}$.

4. (Core) For $y = \log_3(x - 2) + 1$, map the points $\left(\tfrac{1}{3}, -1\right)$, $(1, 0)$, $(3, 1)$, and $(9, 2)$ of $y = \log_3 x$. State the asymptote, domain, and range.

Solution

The mapping rule is $(x, y) \to (x + 2,\ y + 1)$:

$$\left(\tfrac{1}{3}, -1\right) \to \left(\tfrac{7}{3}, 0\right), \quad (1, 0) \to (3, 1), \quad (3, 1) \to (5, 2), \quad (9, 2) \to (11, 3)$$

Asymptote $x = 2$; domain $\{x \in \mathbb{R} \mid x \gt 2\}$; range $\{y \in \mathbb{R}\}$.

5. (Core) For $y = -2\log(x + 1) + 2$:

• (a) Describe the transformations.
• (b) Map $(0.1, -1)$, $(1, 0)$, and $(10, 1)$.
• (c) Find the intercepts and state the asymptote.

Solution

(a) A vertical stretch by a factor of $2$, a reflection in the $x$-axis, and a translation $1$ unit left and $2$ units up.

(b) The rule is $(x, y) \to (x - 1,\ -2y + 2)$:

$$(0.1, -1) \to (-0.9, 4), \qquad (1, 0) \to (0, 2), \qquad (10, 1) \to (9, 0)$$

(c) The $y$-intercept is $2$ (from the point $(0, 2)$). For the $x$-intercept, $-2\log(x + 1) + 2 = 0$ gives $\log(x + 1) = 1$, so $x + 1 = 10$ and $x = 9$. The asymptote is $x = -1$.

6. (Core) Describe the transformations in $y = \log_2(4x + 8)$ and map $(1, 0)$, $(2, 1)$, and $(4, 2)$. Then show that the graph is also a vertical translation of $y = \log_2(x + 2)$.

Solution

Factor: $y = \log_2\big(4(x + 2)\big)$. A horizontal compression by a factor of $\tfrac{1}{4}$, then a translation $2$ units left. The rule is $(x, y) \to \left(\tfrac{x}{4} - 2,\ y\right)$:

$$(1, 0) \to \left(-\tfrac{7}{4}, 0\right), \qquad (2, 1) \to \left(-\tfrac{3}{2}, 1\right), \qquad (4, 2) \to (-1, 2)$$

Since $4 = 2^2$, write $x + 2 = 2^m$. Then $\log_2\big(4(x + 2)\big) = \log_2\left(2^{m + 2}\right) = m + 2 = \log_2(x + 2) + 2$. So the graph is also $y = \log_2(x + 2)$ translated $2$ units up.

Check: at $x = -1$, $\log_2(1) + 2 = 2$, which matches $(-1, 2)$. ✓

7. (Core) A function of the form $y = a\log x + c$ passes through $(1, 4)$ and $(10, 1)$. Find $a$ and $c$.

Solution

At $x = 1$: $a\log 1 + c = 0 + c = 4$, so $c = 4$.

At $x = 10$: $a\log 10 + 4 = a + 4 = 1$, so $a = -3$.

The function is $y = -3\log x + 4$.

8. (Challenge) For $y = \log\big(-2(x - 5)\big)$, describe the transformations, map $(0.1, -1)$, $(1, 0)$, and $(10, 1)$, and state the asymptote and domain.

Solution

$k = -2$ and $d = 5$: a horizontal compression by a factor of $\tfrac{1}{2}$, a reflection in the $y$-axis, and a translation $5$ units right. The rule is $(x, y) \to \left(-\tfrac{x}{2} + 5,\ y\right)$:

$$(0.1, -1) \to (4.95, -1), \qquad (1, 0) \to (4.5, 0), \qquad (10, 1) \to (0, 1)$$

Asymptote $x = 5$. Because $k \lt 0$, the graph is to the left of the asymptote: the domain is $\{x \in \mathbb{R} \mid x \lt 5\}$.

Check: at $x = 0$, $y = \log\big(-2(-5)\big) = \log 10 = 1$. ✓

9. (Challenge) The point $(7, 5)$ is on the graph of $y = 2\log_3(x - 4) + 3$. Which point on $y = \log_3 x$ did it come from?

Solution

The rule is $(x, y) \to (x + 4,\ 2y + 3)$. Work backwards:

$$x + 4 = 7 \Rightarrow x = 3, \qquad 2y + 3 = 5 \Rightarrow y = 1$$

It came from $(3, 1)$, which is on $y = \log_3 x$ since $\log_3 3 = 1$. ✓