Logarithmic graphs, domains and transformations | Year 12

01 / LEARN

Understand the idea.

Start with the basic graph

The graph $y = ln x$ has domain $x > 0$ , range $ℝ$ and vertical asymptote $x = 0$ . Points $(1, 0)$ , $(e, 1)$ and $(e^{- 1}, - 1)$ give exact anchors for a sketch.

It increases while becoming less steep: $y^{'} = \frac{1}{x} > 0$ and $y^{''} = - \frac{1}{x^{2}} < 0$ . As $x$ approaches zero from the positive side, the height tends to negative infinity; the graph has no point on the asymptote.

Read the input before drawing

For $y = a ln (b (x - h)) + k$ with nonzero $a, b$ , solve $b (x - h) > 0$ first. The boundary $x = h$ gives the vertical asymptote, and the sign of $b$ selects which side contains the graph.

The coefficient $a$ scales heights and reflects them if negative; $k$ translates vertically. Find intercepts by solving equations, and check whether $x = 0$ is in the domain before trying to calculate a y-intercept.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

Translate a graph

Give the domain, asymptote and intercepts of $y = ln (x - 1) + 1$ .

Worked solution

1
Find the domain and its logarithmic boundary.
$x - 1 > 0 ⟹ x > 1, vertical asymptote x = 1$
As x approaches 1 from above, the positive input tends to zero and the height tends to negative infinity.
2
Set the height to zero.
$ln (x - 1) + 1 = 0 ⟹ ln (x - 1) = - 1$
An x-intercept is a point where y=0.
3
Exponentiate to find the x-intercept.
$x - 1 = e^{- 1} ⟹ x = 1 + e^{- 1}, (1 + e^{- 1}, 0)$
The exponential gives a positive input, so the resulting x-value satisfies x>1.
4
Check for a y-intercept.
$0 \notin (1, \infty) ⟹ no y-intercept$
A y-intercept requires the function to be defined at x=0.

Answer

Domain $(1, \infty)$ ; asymptote $x = 1$ ; x-intercept $(1 + e^{- 1}, 0)$ ; no y-intercept.

EXAMPLE 02

Reflect and restrict

For $g (x) = 2 - ln (4 - x)$ , find the domain and the point where $g (x) = 2$ .

Worked solution

1
Solve the input restriction.
$4 - x > 0 ⟹ x < 4, domain (- \infty, 4)$
Rearranging the inequality puts the graph to the left of its boundary.
2
Set the requested height.
$2 - ln (4 - x) = 2 ⟹ ln (4 - x) = 0$
Subtracting 2 and changing both signs isolates the logarithm.
3
Reverse the logarithm and solve.
$4 - x = e^{0} = 1 ⟹ x = 3$
A logarithm is zero when its input is 1.
4
Check and give the point.
$3 < 4, g (3) = 2 - ln 1 = 2, (3, 2)$
The candidate belongs to the domain and has the required height.

Answer

Domain $(- \infty, 4)$ and point $(3, 2)$ .

03 / YOUR TURN

Try it yourself.

Check your understanding

What is the vertical asymptote of $y = 3 ln (x + 2) - 5$ ?

Show answer & reasoning

$x = - 2$

The input $x + 2$ approaches zero at $x = - 2$ , from the allowed side $x > - 2$ . Multiplying the logarithm by $3$ and shifting it down by $5$ do not move that vertical asymptote.

Keep practising

6 worksheet questions · 15 marks

Open worksheet Worked solutions

Read the questions online

State the domain and asymptote of $y = {log}_{2} (x - 3)$ .
2 marks
Find the x-intercept of $y = ln x - 2$ .
2 marks
Find the domain and y-intercept of $y = ln (6 - 2 x)$ .
3 marks
Describe the transformation from $y = ln x$ to $y = - 2 ln x + 3$ .
2 marks
Find the inverse of $f (x) = ln (x + 2) - 1$ and its domain.
3 marks
Sketch $y = ln x$ and $y = ln (2 x)$ on the same axes using exact anchor points. State their vertical separation.
3 marks

Read the worked solutions

Equivalent correct methods earn the corresponding marks. Include units where asked.

State the domain and asymptote of $y = {log}_{2} (x - 3)$ .
2 marks
The input $x - 3$ must be positive; its zero gives the asymptote.
$x > 3$ ; $x = 3$
Find the x-intercept of $y = ln x - 2$ .
2 marks
$ln x = 2$ gives $x = e^{2}$ .
$(e^{2}, 0)$
Find the domain and y-intercept of $y = ln (6 - 2 x)$ .
3 marks
$6 - 2 x > 0$ gives $x < 3$ ; substitute $x = 0$ .
$x < 3$ ; $(0, ln 6)$
Describe the transformation from $y = ln x$ to $y = - 2 ln x + 3$ .
2 marks
The multiplier $- 2$ changes output values; the $+ 3$ is applied afterwards.
Reflection in the x-axis, vertical scale factor 2, then translation up 3.
Find the inverse of $f (x) = ln (x + 2) - 1$ and its domain.
3 marks
$y + 1 = ln (x + 2)$ gives $x = e^{y + 1} - 2$ . The original range is all real numbers.
$f^{- 1} (x) = e^{x + 1} - 2$ , $x \in ℝ$
Sketch $y = ln x$ and $y = ln (2 x)$ on the same axes using exact anchor points. State their vertical separation.
3 marks
For $x > 0$ , $ln (2 x) = ln x + ln 2$ . Anchor points include $(1, 0)$ and $(1, ln 2)$ .
The second graph is $ln 2$ units above the first; both have asymptote $x = 0$ .

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Curriculum reference & lesson scope

SACE Board of South Australia

Stage 2 Mathematical Methods: logarithmic functions and their applications; reciprocal-linear antiderivatives are taught in the Integration chapter.

Official curriculum source ↗

In this lesson

Understand the idea.

Start with the basic graph

Read the input before drawing

See the method in action.

Translate a graph

Find the domain and its logarithmic boundary.

Set the height to zero.

Exponentiate to find the x-intercept.

Check for a y-intercept.

Reflect and restrict

Solve the input restriction.

Set the requested height.

Reverse the logarithm and solve.

Check and give the point.

Try it yourself.

Check your understanding

Keep practising