Exponential and logarithmic functions and graphs

IB Mathematics: Analysis and Approaches SL· 25 min read

1. Key Features of Exponential Function Graphs★★☆☆☆⏱ 10 min

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📘 Definition

General Exponential Function

f(x) = ab^{kx} + c, a \neq 0, b > 0, b \neq 1

A function with the independent variable $x$ in the exponent. $b > 1$ gives exponential growth, $0 < b < 1$ gives exponential decay.

Example:

$f (x) = 3 (2)^{x} - 1$ is a growth exponential function

All exponential functions are defined for all real $x$ , so their domain is always $(- \infty, \infty)$ . They have a horizontal asymptote at $y = c$ for the general form, and their range depends on the sign of $a$ .

📐 Worked Example

Sketch $f (x) = 3 (2)^{x} - 1$ , state its domain, range and asymptote.

1
Identify parameters: $a = 3$ , $b = 2 > 1$ , $c = - 1$ . Asymptote is at $y = c = - 1$ .
2
Find the y-intercept by substituting $x = 0$ :
3
$f(0) = 3(2^0) - 1 = 3(1) - 1 = 2$
4
Find the x-intercept by setting $f (x) = 0$ :
5
$0 = 3(2^x) - 1 \implies 2^x = \frac{1}{3} \implies x = \log_2\left(\frac{1}{3}\right) \approx -1.58$
6
State domain and range: $f (x)$ is defined for all real $x$ , so domain is $(- \infty, \infty)$ . Since $3 \cdot 2^{x} > 0$ for all $x$ , $3 \cdot 2^{x} - 1 > - 1$ , so range is $(- 1, \infty)$ .

Exam tip:

Always check the sign of $a$ when finding the range: if $a < 0$ , the range will be $(- \infty, c)$ instead of $(c, \infty)$ . Exclude the asymptote value from the range.

2. Key Features of Logarithmic Function Graphs★★★☆☆⏱ 12 min

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📘 Definition

General Logarithmic Function

f(x) = a\log_b(kx + c) + d, a \neq 0, b > 0, b \neq 1

The inverse of an exponential function, only defined for positive arguments. It has a vertical rather than horizontal asymptote.

Example:

$g (x) = ln (2 x - 4) + 1$ is a natural logarithmic function

Logarithmic functions can only accept positive input values, so their domain is always restricted. The vertical asymptote occurs where the argument of the logarithm equals zero. Their range is always all real numbers.

📐 Worked Example

For $g (x) = ln (2 x - 4) + 1$ , find the asymptote, domain, range and intercepts.

1
The logarithm is only defined when its argument is positive:
2
$2x - 4 > 0 \implies x > 2$
3
The vertical asymptote occurs when the argument equals zero:
4
$2x - 4 = 0 \implies x = 2$
5
State domain and range: Domain is $(2, \infty)$ , range is all real numbers $(- \infty, \infty)$ .
6
Check for intercepts: No y-intercept, since $x = 0$ is not in the domain. Set $g (x) = 0$ to find the x-intercept:
7
$0 = \ln(2x-4) + 1 \implies 2x - 4 = e^{-1} \implies x \approx 2.18$

Exam tip:

Leave intercepts and other values in exact form unless the question explicitly asks for a decimal approximation.

3. Exponential and Logarithmic Functions as Inverses★★☆☆☆⏱ 10 min

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Exponential and logarithmic functions with the same base are inverse functions of one another. This means their graphs are reflections of each other across the line $y = x$ , and their key features swap: domain exchanges with range, and horizontal asymptotes become vertical asymptotes.

📐 Worked Example

Confirm that $f (x) = 2 e^{x + 1}$ and $g (x) = ln (\frac{x}{2}) - 1$ are inverses.

1
To confirm inverses, we check that $f (g (x)) = x$ and $g (f (x)) = x$ .
2
First calculate $f (g (x))$ :
3
$f(g(x)) = 2e^{(\ln(x/2) - 1) + 1} = 2e^{\ln(x/2)} = 2 \cdot \frac{x}{2} = x$
4
Next calculate $g (f (x))$ :
5
$g(f(x)) = \ln\left(\frac{2e^{x+1}}{2}\right) - 1 = \ln(e^{x+1}) - 1 = (x+1) - 1 = x$
6
Since both compositions equal $x$ , the functions are confirmed as inverses.

4. Transformations of Exponential and Logarithmic Graphs★★★☆☆⏱ 15 min

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Standard function transformation rules apply to exponential and logarithmic graphs. The most common mistake comes from misidentifying horizontal translations when the x-term has a coefficient.

📐 Worked Example

Describe the transformation that maps $y = e^{x}$ onto $y = e^{2 x + 4} - 1$ , and state the new asymptote.

1
Rewrite the function to factor out the coefficient of $x$ , to clearly identify transformations:
2
$y = e^{2(x + 2)} - 1$
3
Starting from $y = e^{x}$ : first apply a horizontal compression by scale factor $\frac{1}{2}$ (parallel to the x-axis).
4
Next, translate the graph 2 units to the left.
5
Finally, translate the graph 1 unit down.
6
The original asymptote of $y = e^{x}$ is the horizontal line $y = 0$ . After the 1 unit down translation, the new asymptote is $y = - 1$ .

Exam tip:

Always factor out the coefficient of $x$ before identifying horizontal translations. This avoids the common mistake of shifting 4 units left instead of 2 units left for $e^{2 x + 4}$ .

5. Common Pitfalls

Wrong move:

Writing the range of $f (x) = 2^{x} - 3$ as $[- 3, \infty)$ , including the asymptote.

Why:

$2^{x}$ never equals 0, so $f (x)$ never equals -3. Asymptotes are never included in the range or domain.

Correct move:

Write the range as $(- 3, \infty)$ , using an open bracket for the asymptote value.

Wrong move:

Stating that $f (x) = e^{x + 2}$ is shifted 2 units right from $e^{x}$ .

Why:

Transformation rules state $f (x + a)$ shifts $f (x)$ $a$ units left, because the function reaches the same $y$ -value $a$ units earlier.

Correct move:

Identify a shift of 2 units left from the original graph.

Wrong move:

Claiming $f (x) = (- 3)^{x}$ is a valid exponential function over all real $x$ .

Why:

Negative bases produce undefined values for many $x$ , e.g. $(- 3)^{0.5} = - 3$ is not a real number.

Correct move:

Recognize that exponential functions always have positive bases, not equal to 1, by definition.

Wrong move:

Including the vertical asymptote $x = 3$ in the domain of $f (x) = lo g (3 x - 9)$ .

Why:

The function is undefined at the asymptote, where the argument of the logarithm equals zero.

Correct move:

State the domain as $(3, \infty)$ , excluding the asymptote value $x = 3$ .

6. Quick Reference Cheatsheet

Function Type	Domain	Range	Asymptote
$f (x) = b^{x}, b > 1$	$(- \infty, \infty)$	$(0, \infty)$	Horizontal: $y = 0$
$f (x) = b^{x}, 0 < b < 1$	$(- \infty, \infty)$	$(0, \infty)$	Horizontal: $y = 0$
$f (x) = lo g_{b} x, b > 1$	$(0, \infty)$	$(- \infty, \infty)$	Vertical: $x = 0$
$f (x) = lo g_{b} x, 0 < b < 1$	$(0, \infty)$	$(- \infty, \infty)$	Vertical: $x = 0$
$f (x) = a b^{k x} + c$	$(- \infty, \infty)$	$a > 0 : (c, \infty)$ ; $a < 0 : (- \infty, c)$	Horizontal: $y = c$
$f (x) = a lo g_{b} (k x + c) + d$	$k > 0 : (- \frac{c}{k}, \infty)$	$(- \infty, \infty)$	Vertical: $x = - \frac{c}{k}$

7. Frequently Asked

Do I need to memorize these shapes for the exam?

Yes, paper 1 (non-calculator) frequently asks you to sketch or identify graphs, so knowing base shapes, asymptotes and key intercepts is essential.

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2025 · 1
Identify graph features of log function
2024 · 2
Transform exponential graph, find asymptote
2023 · 1
Confirm inverse relationship of functions

What's Next

Mastery of exponential and logarithmic graphs is a critical foundation for upcoming topics including solving exponential and logarithmic equations, modeling real-world growth and decay, and the differentiation and integration of these functions later in the course. Graph interpretation questions appear frequently on both paper 1 and paper 2 of the IB AA SL exam, often paired with modeling problems or inverse function questions. Building fluency with key features and transformations now will make more complex topics much easier to master.