# Exponential and logarithmic functions and graphs

> IB Mathematics: Analysis and Approaches SL · IB AA SL
> Source: https://www.owlsprep.com/study/ib-math-aa-sl-u2-exponential-and-logarithmic-functions-and/

This subtopic covers the shape, key features and transformations of exponential and logarithmic functions, explores their inverse relationship, and builds core skills for solving graph-based problems common in the IB AA SL exam.

**Prerequisites:** [Function notation and basic properties](https://www.owlsprep.com/study/ib-math-aa-sl-u2-function-basics/); [Inverse function basics](https://www.owlsprep.com/study/ib-math-aa-sl-u2-inverse-functions/); [Laws of logarithms](https://www.owlsprep.com/study/ib-math-aa-sl-u2-logarithm-laws/)

## Learning objectives

- Identify key features of exponential and logarithmic function graphs
- Sketch and transform graphs of exponential and logarithmic functions
- Explain the inverse relationship between exponential and logarithmic functions
- Solve exam problems involving these graphs and their properties

## Key Features of Exponential Function Graphs

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

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

*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.

*Calculator:* forbidden

## Key Features of Logarithmic Function Graphs

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

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

*Example:* $g(x) = \ln(2x - 4) + 1$ 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 $g(x) = \ln(2x - 4) + 1$, 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.

*Calculator:* forbidden

## Exponential and Logarithmic Functions as Inverses

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 $f(x) = 2e^{x+1}$ and $g(x) = \ln\left(\frac{x}{2}\right) - 1$ 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.

> **info**
>
> This inverse relationship is the foundation for solving equations with unknown exponents by taking logarithms of both sides.

*Calculator:* forbidden

## Transformations of Exponential and Logarithmic Graphs

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 $y = e^{2x + 4} - 1$, 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 $e^{2x+4}$.

*Calculator:* forbidden

## Common pitfalls

- **Wrong:** Writing the range of $f(x) = 2^x - 3$ as $[-3, \infty)$, including the asymptote.
  - Why it fails: $2^x$ never equals 0, so $f(x)$ never equals -3. Asymptotes are never included in the range or domain.
  - Correct: Write the range as $(-3, \infty)$, using an open bracket for the asymptote value.
- **Wrong:** Stating that $f(x) = e^{x+2}$ is shifted 2 units right from $e^x$.
  - Why it fails: Transformation rules state $f(x+a)$ shifts $f(x)$ $a$ units left, because the function reaches the same $y$-value $a$ units earlier.
  - Correct: Identify a shift of 2 units left from the original graph.
- **Wrong:** Claiming $f(x) = (-3)^x$ is a valid exponential function over all real $x$.
  - Why it fails: Negative bases produce undefined values for many $x$, e.g. $(-3)^{0.5} = \sqrt{-3}$ is not a real number.
  - Correct: Recognize that exponential functions always have positive bases, not equal to 1, by definition.
- **Wrong:** Including the vertical asymptote $x=3$ in the domain of $f(x) = \log(3x - 9)$.
  - Why it fails: The function is undefined at the asymptote, where the argument of the logarithm equals zero.
  - Correct: State the domain as $(3, \infty)$, excluding the asymptote value $x=3$.

## 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: $y=0$ |
| $f(x) = \log_b x, b>1$ | $(0, \infty)$ | $(-\infty, \infty)$ | Vertical: $x=0$ |
| $f(x) = \log_b x, 0<b<1$ | $(0, \infty)$ | $(-\infty, \infty)$ | Vertical: $x=0$ |
| $f(x) = ab^{kx} + c$ | $(-\infty, \infty)$ | $a>0: (c, \infty)$; $a<0: (-\infty, c)$ | Horizontal: $y=c$ |
| $f(x) = a\log_b(kx + c) + d$ | $k>0: (-\frac{c}{k}, \infty)$ | $(-\infty, \infty)$ | Vertical: $x=-\frac{c}{k}$ |

## 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.

- [Geometry & Trigonometry](https://www.owlsprep.com/study/ib-math-aa-sl-u3-overview/)
- [2D and 3D coordinate geometry](https://www.owlsprep.com/study/ib-math-aa-sl-u3-2d-and-3d-coordinate-geometry/)
- [Right triangle trigonometry and the unit circle](https://www.owlsprep.com/study/ib-math-aa-sl-u3-right-triangle-trigonometry-and-the/)

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