# Exponential and logarithmic functions

> IB Mathematics Applications and Interpretation HL · Unit 2: Functions
> Source: https://www.owlsprep.com/study/ib-math-ai-hl-u2-exponential-and-logarithmic-functions/

This subtopic covers core properties of exponential and logarithmic functions, their relationship as inverse functions, key graph features, algebraic manipulation, and methods for solving equations tested in IB AI HL exams.

**Prerequisites:** [Laws of exponents and indices](https://www.owlsprep.com/study/ib-math-ai-hl-u2-exponent-rules/); [Inverse functions](https://www.owlsprep.com/study/ib-math-ai-hl-u2-inverse-functions/)

## Learning objectives

- Define exponential and logarithmic functions and recognize their inverse relationship
- Apply core laws of logarithms to simplify and manipulate expressions
- Identify key graph features of exponential and logarithmic functions
- Solve exponential and logarithmic equations for unknown variables
- Avoid common algebraic and graphical errors in exam questions

## Core Definitions & Inverse Relationship

**Exponential Function** — A function where the independent variable $x$ appears as the exponent of a constant base $a$. All real numbers are in the domain, and output values are always positive.

*Notation:* f(x) = a^x, \quad a>0, a \neq 1

*Example:* $f(x) = 2^x$, $f(x) = e^x$

**Logarithmic Function** — The inverse of an exponential function with base $a$. By definition, $\log_a x = y$ if and only if $a^y = x$. Only positive input values are defined.

*Notation:* f(x) = \log_a x, \quad a>0, a \neq 1

*Example:* $\log_2 8 = 3$ because $2^3 = 8$

Since exponential and logarithmic functions are inverses, their graphs are reflections of each other over the line $y=x$. This inverse relationship is the foundation for all algebraic manipulation involving these functions.

**Worked example:** Convert $5^3 = 125$ to logarithmic form, and convert $\log_4 \left(\frac{1}{16}\right) = -2$ to exponential form.

1. Recall the core inverse definition: $a^y = x \iff \log_a x = y$
2. For $5^3 = 125$, we identify $a=5$, $y=3$, $x=125$. Substituting into the definition gives:
3. $$\log_5 125 = 3$$
4. For $\log_4 \left(\frac{1}{16}\right) = -2$, identify $a=4$, $y=-2$, $x=\frac{1}{16}$. Substituting gives:
5. $$4^{-2} = \frac{1}{16}$$

> **Exam tip:** Always check your conversion by substituting back into the original form to avoid mixing up the base and exponent.

## Laws of Logarithms

All logarithm laws derive directly from exponent laws, because logarithms are inverses of exponents. The core rules for any valid base $a$ are:

- Product rule: $\log_a (xy) = \log_a x + \log_a y$
- Quotient rule: $\log_a \left(\frac{x}{y}\right) = \log_a x - \log_a y$
- Power rule: $\log_a x^n = n \log_a x$
- Change of base rule: $\log_a b = \frac{\ln b}{\ln a} = \frac{\log_c b}{\log_c a}$ for any positive $c \neq 1$

> **mnemonic**
>
> Product becomes sum, quotient becomes difference, power comes out front

**Worked example:** Simplify $2 \ln 3 + \ln 4 - \ln 12$ into a single logarithm.

1. Apply the power rule to the first term, bringing the 2 out front into the exponent:
2. $$2 \ln 3 = \ln 3^2 = \ln 9$$
3. Apply the product rule to the first two terms, combining the sum of logs into the log of a product:
4. $$\ln 9 + \ln 4 = \ln (9 \times 4) = \ln 36$$
5. Apply the quotient rule, combining the difference of logs into the log of a quotient:
6. $$\ln 36 - \ln 12 = \ln \left(\frac{36}{12}\right) = \ln 3$$
7. The final simplified form is $\ln 3$.

## Graphs and Key Properties

Exponential and logarithmic functions have consistent shapes and key features for bases $a>1$ (the most common case in IB exams). The table below summarizes core properties:

| Feature | Exponential $f(x) = a^x, a>1$ | Logarithmic $f(x) = \log_a x, a>1$ |
| --- | --- | --- |
| Domain | $x \in \mathbb{R}$ | $x > 0$ |
| Range | $y > 0$ | $y \in \mathbb{R}$ |
| Asymptote | Horizontal: $y=0$ | Vertical: $x=0$ |
| Intercept | $y$-intercept at $(0,1)$ | $x$-intercept at $(1,0)$ |
| Monotonicity | Strictly increasing | Strictly increasing |

**Worked example:** Sketch the graph of $f(x) = 2^{x-1} + 3$, label the asymptote and $y$-intercept.

1. This is a transformation of the base $f(x) = 2^x$ graph: shifted 1 unit right, 3 units up.
2. The base graph has a horizontal asymptote at $y=0$. Shifting up 3 units moves the asymptote to:
3. $$y = 3$$
4. Calculate the $y$-intercept by substituting $x=0$:
5. $$f(0) = 2^{-1} + 3 = \frac{1}{2} + 3 = 3.5$$
6. The graph is strictly increasing, approaches $y=3$ as $x \to -\infty$, and increases to infinity as $x \to +\infty$, with $y$-intercept at $(0, 3.5)$.

> **Exam tip:** Exponential functions always have horizontal asymptotes; logarithmic functions always have vertical asymptotes. Mixing these up is a very common exam error.

## Solving Exponential and Logarithmic Equations

To solve equations involving exponential functions with the variable in the exponent, we take logarithms of both sides to bring down the exponent, then isolate the variable. For logarithmic equations, we use logarithm laws to combine terms, then convert to exponential form to solve.

**Worked example:** Solve $3^{2x} = 5^{x+1}$ for $x$, give your answer to 3 significant figures.

1. Take the natural logarithm of both sides to bring down exponents:
2. $$\ln \left(3^{2x}\right) = \ln \left(5^{x+1}\right)$$
3. Apply the power rule to both sides:
4. $$2x \ln 3 = (x+1) \ln 5$$
5. Expand the right-hand side and collect terms with $x$ on the left:
6. $$2x \ln 3 - x \ln 5 = \ln 5 \implies x (2 \ln 3 - \ln 5) = \ln 5$$
7. Isolate $x$ and calculate the numerical value:
8. $$x = \frac{\ln 5}{2 \ln 3 - \ln 5} \approx \frac{1.609}{2.197 - 1.609} \approx 2.74$$
9. Final answer to 3 significant figures is $x \approx 2.74$.

**Check your understanding**

Test your understanding with this quick question:

1. What is the solution to $\log_2 (x + 1) = 3$?

   - A) 2
   - B) 5
   - C) 7
   - D) 8

   *Why:* Use the definition of logarithm: $\log_2 (x+1) = 3 \iff 2^3 = x+1 \implies 8 = x+1 \implies x=7$. Don't forget to subtract 1 after calculating $2^3$!

*Calculator:* allowed

## Common pitfalls

- **Wrong:** Claiming $\ln(x + y) = \ln x + \ln y$
  - Why it fails: The product rule only applies to products of terms, not sums. There is no general rule for the logarithm of a sum.
  - Correct: $\ln(xy) = \ln x + \ln y$, $\ln(x+y)$ cannot be simplified further.
- **Wrong:** Accepting solutions that result in negative arguments for logarithms
  - Why it fails: Logarithms of non-positive numbers are undefined, so extraneous solutions are very common with logarithmic equations.
  - Correct: Always substitute all solutions back into the original equation to check all arguments are positive, and reject any invalid solutions.
- **Wrong:** Stating $f(x) = e^x + 2$ has a vertical asymptote at $x=2$
  - Why it fails: Confusing the type and position of asymptotes for transformed exponential functions.
  - Correct: Exponential functions have horizontal asymptotes. $f(x) = e^x + 2$ has a horizontal asymptote at $y=2$.
- **Wrong:** Claiming $\frac{\log_a x}{\log_a y} = \log_a \left(\frac{x}{y}\right)$
  - Why it fails: Misapplying the quotient rule for logarithms. The quotient rule applies to the logarithm of a quotient, not a quotient of logarithms.
  - Correct: $\log_a \left(\frac{x}{y}\right) = \log_a x - \log_a y$, and $\frac{\log_a x}{\log_a y} = \log_y x$ by the change of base rule.
- **Wrong:** When solving $a^x = b$, writing $x = \frac{\log b}{a}$
  - Why it fails: Common algebraic error when isolating the variable $x$.
  - Correct: If $a^x = b$, then $x = \log_a b = \frac{\ln b}{\ln a}$, there is no division by the base $a$.

## Cheatsheet

| Rule/Relationship | Formula | Key Graph Feature |
| --- | --- | --- |
| Core Inverse Relationship | $a^y = x \iff \log_a x = y$ | Exponential: Horizontal asymptote |
| Product Rule | $\log_a (xy) = \log_a x + \log_a y$ | Logarithm: Vertical asymptote |
| Quotient Rule | $\log_a (x/y) = \log_a x - \log_a y$ | Inverses reflect over $y=x$ |
| Power Rule | $\log_a x^n = n \log_a x$ | Domain of log: $x>0$ |
| Change of Base Rule | $\log_a b = \frac{\ln b}{\ln a}$ | Range of exp: $y>0$ |

## What's next

Exponential and logarithmic functions are the foundation for nearly all continuous modeling topics in IB AI HL. You will use these functions to model real-world scenarios like population growth, radioactive decay, compound interest, and cooling processes. Mastery of algebraic manipulation and graph properties is critical for success on both paper 1 and paper 2 exams, and these concepts will also be extended when you study derivatives and integrals of transcendental functions later in the course. Building a strong understanding now will make more complex application and calculus topics much easier to master.

- [Sinusoidal functions and applications](https://www.owlsprep.com/study/ib-math-ai-hl-u2-sinusoidal-functions-and-applications/)
- [Solving equations: algebraic and graphical methods](https://www.owlsprep.com/study/ib-math-ai-hl-u2-solving-equations-algebraic-and-graphical/)

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