# Exponential and logarithmic functions

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

This module covers core properties of exponential and logarithmic functions, their inverse relationship, solving equations, and modeling real-world growth and decay, all key topics for IB AI SL assessments.

**Prerequisites:** [Laws of exponents and indices](https://www.owlsprep.com/study/ib-math-ai-sl-u1-laws-of-exponents/); [Inverse functions](https://www.owlsprep.com/study/ib-math-ai-sl-u2-inverse-functions/)

## Learning objectives

- Recognize the properties of exponential and logarithmic functions and their inverse relationship
- Convert between exponential and logarithmic forms of expressions and equations
- Apply logarithm laws to simplify expressions and solve equations
- Model exponential growth and decay problems for real-world contexts

## Core Definitions and Inverse Relationship

**Exponential function** — A function where the independent variable $x$ is the exponent of a constant base $a$. Domain is all real numbers, range is all positive real numbers.

*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 the exponential function with base $a$. By definition, $\log_a x = y$ if and only if $a^y = x$. Domain is all positive real numbers, range is all real numbers.

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

*Example:* \log_2 8 = 3$, since $2^3 = 8$

Exponential and logarithmic functions are inverses, so their graphs are reflections of each other across the line $y=x$. This definition is the foundation for all conversions and equation solving.

**Worked example:** Convert $5^3 = 125$ to logarithmic form, and convert $\log_4 16 = 2$ to exponential form.

1. Recall the inverse relationship: $a^y = x \iff \log_a x = y$
2. For $5^3 = 125$, we have $a=5$, $y=3$, $x=125$, so
3. $$\log_5 125 = 3$$
4. For $\log_4 16 = 2$, we have $a=4$, $x=16$, $y=2$, so
5. $$4^2 = 16$$

> **Exam tip:** Always check your conversion by substituting back to confirm you placed the base, exponent, and argument correctly.

## Laws of Logarithms

All logarithm laws are derived directly from the laws of exponents. These rules let you simplify complex logarithmic expressions and solve equations.

| Law Name | Logarithm Form | Exponent Equivalent |
| --- | --- | --- |
| Product Rule | $\log_a(xy) = \log_a x + \log_a y$ | $a^m a^n = a^{m+n}$ |
| Quotient Rule | $\log_a\left(\frac{x}{y}\right) = \log_a x - \log_a y$ | $\frac{a^m}{a^n} = a^{m-n}$ |
| Power Rule | $\log_a(x^n) = n \log_a x$ | $(a^m)^n = a^{mn}$ |
| Change of Base | $\log_a b = \frac{\ln b}{\ln a} = \frac{\log_{10} b}{\log_{10} a}$ | N/A |

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

**Worked example:** Simplify $2 \ln 3 + \ln 4 - \ln 12$

1. Apply the power rule first to the first term:
2. $$2 \ln 3 = \ln 3^2 = \ln 9$$
3. Combine the first two terms with the product rule:
4. $$\ln 9 + \ln 4 = \ln(9 \times 4) = \ln 36$$
5. Apply the quotient rule for the final simplification:
6. $$\ln 36 - \ln 12 = \ln\left(\frac{36}{12}\right) = \ln 3$$

## Solving Exponential and Logarithmic Equations

When the variable is in the exponent, we use logarithms to 'bring down' the exponent and solve. For logarithmic equations, we convert to exponential form after combining terms, then always check for extraneous solutions.

> **warning**
>
> Logarithms can only take positive arguments. Any solution that makes any log argument zero or negative is invalid, even if it comes from correct algebra.

**Worked example:** Solve $\log_2 (x+2) + \log_2 x = 3$

1. Combine the logs using the product rule:
2. $$\log_2 [x(x+2)] = 3$$
3. Convert from logarithmic to exponential form:
4. $$x(x+2) = 2^3 = 8$$
5. Rearrange into standard quadratic form:
6. $$x^2 + 2x - 8 = 0$$
7. Factor and solve the quadratic:
8. $$(x+4)(x-2) = 0 \implies x = -4 \text{ or } x = 2$$
9. Check solutions: $x=-4$ gives undefined logs (negative arguments), so it is extraneous. $x=2$ has positive arguments, so it is valid.
10. Final solution: $x=2$

**Worked example:** Solve $3^{2x -1} = 24$ for $x$, give your answer to 3 significant figures.

1. Take the natural logarithm of both sides:
2. $$\ln\left(3^{2x -1}\right) = \ln 24$$
3. Apply the power rule to bring down the exponent:
4. $$(2x -1) \ln 3 = \ln 24$$
5. Rearrange to isolate the term with $x$:
6. $$2x - 1 = \frac{\ln 24}{\ln 3} \approx 2.893$$
7. Solve for $x$:
8. $$2x \approx 3.893 \implies x \approx 1.95$$

## Exponential Growth and Decay Modeling

**Exponential growth/decay model** — Standard continuous model where $A_0$ is the initial quantity at $t=0$, $k$ is the constant growth/decay rate, $k>0$ for growth and $k<0$ for decay.

*Notation:* A(t) = A_0 e^{kt}

*Example:* Population growth, radioactive decay, cooling

Most applied exponential problems in IB AI SL follow this standard form. The first step is always to use given information to find the unknown constant $k$, then use the model to answer the question.

**Worked example:** A bacterial population grows exponentially. At $t=0$ there are 100 bacteria, after 3 hours there are 800 bacteria. Find $k$ in the model $A(t) = 100e^{kt}$, then find the population after 5 hours.

1. Substitute the known values ($t=3, A=800$) into the model:
2. $$800 = 100 e^{3k}$$
3. Divide both sides by 100 to isolate the exponential:
4. $$8 = e^{3k}$$
5. Take natural log of both sides:
6. $$\ln 8 = 3k$$
7. Solve for $k$:
8. $$k = \frac{\ln 8}{3} = \ln 2 \approx 0.693$$
9. Calculate population at $t=5$:
10. $$A(5) = 100 e^{5 \ln 2} = 100 (e^{\ln 2})^5 = 100 \times 2^5 = 3200$$

> **Exam tip:** Always check that your final population or quantity makes sense: growth should give a larger number than the initial, decay should give a smaller number.

*Calculator:* allowed

## Common pitfalls

- **Wrong:** Forgetting to check for extraneous solutions when solving logarithmic equations
  - Why it fails: Quadratic equations often produce negative solutions that make log arguments negative, which are undefined
  - Correct: Always substitute every solution back into the original equation to confirm all log arguments are positive
- **Wrong:** Writing $\log_a x^n = (\log_a x)^n$ instead of $n\log_a x$
  - Why it fails: Confusion between the exponent of the argument and the exponent of the entire logarithm
  - Correct: Remember the power rule brings the exponent of the argument out front as a multiplier
- **Wrong:** Splitting $\log_a(x + y)$ into $\log_a x + \log_a y$
  - Why it fails: Confusing the product rule for logarithms with addition inside the log
  - Correct: The product rule only applies to products inside the log, not sums. $\log_a(x+y)$ cannot be simplified in general.
- **Wrong:** Taking the logarithm of both sides of $a^x + b = c$ before isolating $a^x$
  - Why it fails: Logarithms do not distribute over addition, so early log application leads to wrong simplification
  - Correct: Always rearrange to isolate the exponential term on one side before taking logarithms
- **Wrong:** Confusing the domain of exponential and logarithmic functions
  - Why it fails: Mixing up which function can accept negative inputs
  - Correct: Exponentials accept any real $x$ and output positive values; logarithms only accept positive $x$ and output any real value

## Cheatsheet

| Concept | Key Formula | Note |
| --- | --- | --- |
| Exponential ↔ Log conversion | $a^y = x \iff \log_a x = y$ | Inverse relationship |
| Product Rule | $\log_a(xy) = \log_a x + \log_a y$ | $x>0, y>0$ |
| Quotient Rule | $\log_a(x/y) = \log_a x - \log_a y$ | $x>0, y>0$ |
| Power Rule | $\log_a(x^n) = n\log_a x$ | $x>0$ |
| Change of Base | $\log_a b = \frac{\ln b}{\ln a}$ | Works for any positive base |
| Continuous Model | $A(t) = A_0 e^{kt}$ | $k>0$ growth, $k<0$ decay |
| Domain Check | $\log_a x$ requires $x>0$ | Always check for extraneous solutions |

## What's next

Exponential and logarithmic functions are one of the most frequently tested topics in IB AI SL, appearing in both pure and applied questions across both papers. They form the foundation for modeling all proportional growth and decay problems, which are core to applications in finance, biology, and physics. Mastering the rules here will make it much easier to tackle applied problems later in the course, where you will use these functions to model real-world phenomena and interpret results in context.

- [Transformations of function graphs](https://www.owlsprep.com/study/ib-math-ai-sl-u2-transformations-of-function-graphs/)
- [Real-world modeling with functions](https://www.owlsprep.com/study/ib-math-ai-sl-u2-real-world-modeling-with-functions/)
- [Geometry and Trigonometry](https://www.owlsprep.com/study/ib-math-ai-sl-u3-overview/)

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