# Logarithms and Exponential Functions

> CIE A-Level Mathematics · 9709 Pure Mathematics 2 & 3 (§2.2 / §3.2)
> Source: https://www.owlsprep.com/study/cie-9709-u2-logarithms-and-exponential-functions/

This subtopic covers core properties of exponential and logarithmic functions, their algebraic rules, inverse relationship, and methods to solve mixed equations, forming a critical foundation for calculus work in P3.

**Prerequisites:** [Indices and laws of exponents](https://www.owlsprep.com/study/cie-9709-u1-indices/); [Quadratic equation solving](https://www.owlsprep.com/study/cie-9709-u1-quadratic-equations/); [Function transformations](https://www.owlsprep.com/study/cie-9709-u1-function-transformations/)

## Learning objectives

- Simplify logarithmic and exponential expressions using core rules
- Solve mixed equations involving logarithms and exponentials
- Identify key features and sketch graphs of these functions
- Apply the inverse relationship between exponentials and logarithms
- Reduce relationships of the form $y = kx^n$ and $y = ka^x$ to linear form using logarithms

## How often is this tested?

Based on OwlsPrep's analysis of official CIE 9709 P2 past papers (2016–2025): **Logarithms and Exponential Functions** appears **163 times** in the last 10 years — **21.5%** of all P2 questions (163 of 455).

Most-tested forms: Logarithm laws and equations (80), e^x and ln x: properties, equations, modelling (60), Reduce y=ax^n or y=Ab^x to linear log form (25), Exponential growth/decay modelling and curve fitting (4).

## Core Definitions and Inverse Relationship

**Logarithm** — The logarithm of $b$ to base $a$ is the power that $a$ must be raised to obtain $b$, for $a>0, a \neq 1, b>0$.

*Notation:* $\log_a b = x \iff a^x = b$

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

**Natural Logarithm** — Logarithm with base equal to the exponential constant $e \approx 2.71828$. Exponential and natural logarithm are inverses of each other, so $e^{\ln x} = x$ and $\ln(e^x) = x$ for $x>0$.

*Notation:* $\ln x = \log_e x$

*Example:* $\ln e^5 = 5$

**Worked example:** Convert $e^{2x} = 7$ to logarithmic form, and $\log_3 (4y - 2) = 2$ to exponential form.

1. For the first equation, base is $e$, so by definition of natural logarithm:
2. $$\ln 7 = 2x$$
3. For the second equation, base is $3$, the right-hand side is the exponent, so by definition:
4. $$3^2 = 4y - 2 \implies 9 = 4y - 2$$

> **Exam tip:** Always remember the domain requirement: the argument of any logarithm must always be positive.

## Laws of Logarithms

All logarithms follow algebraic rules derived directly from the laws of exponents. These rules let you simplify complex expressions and combine multiple logarithmic terms into one.

- 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: $\log_a (x^n) = n \log_a x$
- Identities: $\log_a a = 1$, $a^{\log_a b} = b$, $\log_a 1 = 0$

> **mnemonic**
>
> Logs turn products into sums, powers into multiples.

**Worked example:** Simplify $3 \ln 2 - \frac{1}{2} \ln 36 + \ln 3$ into a single logarithm.

1. Apply the power rule to each term first:
2. $$3 \ln 2 = \ln 2^3 = \ln 8, \quad \frac{1}{2} \ln 36 = \ln 36^{1/2} = \ln 6$$
3. Substitute back into the original expression:
4. $\ln 8 - \ln 6 + \ln 3$
5. Combine terms using product and quotient rules:
6. $$\ln \left(\frac{8 \times 3}{6}\right) = \ln 4 = 2 \ln 2$$

## Graphs of Exponential and Logarithmic Functions

Since exponential and logarithmic functions are inverses, their graphs are reflections of each other over the line $y=x$. The shape depends on whether the base is greater than 1 or between 0 and 1.

| Function | Domain | Range | Asymptote | Key Point |
| --- | --- | --- | --- | --- |
| $y = a^x, a>1$ | $x \in \mathbb{R}$ | $y > 0$ | $y=0$ | $(0, 1)$ |
| $y = a^x, 0<a<1$ | $x \in \mathbb{R}$ | $y > 0$ | $y=0$ | $(0, 1)$ |
| $y = \log_a x, a>1$ | $x > 0$ | $y \in \mathbb{R}$ | $x=0$ | $(1, 0)$ |
| $y = \log_a x, 0<a<1$ | $x > 0$ | $y \in \mathbb{R}$ | $x=0$ | $(1, 0)$ |

**Worked example:** Sketch $y = 3e^{-x} + 2$, label the asymptote and y-intercept.

1. This is a transformation of $y=e^x$: reflect over the y-axis, stretch vertically by factor 3, shift up by 2 units.
2. The original asymptote of $y=e^{-x}$ is $y=0$, so shifting up 2 gives the new asymptote:
3. $$y = 2$$
4. Find the y-intercept when $x=0$:
5. $$y = 3e^{0} + 2 = 3(1) + 2 = 5$$
6. The y-intercept is at $(0,5)$, and the function is decreasing for all $x$, with $y>2$ for all $x$.

## Solving Equations with Exponentials and Logs

**Exam command terms**

Common command terms for this topic have specific exam expectations:

- **Solve** — Find all valid solutions, remember to reject invalid solutions that violate domain rules *(Solve $2\ln x = \ln(x + 6)$)*

- **Express in terms of** — Rewrite the expression using the given variable only *(Express $\log_2 (8x)$ in terms of $\log_2 x$)*

**Check your understanding**

Test your foundational knowledge before proceeding:

1. What is the value of $3^{2\log_3 4}$?

   - 6
   - 8
   - 16
   - 12

   *Why:* Use the power rule and inverse identity: $3^{\log_3 4^2} = 4^2 = 16$

**Worked example:** Solve $e^{2x} - 4e^x - 5 = 0$.

1. This is a quadratic equation in $e^x$. Let $y = e^x$, substitute to get:
2. $$y^2 - 4y - 5 = 0$$
3. Factorise the quadratic:
4. $$(y - 5)(y + 1) = 0 \implies y = 5 \text{ or } y = -1$$
5. Since $y = e^x$ is always positive for all real $x$, reject the negative root $y=-1$.
6. Solve $e^x = 5$ by taking natural logs of both sides:
7. $$x = \ln 5 \approx 1.61 \quad (3 \text{ significant figures})$$

**Worked example:** Solve $2\log_3 x = \log_3 (x + 12)$.

1. Apply the power rule to the left-hand side:
2. $\log_3 x^2 = \log_3 (x + 12)$
3. Since logarithms are one-to-one, equal logs imply equal arguments:
4. $$x^2 = x + 12 \implies x^2 - x - 12 = 0$$
5. Factorise and solve: $(x - 4)(x + 3) = 0 \implies x = 4$ or $x = -3$
6. Check domain: $\log_3 x$ requires $x>0$, so $x=-3$ is invalid. Only solution is $x=4$.

**Worked example:** Solve the inequality $2^x > 5$, and explain how the method changes for $0.6^x < 0.7$.

1. Take natural logarithms of both sides of $2^x > 5$. Because $\ln 2 > 0$, dividing by $\ln 2$ keeps the inequality direction:
2. $$x\ln 2 > \ln 5 \implies x > \frac{\ln 5}{\ln 2} \approx 2.32$$
3. For $0.6^x < 0.7$, taking logs gives $x\ln 0.6 < \ln 0.7$. Here the base is between 0 and 1, so $\ln 0.6 < 0$ and dividing by it flips the inequality:
4. $$x > \frac{\ln 0.7}{\ln 0.6}$$
5. Whenever the base lies in $(0, 1)$ its logarithm is negative, so remember to reverse the inequality sign at the step where you divide by it.

## Transforming to Linear Form

Experimental data believed to follow a power law $y = kx^n$ or an exponential law $y = ka^x$ can be turned into a straight line by taking logarithms of both sides. Once linear, the unknown constants are read directly from the gradient and the vertical-axis intercept of the line.

- Power law $y = kx^n$: taking logs gives $\ln y = n\ln x + \ln k$. Plot $\ln y$ against $\ln x$ (a log-log plot) to get a straight line with gradient $n$ and intercept $\ln k$.
- Exponential law $y = ka^x$: taking logs gives $\ln y = (\ln a)x + \ln k$. Plot $\ln y$ against $x$ (a log-linear plot) to get a straight line with gradient $\ln a$ and intercept $\ln k$.
- In both cases the intercept is $\ln k$, so $k = e^{\text{intercept}}$. For a power law read $n$ straight off the gradient; for an exponential law recover $a = e^{\text{gradient}}$.

> **tip**
>
> Choose the axes from the model: a power of $x$ needs $\ln y$ against $\ln x$; a constant base raised to $x$ needs $\ln y$ against $x$. The intercept always gives $\ln k$.

**Worked example:** A set of data is believed to obey $y = kx^n$. Plotting $\ln y$ against $\ln x$ gives a straight line of gradient $2$ and vertical-axis intercept $1.5$. Find $n$ and $k$, and state the relationship between $y$ and $x$.

1. Take logarithms of $y = kx^n$ to obtain the linear form:
2. $$\ln y = n\ln x + \ln k$$
3. Comparing with $\ln y = (\text{gradient})\ln x + (\text{intercept})$, the gradient equals $n$:
4. $$n = 2$$
5. The intercept equals $\ln k$, so:
6. $$\ln k = 1.5 \implies k = e^{1.5} \approx 4.48$$
7. Therefore $y \approx 4.48x^2$ (3 significant figures).

> **Exam tip:** Read the axis labels first: $\ln y$ against $\ln x$ signals a power law $y = kx^n$, while $\ln y$ against $x$ signals an exponential law $y = ka^x$. The vertical-axis intercept is always $\ln k$.

## Common pitfalls

- **Wrong:** Forgetting to check solutions after solving logarithmic equations
  - Why it fails: Logarithms are only defined for positive arguments, so algebraic solutions may not be valid
  - Correct: Always substitute every solution back into the original equation to confirm all arguments are positive
- **Wrong:** Expanding $\log_a(x + y)$ to $\log_a x + \log_a y$
  - Why it fails: The product rule only applies to products inside the log, not sums
  - Correct: Sums inside logarithms cannot be split into separate logarithms; only products can be split
- **Wrong:** Writing $n\log x = \log(nx)$ instead of $\log x^n$
  - Why it fails: Misremembering the power rule for logarithms
  - Correct: The coefficient of the log becomes the exponent of the argument, not a multiple
- **Wrong:** Keeping negative roots when solving quadratics in $e^x$
  - Why it fails: $e^x$ is always positive for all real $x$, so it can never equal a negative number
  - Correct: Reject any negative roots immediately when solving for $e^x$
- **Wrong:** Swapping the domain and range of exponential and log functions
  - Why it fails: Confusing the inverse relationship between the two function types
  - Correct: Exponentials: domain all real $x$, range $y>0$; Logs: domain $x>0$, range all real $y$

## Cheatsheet

| Rule Type | Logarithm Rule | Exponential Rule |
| --- | --- | --- |
| Core Relationship | $a^x = b \iff x = \log_a b$ | $e^{\ln x} = x, x>0$ |
| Product | $\log(xy) = \log x + \log y$ | $a^m a^n = a^{m+n}$ |
| Quotient | $\log(x/y) = \log x - \log y$ | $a^m / a^n = a^{m-n}$ |
| Power | $\log x^n = n \log x$ | $(a^m)^n = a^{mn}$ |
| Domain/Range | $y = \log_a x: x>0, y \in \mathbb{R}$ | $y = a^x: x \in \mathbb{R}, y>0$ |

## What's next

Logarithms and exponentials are foundational to nearly all remaining topics in CIE Pure Mathematics 2 and 3. Next, you will learn how to differentiate and integrate these functions, a skill that is tested heavily in the P2 and P3 exams. You will also use these tools to solve differential equations, which are a core component of the P3 syllabus (differential equations appear in P3 only). Mastering the algebraic rules in this sub-topic will make all subsequent work with exponentials and logs significantly easier.

- [Trigonometry](https://www.owlsprep.com/study/cie-9709-u2-trigonometry/)
- [Differentiation](https://www.owlsprep.com/study/cie-9709-u2-differentiation/)
- [Integration](https://www.owlsprep.com/study/cie-9709-u2-integration/)

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