Logarithms and Exponential Functions

CIE A-Level Mathematics· Pure Mathematics 2 & 3 (§2.2 / §3.2): Logarithmic and exponential functions· 15 min read

1. Core Definitions and Inverse Relationship★☆☆☆☆⏱ 5 min

📘 Definition

Logarithm

$lo g_{a} b = x ⟺ a^{x} = b$

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

Example:

$2^{3} = 8 ⟹ lo g_{2} 8 = 3$

📘 Definition

Natural Logarithm

$ln x = lo g_{e} x$

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

Example:

$ln e^{5} = 5$

📐 Worked Example

Convert $e^{2 x} = 7$ to logarithmic form, and $lo g_{3} (4 y - 2) = 2$ 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.

2. Laws of Logarithms★★☆☆☆⏱ 8 min

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: $lo g_{a} (x y) = lo g_{a} x + lo g_{a} y$
Quotient rule: $lo g_{a} (\frac{x}{y}) = lo g_{a} x - lo g_{a} y$
Power rule: $lo g_{a} (x^{n}) = n lo g_{a} x$
Identities: $lo g_{a} a = 1$ , $a^{l o g_{a} b} = b$ , $lo g_{a} 1 = 0$

📐 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
$\ln \left(\frac{8 \times 3}{6}\right) = \ln 4 = 2 \ln 2$

3. Graphs of Exponential and Logarithmic Functions★★☆☆☆⏱ 7 min

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
$y = a^{x}, a > 1$	$x \in R$	$y > 0$	$y = 0$	$(0, 1)$
$y = a^{x}, 0 < a < 1$	$x \in R$	$y > 0$	$y = 0$	$(0, 1)$
$y = lo g_{a} x, a > 1$	$x > 0$	$y \in R$	$x = 0$	$(1, 0)$
$y = lo g_{a} x, 0 < a < 1$	$x > 0$	$y \in R$	$x = 0$	$(1, 0)$

📐 Worked Example

Sketch $y = 3 e^{- x} + 2$ , 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 $x$ .

4. Solving Equations with Exponentials and Logs★★★☆☆⏱ 10 min

✓ Quick check

Test your foundational knowledge before proceeding:

What is the value of $3^{2 l o g_{3} 4}$ ?
- 6
- 8
- 16
- 12
Reveal answer
16 —
Use the power rule and inverse identity: $3^{l o g_{3} 4^{2}} = 4^{2} = 16$

📐 Worked Example

Solve $e^{2 x} - 4 e^{x} - 5 = 0$ .

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 $2 lo g_{3} x = lo g_{3} (x + 12)$ .

1
Apply the power rule to the left-hand side:
2
$lo g_{3} x^{2} = lo g_{3} (x + 12)$
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: $(x - 4) (x + 3) = 0 ⟹ x = 4$ or $x = - 3$
6
Check domain: $lo g_{3} x$ 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.

5. Transforming to Linear Form★★★☆☆⏱ 8 min

Experimental data believed to follow a power law $y = k x^{n}$ or an exponential law $y = k a^{x}$ 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 $y = k x^{n}$ : 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 $y = k a^{x}$ : 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 $k = e^{intercept}$ . For a power law read $n$ straight off the gradient; for an exponential law recover $a = e^{gradient}$ .

📐 Worked Example

A set of data is believed to obey $y = k x^{n}$ . 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 $y = k x^{n}$ to obtain the linear form:
2
$\ln y = n\ln x + \ln k$
3
Comparing with $ln y = (gradient) ln x + (intercept)$ , 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 $y \approx 4.48 x^{2}$ (3 significant figures).

Exam tip:

Read the axis labels first: $ln y$ against $ln x$ signals a power law $y = k x^{n}$ , while $ln y$ against $x$ signals an exponential law $y = k a^{x}$ . The vertical-axis intercept is always $ln k$ .

6. Common Pitfalls

Wrong move:

Forgetting to check solutions after solving logarithmic equations

Why:

Logarithms are only defined for positive arguments, so algebraic solutions may not be valid

Correct move:

Always substitute every solution back into the original equation to confirm all arguments are positive

Wrong move:

Expanding $lo g_{a} (x + y)$ to $lo g_{a} x + lo g_{a} y$

Why:

The product rule only applies to products inside the log, not sums

Correct move:

Sums inside logarithms cannot be split into separate logarithms; only products can be split

Wrong move:

Writing $n lo g x = lo g (n x)$ instead of $lo g x^{n}$

Why:

Misremembering the power rule for logarithms

Correct move:

The coefficient of the log becomes the exponent of the argument, not a multiple

Wrong move:

Keeping negative roots when solving quadratics in $e^{x}$

Why:

$e^{x}$ is always positive for all real $x$ , so it can never equal a negative number

Correct move:

Reject any negative roots immediately when solving for $e^{x}$

Wrong move:

Swapping the domain and range of exponential and log functions

Why:

Confusing the inverse relationship between the two function types

Correct move:

Exponentials: domain all real $x$ , range $y > 0$ ; Logs: domain $x > 0$ , range all real $y$

7. Quick Reference Cheatsheet

Rule Type	Logarithm Rule	Exponential Rule
Core Relationship	$a^{x} = b ⟺ x = lo g_{a} b$	$e^{l n x} = x, x > 0$
Product	$lo g (x y) = lo g x + lo g y$	$a^{m} a^{n} = a^{m + n}$
Quotient	$lo g (x / y) = lo g x - lo g y$	$a^{m} / a^{n} = a^{m - n}$
Power	$lo g x^{n} = n lo g x$	$(a^{m})^{n} = a^{mn}$
Domain/Range	$y = lo g_{a} x : x > 0, y \in R$	$y = a^{x} : x \in R, y > 0$

8. Frequently Asked

Why do I need to check solutions for logarithmic equations?

Logarithms are only defined for positive arguments. Any solution that produces a non-positive argument is invalid and must be rejected, even if it satisfies the algebraic equation after manipulating logs.

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.