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.

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

Commonly tested: solve an exponential equation using logarithms
Commonly tested: simplify and combine logarithmic expressions
Commonly tested: solve a quadratic in $e^{x}$
Commonly tested: reduce $y = k x^{n}$ or $y = k a^{x}$ to linear form

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.