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A-Level Mathematics – Chapter 5: Exponentials and Logarithms

Edexcel · AQA · OCR A-Level Mathematics · Updated March 2026

Exponentials and logarithms appear throughout pure mathematics, statistics, mechanics, and the physical sciences. This chapter develops a deep understanding of exponential functions — from the basic graph of $y = a^x$ through to Euler's number $e$ — and establishes the laws of logarithms needed to solve a wide range of equations. Real-world applications from radioactive decay to data linearisation are woven throughout.

Specification Note

Content labelled Year 2 is A2-level only and is not assessed at AS. Sections 5.1–5.4 are required for both AS and full A-Level. Section 5.5 is Year 2 only.

Contents

  1. 5.1 Exponential Functions
  2. 5.2 Logarithms
  3. 5.3 Natural Logarithms
  4. 5.4 Exponential Models
  5. 5.5 Data Linearisation Year 2
  6. Practice Problems

5.1 Exponential Functions

An exponential function has the variable in the exponent rather than the base. The standard form is $f(x) = a^x$, where $a > 0$ and $a \neq 1$. These functions grow (or decay) more rapidly than any polynomial for large $|x|$.

Definition: Exponential Function

A function of the form $f(x) = a^x$, where $a > 0$ and $a \neq 1$, is called an exponential function with base $a$.

The Natural Base $e$

The most important base in mathematics is Euler's number $e \approx 2.71828\ldots$, defined by:

$$e = \lim_{n \to \infty}\!\left(1 + \frac{1}{n}\right)^n$$

The function $y = e^x$ is uniquely self-referential under differentiation: $\dfrac{d}{dx}(e^x) = e^x$. Its gradient at $(0, 1)$ is exactly $1$, and it is the natural choice for all continuous-change models.

The interactive figure below compares $y = 2^x$, $y = e^x$, and $y = 3^x$. All three pass through $(0,1)$; a larger base produces a steeper graph for $x > 0$ and a shallower decay for $x < 0$.

Figure 5.1 — Comparing $y = 2^x$ (blue), $y = e^x$ (purple), and $y = 3^x$ (red). All pass through $(0,1)$.

Example 5.1.1 — Evaluating Exponential Expressions

Evaluate, giving exact values where possible: (a) $e^0$   (b) $e^{-1}$   (c) $e^{0.5}$   (d) $2e^2$.

a $e^0 = 1$

b $e^{-1} = \dfrac{1}{e} \approx 0.368$

c $e^{0.5} = \sqrt{e} \approx 1.649$

d $2e^2 \approx 2 \times 7.389 = 14.78$

Example 5.1.2 — Key Properties from a Graph

State the $y$-intercept, asymptote, and whether the following represent growth or decay.

(a) $y = 5^x$   (b) $y = (0.3)^x$   (c) $y = e^{-x}$

a $y$-intercept $(0,1)$; asymptote $y=0$; base $5 > 1$ so growth.

b $y$-intercept $(0,1)$; asymptote $y=0$; base $0.3 < 1$ so decay.

c $y$-intercept $(0,1)$; asymptote $y=0$; since the exponent is negative, this is decay (same as base $1/e < 1$).

Example 5.1.3 — Transformations of $y = e^x$

Describe each transformation from $y = e^x$ and state the asymptote.

(a) $y = e^{x+3}$   (b) $y = e^x - 4$   (c) $y = 2e^x$   (d) $y = -e^x$

a Translation $\begin{pmatrix}-3\\0\end{pmatrix}$ (left 3). Asymptote: $y = 0$.

b Translation $\begin{pmatrix}0\\-4\end{pmatrix}$ (down 4). Asymptote: $y = -4$. The $y$-intercept moves to $(0,-3)$.

c Stretch in the $y$-direction by factor 2. $y$-intercept becomes $(0,2)$. Asymptote: $y = 0$.

d Reflection in the $x$-axis; graph lies entirely below $y=0$. Asymptote: $y = 0$.

Example 5.1.4 — Sketching with a Shifted Asymptote

Sketch $y = 3e^{-x} + 1$, stating the asymptote, $y$-intercept, and behaviour as $x \to +\infty$ and $x \to -\infty$.

1 Asymptote: as $x \to +\infty$, $e^{-x} \to 0$, so $y \to 1$. Asymptote: $y = 1$.

2 $y$-intercept: $x=0 \Rightarrow y = 3(1)+1 = 4$. Point $(0,4)$.

3 As $x \to -\infty$: $e^{-x} \to +\infty$, so $y \to +\infty$. The curve rises steeply to the left.

4 The graph is a reflection of $y = 3e^x$ in the $y$-axis, then shifted up 1 unit.

Exam Tip

When sketching any exponential, mark three things: the $y$-intercept, the equation of the asymptote, and the direction of approach. Examiners award a mark specifically for a correctly stated asymptote — do not omit it.

5.2 Logarithms

The logarithm is the inverse operation of exponentiation. If $a^y = x$, we write $y = \log_a x$. Logarithms allow us to "bring down" exponents, which is essential for solving equations where the unknown is in a power.

Definition: Logarithm

For $a > 0$, $a \neq 1$, and $x > 0$:

$$\log_a x = y \iff a^y = x$$

Key identities: $\log_a 1 = 0$ (since $a^0 = 1$) and $\log_a a = 1$ (since $a^1 = a$).

The common logarithm $\log_{10} x$ is written $\lg x$ or simply $\log x$.

Laws of Logarithms

For $a > 0$, $a \neq 1$, and positive values $x$, $y$:

Example 5.2.1 — Applying the Log Laws

Write $3\log 2 + \log 5 - \log 4$ as a single logarithm and evaluate it.

1 Power rule: $3\log 2 = \log 8$

2 Product rule: $\log 8 + \log 5 = \log 40$

3 Quotient rule: $\log 40 - \log 4 = \log\!\left(\tfrac{40}{4}\right) = \log 10 = 1$

Example 5.2.2 — Solving $a^x = b$ Using Logs

Solve $3^x = 20$, giving your answer to 3 significant figures.

1 Take $\log$ of both sides: $\log(3^x) = \log 20$

2 Power rule: $x\log 3 = \log 20$

3 Divide: $x = \dfrac{\log 20}{\log 3} \approx \dfrac{1.3010}{0.4771} \approx 2.73$

Example 5.2.3 — Change of Base

Evaluate $\log_5 17$ to 4 decimal places.

1 $\log_5 17 = \dfrac{\log 17}{\log 5} = \dfrac{1.2304}{0.6990} \approx 1.7603$

Alternatively using natural log: $\dfrac{\ln 17}{\ln 5} = \dfrac{2.8332}{1.6094} \approx 1.7603$ ✓

Example 5.2.4 — Solving a Log Equation (Disguised Quadratic)

Solve $\log_3 x = 4 - \log_3(x-8)$.

1 Rearrange: $\log_3 x + \log_3(x-8) = 4$

2 Product rule: $\log_3[x(x-8)] = 4$

3 Exponentiate: $x(x-8) = 3^4 = 81$

4 Expand: $x^2 - 8x - 81 = 0 \Rightarrow (x-9)(x+9) = 0$

5 Domain check: $x > 0$ and $x > 8$, so reject $x = -9$. Answer: $x = 9$.

Example 5.2.5 — Simultaneous Log Equations

Given $\log_2 x + \log_2 y = 5$ and $\log_2 x - \log_2 y = 1$, find $x$ and $y$.

1 Add: $2\log_2 x = 6 \Rightarrow \log_2 x = 3 \Rightarrow x = 8$

2 Subtract: $2\log_2 y = 4 \Rightarrow \log_2 y = 2 \Rightarrow y = 4$

Exam Tip

Always check that solutions satisfy the domain conditions for the original logarithms — each argument must be strictly positive. Examiners regularly award a mark for correctly rejecting an invalid solution.

5.3 Natural Logarithms

The natural logarithm $\ln x = \log_e x$ is the logarithm to base $e$. Because $\ln$ and $e^x$ are inverse functions, each "undoes" the other. This makes $\ln x$ indispensable when solving equations involving $e^x$, and it arises naturally in calculus.

Definition: Natural Logarithm

$$\ln x = \log_e x \iff e^y = x \quad (y = \ln x)$$

Key identities and properties:

Calculus Connection

Two fundamental results (developed fully in Chapters 6 and 7) are:

Example 5.3.1 — Simplifying with $\ln$ and $e$

Simplify: (a) $\ln e^5$   (b) $e^{3\ln 2}$   (c) $\ln\sqrt{e}$   (d) $e^{\ln 7 - \ln 2}$

a $\ln e^5 = 5$

b $e^{3\ln 2} = e^{\ln 2^3} = e^{\ln 8} = 8$

c $\ln\sqrt{e} = \ln e^{1/2} = \tfrac{1}{2}$

d $e^{\ln 7 - \ln 2} = e^{\ln(7/2)} = \tfrac{7}{2}$

Example 5.3.2 — Solving $Ae^{kt} = B$

A radioactive sample has initial mass $200\,\text{g}$ and decays by $m = 200e^{-0.03t}$, where $t$ is in years. Find the time for the mass to fall to $80\,\text{g}$.

1 Set $m = 80$: $200e^{-0.03t} = 80$

2 Divide: $e^{-0.03t} = 0.4$

3 Take $\ln$: $-0.03t = \ln 0.4$

4 Solve: $t = \dfrac{-\ln 0.4}{0.03} = \dfrac{0.9163}{0.03} \approx 30.5\,\text{years}$

Example 5.3.3 — Solving a Natural Log Equation

Solve $\ln(x+3) - \ln x = \ln 4$.

1 Quotient rule: $\ln\!\left(\dfrac{x+3}{x}\right) = \ln 4$

2 Exponentiate: $\dfrac{x+3}{x} = 4$

3 $x + 3 = 4x \Rightarrow 3 = 3x \Rightarrow x = 1$

4 Check: $\ln 4 - \ln 1 = \ln 4$ ✓ and both arguments are positive ✓

Exam Tip

The identity $e^{\ln f(x)} = f(x)$ is extremely useful for simplification. Similarly, $\ln(e^{f(x)}) = f(x)$. When you see either form, simplify immediately rather than leaving it unsimplified in your final answer.

5.4 Exponential Models

Exponential functions model many natural processes: population growth, radioactive decay, drug absorption, compound interest, and cooling. The key feature is that the rate of change is proportional to the current value — which leads directly to the exponential form.

General Exponential Model

$$P = P_0\,e^{kt} \quad \text{or equivalently} \quad P = P_0\,a^t \text{ where } a = e^k$$

The figure below shows $P = 100e^{-kt}$ with an adjustable decay constant. Move the slider to observe how the half-life changes with $k$.

Figure 5.2 — Exponential decay $P = 100e^{-kt}$. The dashed line at $P = 50$ marks the half-life. Adjust $k$ to change the rate of decay.

Example 5.4.1 — Population Growth

A bacterial population is modelled by $N = 500e^{0.4t}$, where $t$ is in hours. Find: (a) the initial population, (b) the population after 3 hours, (c) the doubling time.

a At $t=0$: $N = 500e^0 = 500$ bacteria.

b $N = 500e^{1.2} \approx 500 \times 3.320 \approx 1660$ bacteria.

c $T_2 = \dfrac{\ln 2}{0.4} \approx \dfrac{0.6931}{0.4} \approx 1.73$ hours.

Example 5.4.2 — Radioactive Decay and Half-Life

Carbon-14 has a half-life of 5730 years. Find the decay constant and hence the percentage remaining after 10 000 years.

1 $5730 = \dfrac{\ln 2}{k} \Rightarrow k = \dfrac{\ln 2}{5730} \approx 1.210 \times 10^{-4}$

2 Model: $m = m_0 e^{-1.210 \times 10^{-4}\,t}$

3 At $t = 10\,000$: $\dfrac{m}{m_0} = e^{-1.210} \approx 0.298$, i.e. approximately $\mathbf{29.8\%}$ remains.

Example 5.4.3 — Fitting an Exponential Curve $y = ab^x$

A model $y = ab^x$ passes through $(2, 12)$ and $(5, 96)$. Find $a$ and $b$.

1 Substitute: $ab^2 = 12$ and $ab^5 = 96$.

2 Divide: $b^3 = \dfrac{96}{12} = 8 \Rightarrow b = 2$.

3 Back-substitute: $a = \dfrac{12}{4} = 3$. So $y = 3 \times 2^x$.

Example 5.4.4 — Newton's Law of Cooling

A cup of coffee at $90\,^\circ\text{C}$ cools in a room at $20\,^\circ\text{C}$. The excess temperature follows $\theta = 70e^{-0.05t}$ (°C), where $t$ is in minutes. Find when the coffee reaches $50\,^\circ\text{C}$.

1 Excess temperature at $50\,^\circ\text{C}$: $\theta = 50 - 20 = 30$.

2 $30 = 70e^{-0.05t} \Rightarrow e^{-0.05t} = \dfrac{3}{7}$

3 $-0.05t = \ln\!\left(\tfrac{3}{7}\right) \Rightarrow t = \dfrac{-\ln(3/7)}{0.05} \approx \dfrac{0.8473}{0.05} \approx 16.9\,\text{min}$

Example 5.4.5 — Compound Interest

£3000 is invested at $4\%$ per annum, compounded continuously (i.e. $A = Pe^{rt}$). Find: (a) the amount after 10 years; (b) the time to double the investment.

a $A = 3000e^{0.04 \times 10} = 3000e^{0.4} \approx 3000 \times 1.4918 = \pounds 4475$.

b $6000 = 3000e^{0.04t} \Rightarrow e^{0.04t} = 2 \Rightarrow t = \dfrac{\ln 2}{0.04} \approx 17.3\,\text{years}$.

Exam Tip

In modelling questions, identify $P_0$ (the value at $t = 0$) first — it often falls out immediately without algebra. Also, always state the units of $t$ in your final answer.

5.5 Data Linearisation Year 2

Real data rarely fits a perfect exponential or power law. The technique of linearisation transforms a non-linear relationship into a straight-line form, allowing the parameters to be read directly from a graph. Two common models are treated below.

Linearising $y = ax^n$ (Power Law)

Take $\log_{10}$ of both sides:

$$\log y = n\log x + \log a$$

Plot $Y = \log y$ against $X = \log x$. The result is a straight line:

Linearising $y = ab^x$ (Exponential)

Take $\log_{10}$ of both sides:

$$\log y = x\log b + \log a$$

Plot $Y = \log y$ against $X = x$. The result is a straight line:

Example 5.5.1 — Power Law from a Log-Log Graph

A $\log y$ vs $\log x$ graph passes through $(0.3, 0.9)$ and $(1.1, 2.5)$. Find $a$ and $n$ in $y = ax^n$.

1 Gradient: $n = \dfrac{2.5 - 0.9}{1.1 - 0.3} = \dfrac{1.6}{0.8} = 2$

2 Line equation: $\log y = 2\log x + c$. Substitute $(0.3, 0.9)$: $c = 0.9 - 0.6 = 0.3$.

3 $\log a = 0.3 \Rightarrow a = 10^{0.3} \approx 2.00$. So $y \approx 2x^2$.

Example 5.5.2 — Exponential Law from a Log-Linear Graph

A $\log y$ vs $x$ graph has gradient $0.15$ and $Y$-intercept $1.2$. Find $a$ and $b$ in $y = ab^x$.

1 $\log b = 0.15 \Rightarrow b = 10^{0.15} \approx 1.41$

2 $\log a = 1.2 \Rightarrow a = 10^{1.2} \approx 15.8$

So $y \approx 15.8 \times 1.41^x$.

Example 5.5.3 — Verifying a Power Law from Data

Data pairs $(t, P)$: $(2, 5.66)$, $(4, 16.0)$, $(8, 45.3)$. Verify a power law $P = at^n$ and estimate $a$ and $n$.

1 Compute $(\log t, \log P)$: $(0.301, 0.753)$, $(0.602, 1.204)$, $(0.903, 1.656)$.

2 Gradient between 1st and 3rd: $\dfrac{1.656 - 0.753}{0.903 - 0.301} = \dfrac{0.903}{0.602} \approx 1.5$ — consistent, confirming a power law.

3 $n \approx 1.5$. Using $(0.301, 0.753)$: $0.753 = 1.5(0.301) + \log a \Rightarrow \log a \approx 0.302 \Rightarrow a \approx 2.00$.

Model: $P \approx 2t^{1.5}$.

Exam Tip

In data linearisation questions, write down the linearised equation explicitly (e.g. $\log y = n\log x + \log a$) before reading off the gradient and intercept. Mixing up which quantity gives $n$ versus $\log a$ is the most common error in this topic.

Practice Problems

Problem 1

Sketch $y = 2e^x - 1$, labelling the $y$-intercept and the equation of the asymptote.

Show Solution

$y$-intercept: $x=0 \Rightarrow y = 2(1)-1 = 1$. Point $(0,1)$.

Asymptote: as $x \to -\infty$, $e^x \to 0$, so $y \to -1$. Asymptote: $y = -1$.

The graph is a vertical stretch of $y = e^x$ by 2, then translated down by 1.

Problem 2

Solve $5^{2x-1} = 40$, giving your answer to 3 significant figures.

Show Solution

$(2x-1)\log 5 = \log 40$

$2x - 1 = \dfrac{\log 40}{\log 5} \approx \dfrac{1.6021}{0.6990} \approx 2.292$

$x \approx \dfrac{3.292}{2} \approx 1.65$

Problem 3

Write $\log_6 9 + 2\log_6 4 - \log_6 16$ as a single logarithm in its simplest form.

Show Solution

$\log_6 9 + \log_6 16 - \log_6 16 = \log_6 9$

(Note: $2\log_6 4 = \log_6 16$.)

Problem 4

A population of insects grows according to $P = 300e^{0.12t}$, where $t$ is in weeks. Find the number of weeks for the population to reach 2000, giving your answer to the nearest week.

Show Solution

$2000 = 300e^{0.12t} \Rightarrow e^{0.12t} = \dfrac{20}{3}$

$0.12t = \ln\!\left(\dfrac{20}{3}\right) \approx 1.897$

$t \approx 15.8 \Rightarrow$ 16 weeks

Problem 5

Solve $2\ln x - \ln(x+6) = \ln 3$.

Show Solution

$\ln\!\left(\dfrac{x^2}{x+6}\right) = \ln 3 \Rightarrow \dfrac{x^2}{x+6} = 3$

$x^2 = 3x + 18 \Rightarrow x^2 - 3x - 18 = 0 \Rightarrow (x-6)(x+3) = 0$

Reject $x = -3$ (domain). So $x = 6$.

Problem 6

A curve has equation $y = Ae^{kt}$. Given $y = 4$ when $t = 0$ and $y = 10$ when $t = 3$, find $A$ and $k$ exactly.

Show Solution

$t=0$: $A = 4$.

$t=3$: $10 = 4e^{3k} \Rightarrow 3k = \ln\!\left(\tfrac{5}{2}\right) \Rightarrow k = \dfrac{\ln(5/2)}{3} = \dfrac{\ln 2.5}{3}$

Problem 7

Solve $e^{2x} - 5e^x + 6 = 0$, giving exact answers.

Show Solution

Let $u = e^x$: $u^2 - 5u + 6 = 0 \Rightarrow (u-2)(u-3) = 0$

$e^x = 2 \Rightarrow x = \ln 2$   or   $e^x = 3 \Rightarrow x = \ln 3$

Problem 8

The value $V$ (in £thousands) of a car after $t$ years is $V = 18 \times 0.82^t$. Find the number of years until the car falls below £5000.

Show Solution

$18 \times 0.82^t < 5 \Rightarrow 0.82^t < \dfrac{5}{18}$

$t\log 0.82 < \log\!\left(\tfrac{5}{18}\right)$. Since $\log 0.82 < 0$, the inequality reverses when dividing:

$t > \dfrac{\log(5/18)}{\log 0.82} \approx \dfrac{-0.5563}{-0.0862} \approx 6.45$

So after 7 years.

Problem 9 Year 2

Data follows $y = ax^n$. A plot of $\log y$ against $\log x$ gives a straight line through $(0, 1.5)$ and $(2, 4.1)$. Find $a$ and $n$.

Show Solution

Gradient $= n = \dfrac{4.1 - 1.5}{2 - 0} = 1.3$

$Y$-intercept $= \log a = 1.5 \Rightarrow a = 10^{1.5} \approx 31.6$

$y \approx 31.6\,x^{1.3}$

Problem 10 Year 2

It is given that $y = ab^x$. A plot of $\log y$ against $x$ has gradient $0.2$ and passes through $(0, 0.6)$. Find $a$ and $b$ to 3 significant figures.

Show Solution

$\log b = 0.2 \Rightarrow b = 10^{0.2} \approx 1.58$

$\log a = 0.6 \Rightarrow a = 10^{0.6} \approx 3.98$