What are the key properties of exponential functions in Pure Mathematics 3 for Edexcel International A Levels?

Exponential functions, typically expressed as f(x) = a^x where a > 0 and a ≠ 1, have several key properties: they are continuous and smooth, have a horizontal asymptote at y = 0, and are strictly increasing if a > 1 or strictly decreasing if 0 < a < 1. These properties are crucial for solving complex problems in Pure Mathematics 3.

How do you solve exponential equations using logarithms in the context of Edexcel International A Levels?

To solve exponential equations using logarithms, you can take the logarithm of both sides of the equation. For example, if you have an equation like a^x = b, you can apply the logarithm to get x * log(a) = log(b). Then, solve for x by dividing both sides by log(a). This technique is essential for solving exponential equations in Pure Mathematics 3.

What is the relationship between exponential and logarithmic functions in Pure Mathematics 3?

Exponential and logarithmic functions are inverses of each other. If y = a^x, then x = log_a(y). This inverse relationship means that the logarithm of a number is the exponent to which the base must be raised to produce that number. Understanding this relationship is fundamental in solving problems involving these functions in Edexcel International A Levels.

How do you differentiate exponential and logarithmic functions in Pure Mathematics 3?

To differentiate exponential functions like f(x) = e^x, the derivative is f'(x) = e^x. For logarithmic functions, such as f(x) = ln(x), the derivative is f'(x) = 1/x. These differentiation rules are crucial for solving calculus problems involving exponential and logarithmic functions in the Edexcel International A Levels curriculum.

What are common applications of exponential and logarithmic functions in real-world scenarios covered in Pure Mathematics 3?

Exponential and logarithmic functions are used in various real-world applications, such as modeling population growth, radioactive decay, and compound interest. In Pure Mathematics 3, students learn to apply these functions to solve practical problems, enhancing their understanding of how mathematics can be used to describe and predict real-world phenomena.

Why is "Quoting $x = 0.473$ when the question asks for exact form" a common mistake in Exponential and logarithms 2?

Why it happens: Calculator habit. How to avoid it: READ the question. 'Exact form' / 'in terms of ' / 'leave as a logarithm' — all mean NO decimals. Final answer should be 7 - 1/2 or 12 7 - 12. Source: WMA13/01 examiner reports — flagged annually

Why is "$\ln(a + b) = \ln a + \ln b$" a common mistake in Exponential and logarithms 2?

Why it happens: Mixing up the log of a SUM with log of a PRODUCT. How to avoid it: (ab) = a + b (product rule). (a + b) does NOT simplify. There's no log law for sums.

Why is "Plotting $\ln y$ vs $\ln x$ when the model is $y = ab^{x}$" a common mistake in Exponential and logarithms 2?

Why it happens: Defaulting to log-log axes. How to avoid it: Look at what's exponential: y = a x^n — power law — log BOTH axes. y = a b^x — exponential — log y only. The variable in the exponent dictates whether you log the x-axis.

Why is "Writing $\ln(-2)$ as a real number" a common mistake in Exponential and logarithms 2?

Why it happens: Forgetting has domain x > 0. How to avoid it: Always check that any argument of is positive. If e^x = -2 arises in a substitution, REJECT — there is no real solution because e^x > 0 always.

Why is "Stating half-life $= 0.5 / k$" a common mistake in Exponential and logarithms 2?

Why it happens: Confusing 'half' the constant with the half-life calculation. How to avoid it: Half-life T = 2/|k| 0.693/|k|. Derive each time from A e^-kT = A/2.

Pure Mathematics 3Edexcel International A Levels Mathematics (XMA01-YMA01)

Exponential and logarithms 2

Q: How do you solve exponential equations using logarithms in the context of Edexcel International A Levels?

To solve exponential equations using logarithms, you can take the logarithm of both sides of the equation. For example, if you have an equation like a^x = b, you can apply the logarithm to get x * log(a) = log(b). Then, solve for x by dividing both sides by log(a). This technique is essential for solving exponential equations in Pure Mathematics 3.

Q: What is the relationship between exponential and logarithmic functions in Pure Mathematics 3?

Exponential and logarithmic functions are inverses of each other. If y = a^x, then x = log_a(y). This inverse relationship means that the logarithm of a number is the exponent to which the base must be raised to produce that number. Understanding this relationship is fundamental in solving problems involving these functions in Edexcel International A Levels.

Q: How do you differentiate exponential and logarithmic functions in Pure Mathematics 3?

To differentiate exponential functions like f(x) = e^x, the derivative is f'(x) = e^x. For logarithmic functions, such as f(x) = ln(x), the derivative is f'(x) = 1/x. These differentiation rules are crucial for solving calculus problems involving exponential and logarithmic functions in the Edexcel International A Levels curriculum.

Q: What are common applications of exponential and logarithmic functions in real-world scenarios covered in Pure Mathematics 3?

Exponential and logarithmic functions are used in various real-world applications, such as modeling population growth, radioactive decay, and compound interest. In Pure Mathematics 3, students learn to apply these functions to solve practical problems, enhancing their understanding of how mathematics can be used to describe and predict real-world phenomena.

Q: Why is "Quoting $x = 0.473$ when the question asks for exact form" a common mistake in Exponential and logarithms 2?

Why it happens: Calculator habit. How to avoid it: READ the question. 'Exact form' / 'in terms of ' / 'leave as a logarithm' — all mean NO decimals. Final answer should be 7 - 1/2 or 12 7 - 12. Source: WMA13/01 examiner reports — flagged annually

Q: Why is "$\ln(a + b) = \ln a + \ln b$" a common mistake in Exponential and logarithms 2?

Why it happens: Mixing up the log of a SUM with log of a PRODUCT. How to avoid it: (ab) = a + b (product rule). (a + b) does NOT simplify. There's no log law for sums.

Q: Why is "Plotting $\ln y$ vs $\ln x$ when the model is $y = ab^{x}$" a common mistake in Exponential and logarithms 2?

Why it happens: Defaulting to log-log axes. How to avoid it: Look at what's exponential: y = a x^n — power law — log BOTH axes. y = a b^x — exponential — log y only. The variable in the exponent dictates whether you log the x-axis.

Q: Why is "Writing $\ln(-2)$ as a real number" a common mistake in Exponential and logarithms 2?

Why it happens: Forgetting has domain x > 0. How to avoid it: Always check that any argument of is positive. If e^x = -2 arises in a substitution, REJECT — there is no real solution because e^x > 0 always.

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Exponentials and Logarithms 2 Study Notes — Edexcel IAL Mathematics P3 (WMA13/01, 2018 spec — 2026 onwards)

The natural exponential $e^{x}$ and natural logarithm $\ln x$ — inverse functions of each other. Exponential growth/decay modelling. Linearising a power law $y = ax^{n}$ or an exponential law $y = ab^{x}$ by taking logs to convert into a straight-line plot. Reading $a$ , $b$ , $n$ , $k$ from the gradient and intercept.

What you’ll learn

Mapped to the Cambridge IGCSE XMA01-YMA01 syllabus (2026 onwards).

P3 5.1 — Use $e^{x}$ and $\ln x$ to solve equations and model real-world processes.
P3 5.2 — Apply exponential growth and decay models $N = Ae^{kt}$ .
P3 5.3 — Linearise non-linear data using logarithms; read parameters from a straight-line plot.

Natural exponential and natural logarithm

$e^{x}$ and $\ln x$ are inverses. They unlock all exponential equations.

Definitions.

$e$ is Euler's constant, $e \approx 2.71828$ .
$\ln x = \log_{e} x$ — the logarithm to base $e$ .

Inverse relationships. $e^{\ln x} = x \quad (x > 0), \qquad \ln(e^{x}) = x \quad (x \in \mathbb{R}).$

Graphs.

$y = e^{x}$ : increasing, always positive, asymptote $y = 0$ as $x \to -\infty$ . $(0, 1)$ and $(1, e)$ .
$y = \ln x$ : increasing, defined for $x > 0$ , asymptote $x = 0$ . $(1, 0)$ and $(e, 1)$ .
Graphs are reflections of each other in $y = x$ .

Log laws (same as for any log):

$\ln(xy) = \ln x + \ln y$ .
$\ln(x/y) = \ln x - \ln y$ .
$\ln(x^{p}) = p\ln x$ .
$\ln 1 = 0$ , $\ln e = 1$ .

Solving equations.

$e^{f(x)} = k$ ( $k > 0$ ): take $\ln$ both sides, $f(x) = \ln k$ . $\ln f(x) = c$ : exponentiate, $f(x) = e^{c}$ .

Worked example. Solve $e^{2x + 1} = 7$ .

$2x + 1 = \ln 7 \Rightarrow x = \dfrac{\ln 7 - 1}{2}$ (exact form).

Worked example. Solve $\ln(3x - 1) = 2$ .

$3x - 1 = e^{2} \Rightarrow x = \dfrac{e^{2} + 1}{3}$ .

$e$ is the natural exponential base.
$\ln$ is base- $e$ log; inverse of $e^{x}$ .
Log laws: product → sum, quotient → diff, power → coefficient.
Solve by taking $\ln$ (for $e^{f}$ ) or exponentiating (for $\ln f$ ).

Exponential growth and decay modelling

$N = Ae^{kt}$ . Sign of $k$ determines growth vs decay.

Model. A quantity $N$ whose rate of change is proportional to itself ( $dN/dt \propto N$ ) satisfies $N = A e^{kt}$ where:

$A = N(0)$ — initial value at $t = 0$ .
$k > 0$ : GROWTH. $k < 0$ : DECAY.
$t$ — time.

Doubling time. $N(T) = 2A \Rightarrow e^{kT} = 2 \Rightarrow T = \dfrac{\ln 2}{k}$ (for $k > 0$ ).

Half-life. $N(T) = A/2 \Rightarrow e^{kT} = 1/2 \Rightarrow T = \dfrac{\ln 2}{|k|}$ (for $k < 0$ , write $|k|$ ).

Worked example. A radioactive substance has mass $M(t) = 250 e^{-0.011 t}$ (t in years). Find the half-life.

$250 e^{-0.011 T} = 125 \Rightarrow e^{-0.011 T} = 0.5 \Rightarrow -0.011 T = \ln 0.5 \Rightarrow T = \dfrac{\ln 2}{0.011} \approx 63.0$ years.

Other models. Some modelling problems use:

$N = A(1 - e^{-kt})$ (asymptotic growth from $0$ towards $A$ , e.g. charging a capacitor).
$T - T_{\infty} = (T_{0} - T_{\infty}) e^{-kt}$ (Newton's law of cooling).
$N = A b^{t}$ with $b = e^{k}$ — equivalent re-parametrisation.

Reading the question. Identify:

Variable for the quantity (mass / population / temperature).
Variable for time.
Initial value (constant $A$ ).
Rate constant ( $k$ ).
The unknown (a value at a future time, OR the time at which a target value is reached, OR a rate $dN/dt$ ).

$N = Ae^{kt}$ , $A = N(0)$ .
$k > 0$ growth, $k < 0$ decay.
Half-life $T = \dfrac{\ln 2}{|k|}$ .
Read the question for which quantity is being asked.

Linearisation of multiplicative laws

Power law → log both axes. Exponential law → log y-axis only.

Why linearise? When you suspect data follows $y = ax^{n}$ or $y = ab^{x}$ , taking logs converts the non-linear law to a STRAIGHT LINE. A scatter plot of the log-transformed data is easy to interpret: gradient and intercept give the unknown parameters.

Case 1: $y = ax^{n}$ (power law).

Take $\ln$ of both sides: $\ln y = \ln a + n \ln x.$

This is $Y = mX + c$ with $Y = \ln y$ , $X = \ln x$ , $m = n$ , $c = \ln a$ .

Plot $\ln y$ against $\ln x$ — log-LOG plot.

Gradient = $n$ .
$y$ -intercept (where $\ln x = 0$ , i.e. $x = 1$ ) = $\ln a$ , so $a = e^{\text{intercept}}$ .

Worked example. Data $(x, y)$ plotted with $\ln y$ on the $y$ -axis and $\ln x$ on the $x$ -axis gives a line with gradient $1.5$ and intercept $\ln 2$ . Find the relationship.

$n = 1.5$ .
$\ln a = \ln 2 \Rightarrow a = 2$ .
Model: $y = 2 x^{1.5}$ .

Case 2: $y = ab^{x}$ (exponential law).

Take $\ln$ of both sides: $\ln y = \ln a + x \ln b.$

This is $Y = mX + c$ with $Y = \ln y$ , $X = x$ (NOT $\ln x$ !), $m = \ln b$ , $c = \ln a$ .

Plot $\ln y$ against $x$ — log-LINEAR plot.

Gradient = $\ln b$ , so $b = e^{\text{gradient}}$ .
$y$ -intercept (where $x = 0$ ) = $\ln a$ , so $a = e^{\text{intercept}}$ .

Worked example. Data plotted gives a line with gradient $0.4$ and $y$ -intercept $\ln 3$ . Find the model.

$\ln b = 0.4 \Rightarrow b = e^{0.4} \approx 1.492$ .
$\ln a = \ln 3 \Rightarrow a = 3$ .
Model: $y = 3 \cdot 1.492^{x}$ .

Which axes to log?

Model	Plot
$y = ax^{n}$	$\ln y$ vs $\ln x$
$y = ab^{x}$	$\ln y$ vs $x$
$y = a + bx^{n}$	not log-linear; can't be linearised directly

Power law $y = ax^{n}$ : log both axes.
Exponential law $y = ab^{x}$ : log $y$ only.
Gradient gives $n$ or $\ln b$ .
Intercept gives $\ln a$ .

Disguised quadratics in $e^{x}$

Substitute $u = e^{x}$ to spot a quadratic.

Pattern. Equations like $e^{2x} + \alpha e^{x} + \beta = 0$ are quadratics in $u = e^{x}$ .

Worked example. Solve $e^{2x} - 5 e^{x} + 6 = 0$ .

Substitute $u = e^{x}$ , $u > 0$ : $u^{2} - 5u + 6 = 0 \Rightarrow (u - 2)(u - 3) = 0 \Rightarrow u = 2 \text{ or } u = 3.$

Both positive — both valid.

$e^{x} = 2 \Rightarrow x = \ln 2$ . $e^{x} = 3 \Rightarrow x = \ln 3$ .

Rejecting negative $u$ . If a substitution gives $u = -1$ (say), REJECT because $e^{x} > 0$ always.

Similar with $\ln$ . Equations like $(\ln x)^{2} - 3\ln x + 2 = 0$ are quadratics in $u = \ln x$ .

Worked example. $(\ln x)^{2} - 3\ln x + 2 = 0 \Rightarrow u^{2} - 3u + 2 = 0 \Rightarrow u = 1$ or $u = 2$ .

$\ln x = 1 \Rightarrow x = e$ . $\ln x = 2 \Rightarrow x = e^{2}$ .

Trickier cases. Sometimes the substitution is non-obvious — look for repeated structure (e.g. $e^{x} + e^{-x}$ might suggest $u = e^{x}$ then express $e^{-x} = 1/u$ , giving a rational equation).

Substitute $u = e^{x}$ ( $u > 0$ ) for $e^{2x}$ structure.
Or $u = \ln x$ for $(\ln x)^{2}$ structure.
Solve the resulting quadratic.
Reject negative $u$ when $u = e^{x}$ .

How it’s examined

Exponentials and Logarithms 2 appears on every WMA13/01 paper as a $6$ - $12$ -mark question, often a modelling context. Typical sub-parts: (a) substitute given values to find $k$ ( $3$ - $4$ marks); (b) find the value at a specified time ( $2$ marks); (c) find the time at which a target value is reached ( $3$ - $4$ marks); (d) find half-life / doubling time. Linearisation question: 'A scatter plot of $\ln y$ against $\ln x$ is linear; find $a$ , $n$ .' Always state EXACT form first, decimal form second. Examiner reports flag missing exact forms and rounding errors.

Sources: Pearson Edexcel International Advanced Level Mathematics specification (2018 — current for 2026 onwards); Edexcel IAL Mathematical Formulae and Statistical Tables (2018); WMA13/01 Examiner Reports — January 2023, June 2023, January 2024, June 2024. Last reviewed 2026-05-12.

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Step-by-step worked examples — Exponential and logarithms 2

Step-by-step solutions to past-paper-style questions on exponential and logarithms 2, written exactly the way a tutor would explain them at the board.

1Solve $e^{2x + 1} = 7$ giving an exact answer (3 marks, P3)
Core• Adapted from WMA13/01 January 2024 Q1• exponential equation, natural log
Question
Solve $e^{2x + 1} = 7$ , giving your answer in exact form.
Step-by-step solution
1. Step 1
  Take $\ln$ of both sides:
  $\ln(e^{2x + 1}) = \ln 7$
2. Step 2
  Use $\ln(e^{u}) = u$ :
  $2x + 1 = \ln 7$
3. Step 3
  Solve for $x$ :
  $x = \dfrac{\ln 7 - 1}{2}$
Answer
$x = \dfrac{\ln 7 - 1}{2}$ (or equivalently $\tfrac{1}{2}\ln 7 - \tfrac{1}{2}$ ).
Examiner tip
M1 take logs of both sides. A1 isolate $2x + 1$ . A1 final exact form. Decimal answer ( $x \approx 0.473$ ) loses the final A mark unless the question asks for a decimal. 'Exact form' means EXACT — no decimals.
2Exponential decay — half-life problem (6 marks, P3)
Extended• Adapted from WMA13/01 June 2024 Q7• modelling, exponential decay
Question
The mass $M$ grams of a radioactive substance at time $t$ years is modelled by $M = 250 e^{-kt}$ , where $k > 0$ . After $20$ years the mass has decreased to $200$ g. (a) Find $k$ , giving your answer to 3 d.p. (b) Find the half-life of the substance.
Step-by-step solution
1. Step 1
  (a) Substitute $t = 20$ , $M = 200$ :
  $200 = 250 e^{-20k} \;\Rightarrow\; e^{-20k} = \dfrac{200}{250} = 0.8$
2. Step 2
  Take logs:
  $-20k = \ln 0.8 \;\Rightarrow\; k = -\dfrac{\ln 0.8}{20}$
3. Step 3
  $k = -\dfrac{-0.2231}{20} \approx 0.01116$ , so $k \approx 0.011$ (3 d.p.).
4. Step 4
  (b) Half-life $T$ : $M(T) = 125 = 250 e^{-kT} \Rightarrow e^{-kT} = 0.5$ . Take logs:
  $-kT = \ln 0.5 \;\Rightarrow\; T = \dfrac{\ln 2}{k} \approx \dfrac{0.693}{0.01116} \approx 62.1 \text{ years}$
Answer
(a) $k \approx 0.011$ . (b) Half-life $\approx 62.1$ years.
Examiner tip
(a) M1 substitution. M1 take logs. A1 value of $k$ (3 d.p. or exact). (b) M1 setting up $M = 125$ . M1 solve for $T$ . A1 final value (accept $T = \dfrac{20 \ln 2}{\ln(5/4)}$ as exact form). Always quote both exact and decimal forms when both are valid — saves marks if you mis-round.
3Linearise $y = a x^{n}$ via a $\ln y$ vs $\ln x$ plot (6 marks, P3)
Extended• Adapted from WMA13/01 January 2023 Q5• linearisation, log-log plot
Question
Data points $(x, y)$ are modelled by $y = a x^{n}$ . A scatter plot of $\ln y$ against $\ln x$ gives a straight line with gradient $1.5$ and $y$ -intercept $0.69$ . Find $a$ and $n$ .
Step-by-step solution
1. Step 1
  Take $\ln$ of both sides of $y = a x^{n}$ :
  $\ln y = \ln a + n \ln x$
2. Step 2
  Compare with straight-line form $Y = mX + c$ where $Y = \ln y$ , $X = \ln x$ , $m = n$ , $c = \ln a$ .
3. Step 3
  From the plot: gradient $m = 1.5 \Rightarrow n = 1.5$ .
4. Step 4
  Intercept $c = 0.69 \Rightarrow \ln a = 0.69 \Rightarrow a = e^{0.69} \approx 1.99$ . Recognising $0.69 \approx \ln 2$ , we get $a = 2$ exactly.
Answer
$a = 2$ , $n = 1.5$ (or $3/2$ ).
Examiner tip
M1 for taking logs of both sides. A1 for the linearised form. M1 for matching gradient to $n$ . A1 for $n = 1.5$ . M1 for $\ln a$ from intercept. A1 for $a = 2$ (or $a \approx 1.99$ ). Recognising 'famous' decimal logs ( $\ln 2 \approx 0.693$ , $\ln 3 \approx 1.099$ , $\ln 10 \approx 2.303$ ) speeds these questions up.
4Linearise $y = ab^{x}$ from a straight-line plot (6 marks, P3)
Extended• Adapted from WMA13/01 June 2023 Q6• linearisation, log-linear plot
Question
Data points $(x, y)$ are modelled by $y = a b^{x}$ . A scatter plot of $\ln y$ against $x$ (NOT $\ln x$ ) gives a straight line with gradient $0.4$ and $y$ -intercept $\ln 3$ . Find $a$ and $b$ (give $b$ to 3 d.p.).
Step-by-step solution
1. Step 1
  Take $\ln$ of $y = a b^{x}$ :
  $\ln y = \ln a + x \ln b$
2. Step 2
  Compare with $Y = mX + c$ , $Y = \ln y$ , $X = x$ , $m = \ln b$ , $c = \ln a$ .
3. Step 3
  Gradient $= 0.4 = \ln b \Rightarrow b = e^{0.4} \approx 1.492$ .
4. Step 4
  Intercept $= \ln 3 = \ln a \Rightarrow a = 3$ .
Answer
$a = 3$ , $b \approx 1.492$ .
Examiner tip
M1 take logs. A1 linearised form. M1 compare. A1 $a = 3$ . M1 $b = e^{0.4}$ . A1 $b \approx 1.492$ . Key insight for the difference between $y = ax^{n}$ (use $\ln y$ vs $\ln x$ ) and $y = ab^{x}$ (use $\ln y$ vs $x$ ). Pick the right axes by looking at the model.
5Solve $e^{2x} - 5 e^{x} + 6 = 0$ (5 marks, P3)
Extended• Adapted from WMA13/01 January 2024 Q6• natural log, quadratic substitution
Question
Solve $e^{2x} - 5 e^{x} + 6 = 0$ , giving exact values.
Step-by-step solution
1. Step 1
  Notice $e^{2x} = (e^{x})^{2}$ . Substitute $u = e^{x}$ (with $u > 0$ ).
  $u^{2} - 5u + 6 = 0$
2. Step 2
  Factorise: $(u - 2)(u - 3) = 0$ . So $u = 2$ or $u = 3$ .
3. Step 3
  Substitute back: $e^{x} = 2 \Rightarrow x = \ln 2$ , $e^{x} = 3 \Rightarrow x = \ln 3$ .
Answer
$x = \ln 2$ or $x = \ln 3$ .
Examiner tip
M1 spot the quadratic structure / substitute $u = e^{x}$ . A1 quadratic in $u$ . M1 factorise. A1 both values of $u$ . A1 both exact $x$ values. Both $u$ values are positive (good — $e^{x} > 0$ always). Reject any negative $u$ that arises.

Key Formulae — Exponential and logarithms 2

The formulae you need to memorise for exponential and logarithms 2 on the Pearson Edexcel IAL XMA01 / YMA01 paper, with every variable defined in plain English and a note on when to use it.

Natural exponential / log inverses
$e^{\ln x} = x \;\; (x > 0), \qquad \ln(e^{x}) = x \;\; (x \in \mathbb{R})$
$e$
$\approx 2.71828$ , base of natural log
$\ln x$
natural log of $x$
When to use
Whenever solving an equation involving $e^{f(x)} = k$ or $\ln f(x) = k$ — take the inverse of both sides. Definitions are in the IAL formula booklet implicitly (under log laws).
Example
$e^{2x + 1} = 7 \Rightarrow 2x + 1 = \ln 7$ .
Exponential growth/decay model
$N = A e^{kt}$
$N$
quantity at time $t$
$A$
initial value at $t = 0$ ( $N(0) = A$ )
$k$
growth (if $k > 0$ ) or decay (if $k < 0$ ) constant
$t$
time
When to use
Population growth, radioactive decay, cooling, account interest (continuous compounding), and any process where the rate is proportional to the current amount. NOT in the formula booklet — memorise the form and adjust the sign of $k$ .
Example
Half-life $T = \dfrac{\ln 2}{|k|}$ .
Linearise power law
$y = a x^{n} \;\Leftrightarrow\; \ln y = \ln a + n \ln x$
$n$
gradient of $\ln y$ vs $\ln x$ plot
$\ln a$
$y$ -intercept of the plot
When to use
When data is suspected to follow $y = a x^{n}$ — plot $\ln y$ against $\ln x$ . NOT in the formula booklet — derive by taking logs.
Example
Straight line $\ln y = 1.5 \ln x + \ln 2 \Rightarrow y = 2 x^{1.5}$ .
Linearise exponential law
$y = a b^{x} \;\Leftrightarrow\; \ln y = \ln a + x \ln b$
$\ln b$
gradient of $\ln y$ vs $x$ plot
$\ln a$
$y$ -intercept of the plot
When to use
When data follows $y = a b^{x}$ — plot $\ln y$ against $x$ (not $\ln x$ ). The base of log doesn't matter so long as it's consistent. NOT in the formula booklet.
Example
$\ln y = 0.4 x + \ln 3 \Rightarrow y = 3 \cdot e^{0.4 x} = 3 \cdot 1.492^{x}$ .

Key Definitions and Keywords — Exponential and logarithms 2

Definitions to memorise and the exact keywords mark schemes credit for exponential and logarithms 2 answers — sharpened from recent examiner reports for the 2026 Pearson Edexcel IAL XMA01 / YMA01 sitting.

Euler's number $e$
Examiner keyword
An irrational constant $e \approx 2.71828$ defined as $\lim_{n \to \infty}(1 + 1/n)^{n}$ , or equivalently as the unique base such that the derivative of $a^{x}$ equals $a^{x}$ when $a = e$ .
Natural logarithm ( $\ln x$ )
Examiner keyword
$\ln x = \log_{e} x$ — the logarithm to base $e$ . Domain: $x > 0$ . Range: all reals. $\ln 1 = 0$ , $\ln e = 1$ .
Exponential growth / decay
Examiner keyword
A process where the rate of change is proportional to the current amount, modelled by $N = A e^{kt}$ . $k > 0$ gives growth (doubling time = $\ln 2 / k$ ); $k < 0$ gives decay (half-life = $\ln 2 / |k|$ ).
Linearisation
Examiner keyword
The process of transforming a non-linear relation into a straight-line form by taking logs or another change of variables. Allows gradient and intercept to read off the parameters of the underlying model.
Half-life
The time required for the quantity in an exponential decay to reduce to half its current value: $T = \dfrac{\ln 2}{|k|}$ . Constant for a given $k$ — independent of starting value.

Common Mistakes and Misconceptions — Exponential and logarithms 2

The traps other students keep falling into on exponential and logarithms 2 questions — taken from recent Pearson Edexcel IAL XMA01 / YMA01 examiner reports and mark schemes — and how to avoid them.

✕Quoting $x = 0.473$ when the question asks for exact form
WMA13/01 examiner reports — flagged annually
Why it happens
Calculator habit.
How to avoid it
READ the question. 'Exact form' / 'in terms of $\ln$ ' / 'leave as a logarithm' — all mean NO decimals. Final answer should be $\dfrac{\ln 7 - 1}{2}$ or $\tfrac{1}{2}\ln 7 - \tfrac{1}{2}$ .
✕ $\ln(a + b) = \ln a + \ln b$
Why it happens
Mixing up the log of a SUM with log of a PRODUCT.
How to avoid it
$\ln(ab) = \ln a + \ln b$ (product rule). $\ln(a + b)$ does NOT simplify. There's no log law for sums.
✕Plotting $\ln y$ vs $\ln x$ when the model is $y = ab^{x}$
Why it happens
Defaulting to log-log axes.
How to avoid it
Look at what's exponential: $y = a x^{n}$ — power law — log BOTH axes. $y = a b^{x}$ — exponential — log $y$ only. The variable in the exponent dictates whether you log the $x$ -axis.
✕Writing $\ln(-2)$ as a real number
Why it happens
Forgetting $\ln$ has domain $x > 0$ .
How to avoid it
Always check that any argument of $\ln$ is positive. If $e^{x} = -2$ arises in a substitution, REJECT — there is no real solution because $e^{x} > 0$ always.
✕Stating half-life $= 0.5 / k$
Why it happens
Confusing 'half' the constant with the half-life calculation.
How to avoid it
Half-life $T = \dfrac{\ln 2}{|k|} \approx \dfrac{0.693}{|k|}$ . Derive each time from $A e^{-kT} = A/2$ .

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