What is the relationship between exponential functions and logarithms in Pure Mathematics 2?

In Pure Mathematics 2, exponential functions and logarithms are closely related as inverse functions. An exponential function is of the form y = a^x, where a is a positive constant. The logarithm function, log_a(y) = x, is the inverse, meaning it undoes the exponential function. This relationship is fundamental in solving equations involving exponentials and logarithms, which is a key concept in the Edexcel International A Levels curriculum.

How do you solve exponential equations using logarithms?

To solve exponential equations using logarithms, you first isolate the exponential expression. Then, apply the logarithm to both sides of the equation. For example, if you have 2^x = 8, you can take the logarithm base 2 of both sides to get x = log_2(8). This simplifies to x = 3, since 2^3 = 8. This method is crucial for solving complex exponential equations in Pure Mathematics 2.

What are the properties of logarithms that are important for Edexcel International A Levels?

The key properties of logarithms include the product rule (log_a(xy) = log_a(x) + log_a(y)), the quotient rule (log_a(x/y) = log_a(x) - log_a(y)), and the power rule (log_a(x^b) = b*log_a(x)). These properties are essential for simplifying logarithmic expressions and solving logarithmic equations, which are important skills in the Edexcel International A Levels Mathematics curriculum.

How do you convert between different logarithmic bases in Pure Mathematics 2?

To convert between different logarithmic bases, you can use the change of base formula: log_a(b) = log_c(b) / log_c(a), where c is any positive number. This formula is particularly useful when you need to calculate logarithms with a base that is not available on a standard calculator, such as converting log_2(8) to a base 10 logarithm. Understanding this conversion is important for solving problems in the Edexcel International A Levels syllabus.

Why are exponential and logarithmic functions important in real-world applications?

Exponential and logarithmic functions are crucial in modeling real-world phenomena such as population growth, radioactive decay, and interest calculations. In the context of Edexcel International A Levels, understanding these functions helps students apply mathematical concepts to practical situations, enhancing their problem-solving skills and preparing them for further studies in mathematics and related fields.

Why is "Writing $\log(M + N) = \log M + \log N$" a common mistake in Exponential and logarithms?

Why it happens: Over-extending the product law (MN) = M + N to sums. How to avoid it: Test with numbers: \log_10(10 + 100) = \log_10 110 2.04, NOT 1 + 2 = 3. Log distributes over multiplication and division ONLY. Source: WMA12/01 examiner reports — flagged annually

Why is "Treating $2 \log x$ as $\log(2x)$" a common mistake in Exponential and logarithms?

Why it happens: Confusing the power law with a multiplicative shortcut. How to avoid it: n \log_a M = \log_a(M^n). So 2 x = (x^2), NOT (2x). Coefficients go INSIDE as POWERS, not as multipliers.

Why is "Accepting $\log_{a}(\text{negative})$ as a valid value" a common mistake in Exponential and logarithms?

Why it happens: Forgetting that the argument of a log must be positive. How to avoid it: After solving a log equation, always check that EVERY \log_a() has a positive argument. Discard any root that makes an argument zero or negative.

Why is "Solving $3^{x} = 17$ as $x = 17 / 3$" a common mistake in Exponential and logarithms?

Why it happens: Treating the exponential as if it were a coefficient. How to avoid it: 3^x is 3 raised to the power x, not 3 times x. To solve, take logs: x 3 = 17, so x = 17 / 3 2.58.

Why is "Computing half-life as $(\ln 0.5) / k$ and reporting a negative time" a common mistake in Exponential and logarithms?

Why it happens: Sign confusion with decay constants. How to avoid it: If A = A_0 e^-kt (with k > 0), half-life t_1/2 = ( 2) / k. The minus in the exponent cancels with the minus from (1/2) = - 2. Time must be positive.

Pure Mathematics 2Edexcel International A Levels Mathematics (XMA01-YMA01)

Exponential and logarithms

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Exponentials and Logarithms Study Notes — Edexcel IAL Mathematics P2 (WMA12/01, 2018 spec — 2026 onwards)

Exponential function $y = a^{x}$ , the natural exponential $e^{x}$ , logarithms (the inverse of exponentials), the natural log $\ln$ , the laws of logs, solving exponential equations, and modelling exponential growth and decay. The bridge from algebra into calculus of transcendental functions.

At a glance

Logarithm definition: $y = \log_{a} x \Leftrightarrow a^{y} = x$ (with $a > 0$ , $a \neq 1$ , $x > 0$ ).
Laws of logs: $\log(MN) = \log M + \log N$ ; $\log(M/N) = \log M - \log N$ ; $\log(M^{n}) = n \log M$ .
Special values: $\log_{a} 1 = 0$ , $\log_{a} a = 1$ , $\log_{a}(a^{x}) = x$ , $a^{\log_{a} x} = x$ .
Natural log: $\ln = \log_{e}$ , inverse of $e^{x}$ . $\ln 1 = 0$ , $\ln e = 1$ .
Solving $a^{x} = b$ : take logs of both sides, $x = \log_{a} b = \dfrac{\ln b}{\ln a}$ .
Growth/decay model: $A = A_{0} e^{kt}$ ; half-life $= (\ln 2)/|k|$ .
Domain warning: $\log_{a}(\text{negative})$ does NOT exist — discard such roots.

What you’ll learn

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

P2 3.1 — Understand and use the function $a^{x}$ and its graph, where $a$ is positive.
P2 3.2 — Understand and use the function $e^{x}$ and its graph.
P2 3.3 — Know and use the definition of $\log_{a} x$ as the inverse of $a^{x}$ , where $a$ is positive and $x \geq 0$ .
P2 3.4 — Know and use the function $\ln x$ and its graph; know and use $\ln x$ as the inverse function of $e^{x}$ .
P2 3.5 — Understand and use the laws of logarithms.
P2 3.6 — Solve equations of the form $a^{x} = b$ .
P2 3.7 — Use logarithmic graphs to estimate parameters in relationships of the form $y = a x^{n}$ and $y = k b^{x}$ .
P2 3.8 — Understand and use exponential growth and decay; use in modelling.

Exponential functions $y = a^{x}$ and $y = e^{x}$

Graphs through $(0, 1)$ . $e$ is the natural base where derivative equals the function.

The exponential function $y = a^{x}$ with $a > 0$ , $a \neq 1$ :

Passes through $(0, 1)$ since $a^{0} = 1$ .
Always positive: $a^{x} > 0$ for all real $x$ .
If $a > 1$ : increasing, asymptotic to the $x$ -axis as $x \to -\infty$ .
If $0 < a < 1$ : decreasing, asymptotic to the $x$ -axis as $x \to +\infty$ .

The natural exponential $y = e^{x}$ where $e \approx 2.71828$ :

Special property: the gradient at any point equals the $y$ -value. Derivative $\dfrac{d}{dx} e^{x} = e^{x}$ .
All exponentials can be re-expressed using $e$ : $a^{x} = e^{x \ln a}$ .

Key graph features.

Feature	$y = a^{x}$ , $a > 1$	$y = a^{x}$ , $0 < a < 1$
Through	$(0, 1)$	$(0, 1)$
As $x \to \infty$	$y \to \infty$	$y \to 0^{+}$
As $x \to -\infty$	$y \to 0^{+}$	$y \to \infty$
Asymptote	$x$ -axis	$x$ -axis
Monotonicity	Increasing	Decreasing

Transformations. $y = e^{kx}$ stretches horizontally by $1/k$ . $y = e^{x} + c$ shifts up by $c$ (asymptote becomes $y = c$ ). $y = a e^{kx} + c$ is the general modelling form.

$a^{0} = 1$ for any $a > 0$ .
$a^{x} > 0$ always.
Horizontal asymptote at $y = 0$ .
$e^{x}$ is its own derivative.

Logarithms as inverse of exponentials

$\log_{a} x$ asks: 'to what power do I raise $a$ to get $x$ ?'

Definition. For $a > 0$ , $a \neq 1$ , $x > 0$ :

$y = \log_{a} x \quad \Leftrightarrow \quad a^{y} = x.$

So $\log_{a}$ is the inverse function of $a^{x}$ .

Key consequences.

$\log_{a} 1 = 0$ (because $a^{0} = 1$ ).
$\log_{a} a = 1$ (because $a^{1} = a$ ).
$\log_{a}(a^{x}) = x$ and $a^{\log_{a} x} = x$ (inverse functions cancel).

Natural log. $\ln x = \log_{e} x$ is the inverse of $e^{x}$ .

$\ln 1 = 0$ , $\ln e = 1$ .
$\ln(e^{x}) = x$ and $e^{\ln x} = x$ .

Common log. $\log x$ (no base) usually means $\log_{10} x$ in IAL contexts.

Domain of $\log_{a}$ . The argument MUST be positive. $\log_{a}(\text{negative})$ and $\log_{a}(0)$ do not exist.

$y = \log_{a} x \Leftrightarrow a^{y} = x$ .
$\log_{a} 1 = 0$ , $\log_{a} a = 1$ .
$\ln = \log_{e}$ , inverse of $e^{x}$ .
Argument of any log must be positive.

Laws of logarithms

Multiplication $\leftrightarrow$ addition. Division $\leftrightarrow$ subtraction. Power $\leftrightarrow$ multiplication.

For $a > 0$ , $a \neq 1$ , and positive $M, N$ :

Law	Statement	Example
Product	$\log_{a}(MN) = \log_{a} M + \log_{a} N$	$\log_{2}(8 \cdot 4) = 3 + 2 = 5$
Quotient	$\log_{a}(M/N) = \log_{a} M - \log_{a} N$	$\log_{2}(8/4) = 3 - 2 = 1$
Power	$\log_{a}(M^{n}) = n \log_{a} M$	$\log_{2}(8^{3}) = 3 \cdot 3 = 9$
Reciprocal	$\log_{a}(1/M) = -\log_{a} M$	$\log_{2}(1/8) = -3$
Change of base	$\log_{a} x = \dfrac{\log_{b} x}{\log_{b} a}$	$\log_{3} 17 = \dfrac{\ln 17}{\ln 3} \approx 2.58$

Worked example. Express $3 \log_{2} x - 2 \log_{2} 5 + \log_{2} 7$ as a single log.

$3 \log_{2} x = \log_{2}(x^{3})$ (power law).
$2 \log_{2} 5 = \log_{2}(25)$ (power law).
Combine: $\log_{2}(x^{3}) + \log_{2} 7 - \log_{2} 25 = \log_{2}\!\left(\dfrac{7 x^{3}}{25}\right)$ .

Reverse direction. Often a question asks you to SPLIT a single log into a sum/difference of simpler logs — same rules, applied in reverse.

Multiplication $\to$ addition (and back).
Division $\to$ subtraction.
Power $\to$ multiplication.
Change of base via the ratio of any common base.

Solving exponential equations

Isolate the exponential. Take logs. Solve linearly.

Standard form $a^{x} = b$ . Take logs:

$x \ln a = \ln b \;\;\Rightarrow\;\; x = \dfrac{\ln b}{\ln a}.$

Worked example. Solve $5 \cdot 3^{2x} = 80$ .

Divide by $5$ : $3^{2x} = 16$ .
Take $\ln$ : $2x \ln 3 = \ln 16$ , so $x = \dfrac{\ln 16}{2 \ln 3} \approx 1.26$ .

Quadratic-in-exponential. Equations of the form $A e^{2x} + B e^{x} + C = 0$ become quadratics under the substitution $u = e^{x}$ .

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

$u = e^{x}$ : $u^{2} - 5u + 6 = 0$ , so $(u - 2)(u - 3) = 0$ , giving $u = 2$ or $3$ .
$e^{x} = 2 \Rightarrow x = \ln 2$ ; $e^{x} = 3 \Rightarrow x = \ln 3$ .

(If a $u$ -value comes out negative, REJECT it: $e^{x} > 0$ always.)

Log equations. Equations involving $\log_{a}(\ldots)$ are usually solved by combining logs on each side to get $\log_{a}(\text{expr}_{1}) = \log_{a}(\text{expr}_{2})$ , then equating arguments. ALWAYS check both arguments are positive at the solution.

Worked example. Solve $\log_{2}(x + 3) + \log_{2}(x - 1) = 5$ .

Combine: $\log_{2}[(x + 3)(x - 1)] = 5$ .
Exponentiate: $(x + 3)(x - 1) = 2^{5} = 32$ .
Expand: $x^{2} + 2x - 3 = 32$ , so $x^{2} + 2x - 35 = 0$ , giving $(x - 5)(x + 7) = 0$ .
$x = 5$ or $x = -7$ . Check: $x = -7$ makes $\log_{2}(-4)$ undefined — REJECT.
Answer: $x = 5$ .

Isolate the exponential before taking logs.
Use $\ln$ for $e^{x}$ , any log otherwise.
Quadratic-in-exponential: substitute $u = e^{x}$ , then solve.
Log equations: combine, exponentiate, CHECK domain.

Exponential growth and decay modelling

$A = A_{0} e^{kt}$ . Fit $k$ from a data point; use for predictions and half-life.

Standard model. Many real-world processes follow

$A = A_{0} e^{kt}$

where $A_{0}$ is the initial value (at $t = 0$ ), $k$ is the growth/decay rate constant ( $k > 0$ growth, $k < 0$ decay), and $t$ is time.

Setting up the model.

Identify $A_{0}$ from the initial condition.
Use a known data point $(t_{1}, A_{1})$ to solve for $k$ : $k = \dfrac{1}{t_{1}} \ln\!\left(\dfrac{A_{1}}{A_{0}}\right)$ .

Worked example. Radioactive mass: $m = 80 e^{-kt}$ . After $5$ days, $m = 60$ . Find $k$ and the half-life.

$60 = 80 e^{-5k} \Rightarrow e^{-5k} = 0.75$ , so $k = -\tfrac{1}{5} \ln 0.75 = \tfrac{\ln(4/3)}{5} \approx 0.0575$ .
Half-life: solve $m = 40$ , i.e. $e^{-kt} = 0.5$ . So $t = (\ln 2)/k \approx 12.0$ days.

Other forms.

Compound interest / population: $A = A_{0} (1 + r)^{t}$ where $r$ is per-period rate.
Newton's cooling (extension): $T - T_{\infty} = (T_{0} - T_{\infty}) e^{-kt}$ .

Linearising for $\log$ graphs. If $y = k b^{x}$ , then $\log y = \log k + x \log b$ : a straight line with intercept $\log k$ and gradient $\log b$ . This converts exponential-fit problems into linear-regression problems.

$A = A_{0} e^{kt}$ is the universal exponential model.
$A_{0}$ from initial condition; $k$ from a second data point.
Half-life: $t_{1/2} = (\ln 2) / |k|$ .
Linearise with logs to fit data on a straight line.

How it’s examined

Exponentials and logs appears in WMA12/01 P2 every sitting (Jan & Jun), typically as Q5-Q11 — worth $8$ - $14$ marks total across multiple questions. Typical questions: combine logs into a single log (3-4 marks); solve $a^{x} = b$ giving an exact and decimal answer (4 marks); solve a quadratic in $e^{x}$ (5-6 marks); exponential decay modelling with half-life (7-10 marks); use logs to linearise data (4-5 marks). EXACT-FORM answers (in $\ln$ , $e$ , or surds) are usually required as an A1 mark before the decimal A1. A* students should aim for $\geq 90\%$ here — this is a high-yield topic.

Sources: Pearson Edexcel International Advanced Level Mathematics specification (2018 — current for 2026 onwards); Edexcel IAL Mathematical Formulae and Statistical Tables (2018); WMA12/01 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

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

1Single log expression from three terms (4 marks, P2)
Core• Adapted from WMA12/01 January 2024 Q5• laws of logs, single log
Question
Express $3 \log_{2} x - 2 \log_{2} 5 + \log_{2} 7$ as a single logarithm.
Step-by-step solution
1. Step 1
  Move coefficients to powers: $3 \log_{2} x = \log_{2} x^{3}$ and $2 \log_{2} 5 = \log_{2} 25$ .
2. Step 2
  Combine the addition: $\log_{2} x^{3} + \log_{2} 7 = \log_{2}(7 x^{3})$ .
3. Step 3
  Combine the subtraction: $\log_{2}(7 x^{3}) - \log_{2} 25 = \log_{2}\!\left(\dfrac{7 x^{3}}{25}\right)$ .
Answer
$\log_{2}\!\left(\dfrac{7 x^{3}}{25}\right)$ .
Examiner tip
M1 for any correct use of the power law. M1 for any correct combination using addition $\to$ multiplication. A1 for the correct numerator $7 x^{3}$ . A1 for the correct denominator $25$ . The most common slip is forgetting that $2 \log 5 = \log 25$ (not $\log 10$ ).
2Solve an exponential equation using logs (4 marks, P2)
Core• Adapted from WMA12/01 June 2024 Q6• solving exponentials
Question
Solve $5 \cdot 3^{2x} = 80$ , giving your answer to $3$ significant figures.
Step-by-step solution
1. Step 1
  Divide both sides by $5$ : $3^{2x} = 16$ .
2. Step 2
  Take $\log$ (any base — using natural log here):
  $2x \ln 3 = \ln 16$
3. Step 3
  Rearrange:
  $x = \dfrac{\ln 16}{2 \ln 3}$
4. Step 4
  Calculator: $\ln 16 \approx 2.7726$ , $\ln 3 \approx 1.0986$ . So $x \approx \dfrac{2.7726}{2.1972} \approx 1.2618$ .
Answer
$x \approx 1.26$ (3 s.f.).
Examiner tip
M1 for isolating the exponential (dividing by $5$ ). M1 for taking logs of both sides. A1 for the exact form $x = \dfrac{\ln 16}{2 \ln 3}$ . A1 for $1.26$ . The exact form usually attracts an explicit A1 in WMA12/01 — don't lose it by jumping straight to the decimal.
3Quadratic-in-exponential (5 marks, P2)
Extended• Adapted from WMA12/01 January 2023 Q9• quadratic substitution, exponential equation
Question
Solve $e^{2x} - 5 e^{x} + 6 = 0$ , giving your answers in exact form.
Step-by-step solution
1. Step 1
  Substitute $u = e^{x}$ , so $e^{2x} = u^{2}$ :
  $u^{2} - 5u + 6 = 0$
2. Step 2
  Factorise: $(u - 2)(u - 3) = 0$ , giving $u = 2$ or $u = 3$ .
3. Step 3
  Convert back: $e^{x} = 2$ gives $x = \ln 2$ ; $e^{x} = 3$ gives $x = \ln 3$ .
Answer
$x = \ln 2$ or $x = \ln 3$ .
Examiner tip
M1 for the substitution $u = e^{x}$ . M1 for solving the resulting quadratic. A1 for $u = 2, 3$ . M1 for taking $\ln$ of both sides. A1 for both exact answers. Both roots are positive, so both are valid — if a root of the quadratic were negative, it would have to be rejected (since $e^{x} > 0$ ).
4Exponential decay model (7 marks, P2)
Extended• Adapted from WMA12/01 June 2023 Q10• modelling, decay
Question
The mass $m$ (in grams) of a radioactive sample after $t$ days is modelled by $m = 80 e^{-kt}$ , where $k$ is a positive constant. After $5$ days the mass is $60$ g. (a) Find the value of $k$ to $3$ s.f. (b) Find the half-life of the sample (the time for the mass to halve), giving your answer in days to $3$ s.f.
Step-by-step solution
1. Step 1
  (a) Substitute $t = 5$ , $m = 60$ :
  $60 = 80 e^{-5k} \;\;\Rightarrow\;\; e^{-5k} = \dfrac{60}{80} = \dfrac{3}{4}$
2. Step 2
  Take $\ln$ : $-5k = \ln(3/4)$ , so $k = -\dfrac{\ln(3/4)}{5} = \dfrac{\ln(4/3)}{5}$ .
3. Step 3
  Compute: $\ln(4/3) \approx 0.2877$ , so $k \approx 0.0575$ .
4. Step 4
  (b) Half-life: solve $m = 40 = 80 e^{-kt}$ , i.e. $e^{-kt} = \tfrac{1}{2}$ .
5. Step 5
  Take $\ln$ : $-kt = \ln(1/2) = -\ln 2$ , so $t = \dfrac{\ln 2}{k}$ .
6. Step 6
  Compute: $t = \dfrac{0.6931}{0.0575} \approx 12.05$ .
Answer
(a) $k \approx 0.0575$ . (b) Half-life $\approx 12.0$ days (3 s.f.).
Examiner tip
(a) M1 for substituting the data point. M1 for taking logs. A1 for $k$ in exact form. A1 for $k \approx 0.0575$ . (b) M1 for the half-life equation. M1 for solving. A1 for $t \approx 12.0$ . Exact-form working ( $\ln(4/3)/5$ , $(\ln 2)/k$ ) usually attracts the A1 — examiners want to see that students can avoid premature rounding.

Key Formulae — Exponential and logarithms

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

Definition of logarithm
$y = \log_{a} x \;\Leftrightarrow\; a^{y} = x \quad (a > 0, a \neq 1, x > 0)$
$a$
base of the logarithm
$x$
argument (must be positive)
$y$
the logarithm itself
When to use
Whenever you need to flip between exponential and logarithmic forms. NOT in the formula booklet.
Example
$\log_{2} 8 = 3$ because $2^{3} = 8$ .
Laws of logarithms
$\log_{a}(MN) = \log_{a} M + \log_{a} N, \quad \log_{a}\!\left(\dfrac{M}{N}\right) = \log_{a} M - \log_{a} N, \quad \log_{a}(M^{n}) = n \log_{a} M$
$a$
base (positive, $\neq 1$ )
$M, N$
positive real numbers
$n$
any real number
When to use
Combining or splitting logs in equations, simplifying log expressions, linearising exponential models. NOT in the IAL formula booklet — memorise.
Example
$\log_{2}(8 \cdot 4) = \log_{2} 8 + \log_{2} 4 = 3 + 2 = 5$ .
Change of base
$\log_{a} x = \dfrac{\log_{b} x}{\log_{b} a}$
$a, b$
bases
$x$
argument
When to use
When a calculator only computes $\log_{10}$ or $\ln$ , or when comparing logs of different bases. NOT in the formula booklet.
Example
$\log_{3} 17 = \dfrac{\ln 17}{\ln 3} \approx \dfrac{2.833}{1.099} \approx 2.579$ .
Exponential growth / decay model
$A = A_{0} e^{kt}$
$A$
quantity at time $t$
$A_{0}$
initial quantity (at $t = 0$ )
$k$
growth rate constant ( $k > 0$ growth, $k < 0$ decay)
$t$
time
When to use
Modelling populations, radioactive decay, cooling, interest, etc. NOT in the formula booklet (but the form is universal).
Example
$m = 80 e^{-0.0575 t}$ models radioactive mass with half-life $\approx 12$ days.
Natural logarithm and $e$
$\ln x = \log_{e} x, \quad e \approx 2.71828, \quad e^{\ln x} = x, \quad \ln(e^{x}) = x$
$e$
the natural-log base, $\approx 2.71828$
$x$
positive real
When to use
Solving equations involving $e^{x}$ ; differentiation and integration of exponentials in P2.7-P2.8. NOT in the formula booklet (but $e$ is a calculator constant).

Key Definitions and Keywords — Exponential and logarithms

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

Exponential function
Examiner keyword
A function of the form $f(x) = a^{x}$ with $a > 0$ , $a \neq 1$ . The graph passes through $(0, 1)$ , is always positive, increasing for $a > 1$ and decreasing for $0 < a < 1$ . The $x$ -axis is a horizontal asymptote.
Logarithm
Examiner keyword
$\log_{a} x$ is the exponent to which $a$ must be raised to give $x$ . Equivalently, $\log_{a}$ is the inverse function of $a^{x}$ .
Example
$\log_{10} 1000 = 3$ because $10^{3} = 1000$ .
Natural logarithm
Examiner keyword
Logarithm to the base $e$ , written $\ln x$ (so $\ln x = \log_{e} x$ ). The inverse of $e^{x}$ . $\ln 1 = 0$ and $\ln e = 1$ .
Euler's number $e$
An irrational constant approximately equal to $2.71828$ . Defined by $\lim_{n \to \infty}\!(1 + 1/n)^{n}$ or by $e = \sum 1/n!$ . The unique base such that $\dfrac{d}{dx} e^{x} = e^{x}$ .
Half-life
The time for a decaying quantity to halve. If $A = A_{0} e^{-kt}$ then half-life $= (\ln 2)/k$ .

Common Mistakes and Misconceptions — Exponential and logarithms

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

✕Writing $\log(M + N) = \log M + \log N$
WMA12/01 examiner reports — flagged annually
Why it happens
Over-extending the product law $\log(MN) = \log M + \log N$ to sums.
How to avoid it
Test with numbers: $\log_{10}(10 + 100) = \log_{10} 110 \approx 2.04$ , NOT $1 + 2 = 3$ . Log distributes over multiplication and division ONLY.
✕Treating $2 \log x$ as $\log(2x)$
Why it happens
Confusing the power law with a multiplicative shortcut.
How to avoid it
$n \log_{a} M = \log_{a}(M^{n})$ . So $2 \log x = \log(x^{2})$ , NOT $\log(2x)$ . Coefficients go INSIDE as POWERS, not as multipliers.
✕Accepting $\log_{a}(\text{negative})$ as a valid value
Why it happens
Forgetting that the argument of a log must be positive.
How to avoid it
After solving a log equation, always check that EVERY $\log_{a}(\ldots)$ has a positive argument. Discard any root that makes an argument zero or negative.
✕Solving $3^{x} = 17$ as $x = 17 / 3$
Why it happens
Treating the exponential as if it were a coefficient.
How to avoid it
$3^{x}$ is $3$ raised to the power $x$ , not $3$ times $x$ . To solve, take logs: $x \log 3 = \log 17$ , so $x = \log 17 / \log 3 \approx 2.58$ .
✕Computing half-life as $(\ln 0.5) / k$ and reporting a negative time
Why it happens
Sign confusion with decay constants.
How to avoid it
If $A = A_{0} e^{-kt}$ (with $k > 0$ ), half-life $t_{1/2} = (\ln 2) / k$ . The minus in the exponent cancels with the minus from $\ln(1/2) = -\ln 2$ . Time must be positive.

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