Logarithms and Exponentials — A-Level Mathematics Pure 3 Study Guide

For: A-Level Mathematics candidates sitting Paper 3 (Pure Mathematics 3).

Covers: Properties of $a^x$ and $e^x$, core logarithm laws, natural log inverse relationships, solving exponential equations, linearising $y=a{x}^n$ and $y=a{b}^x$ models, and exponential growth/decay applications.

You should already know: A-Level Mathematics Pure 1 (functions, calculus, trigonometry).

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the A-Level Mathematics style for educational use. They are not reproductions of past Cambridge International examination papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official Cambridge mark schemes for grading conventions.

1. What Is Logarithms and Exponentials?

Logarithms and exponentials are paired inverse functions that describe multiplicative change, a core concept for modelling real-world processes and solving algebraic equations with unknown exponents. This topic accounts for 10-15% of total Paper 3 marks, appearing in standalone algebra questions, calculus problems, and data modelling tasks. It is closely tied to later syllabus content including differential equations and numerical methods.

2. Properties of $a^x$ and $e^x$

The general exponential function is defined as $f(x)=a^x$, where $a>0$ and $a\ne 1$ (a value of $a=1$ gives a flat constant function, which is not considered exponential). Its domain is all real values of $x$, and its range is strictly positive: $f(x)>0$ for all $x$. For $a>1$, the function is strictly increasing, and for $0<a<1$, it is strictly decreasing.

The natural exponential function $e^x$ uses the irrational constant $e\approx 2.71828$ as its base, and is unique because its derivative is equal to itself: $\frac{d}{dx}{e}^x=e^x$. This property makes it the standard base for calculus-based modelling in Pure Math 3.

Worked Example: Calculate the value of $e^{1.5}$ to 3 significant figures, and state the range of $f(x)=2e^{x-1}$.

• $e^{1.5}\approx 4.48$
• Since $e^{x-1}>0$ for all $x$, the range of $f(x)$ is $f(x)>0$.

Exam Tip: Examiners regularly test that you recognize $e^x$ is never negative, so you can discard negative solutions when solving equations with exponential terms.

3. Logarithm laws — $\log (ab),\log (a/b),\log (a^n)$

A logarithm is the inverse of an exponential: ${\log }_{b}a=c$ is equivalent to $b^c=a$, where $b$ is the base, $a$ is the argument, and $c$ is the exponent. The three core logarithm laws are derived directly from exponent rules:

1. Product rule: ${\log }_b(ab)={\log }_{b}a+{\log }_{b}b$, from $b^m\times b^n=b^{m+n}$
2. Quotient rule: ${\log }_b\left(\frac{a}{b}\right)={\log }_{b}a-{\log }_{b}b$, from $\frac{b^m}{b^n}=b^{m-n}$
3. Power rule: ${\log }_b(a^n)=n{\log }_{b}a$, from $(b^m{)}^n=b^{mn}$

Note that these laws only apply when the arguments of the logs are positive, as logs of non-positive values are undefined.

Worked Example: Simplify ${\log }_{5}125+{\log }_{5}5-{\log }_{5}25$, and expand ${\log }_{10}\left(\frac{1000x^2}{\sqrt{y}}\right)$.

• Simplification: ${\log }_{5}125=3$, ${\log }_{5}5=1$, ${\log }_{5}25=2$, so total $3+1-2=2$
• Expansion: ${\log }_{10}1000+2{\log }_{10}x-\frac{1}{2}{\log }_{10}y=3+2\log x-\frac{1}{2}\log y$

4. Natural log $\ln$ and the relationship $e^{\ln x}=x$

The natural logarithm, written as $\ln x$, is a logarithm with base $e$, so $\ln x={\log }_{e}x$. Its domain is $x>0$, and its range is all real numbers. As inverse functions, $\ln x$ and $e^x$ cancel each other out when composed:

• $e^{\ln x}=x$ for all $x>0$
• $\ln (e^x)=x$ for all real $x$

This inverse relationship is the most frequently used tool for solving equations involving $e$ or $\ln$ in Paper 3.

Worked Example: Solve $\ln (3x+4)=2.5$, giving your answer to 3 significant figures.

1. Exponentiate both sides with base $e$ to eliminate the ln: $e^{\ln (3x+4)}=e^{2.5}$
2. Simplify using the inverse rule: $3x+4=e^{2.5}\approx 12.182$
3. Rearrange to solve for x: $3x=8.182⟹x\approx 2.73$
4. Check domain: $3(2.73)+4=12.19>0$, so the solution is valid.

5. Solving exponential equations using logs

When you cannot rewrite both sides of an exponential equation with the same base, take the logarithm of both sides (you can use base 10 or natural log, both give the same result). The change of base formula ${\log }_{a}b=\frac{\ln b}{\ln a}$ is used to rearrange for unknown exponents.

For quadratic equations in $e^x$, use the substitution $y=e^x$ to convert to a standard quadratic form, then solve and discard any negative solutions (since $e^x>0$).

Worked Example: Solve $e^{2x}-7e^x+12=0$, giving exact answers.

1. Let $y=e^x$, so the equation becomes $y^2-7y+12=0$
2. Factor: $(y-3)(y-4)=0$, so $y=3$ or $y=4$
3. Convert back to x: $e^x=3⟹x=\ln 3$, $e^x=4⟹x=\ln 4$
4. Both solutions are valid, as $e^x$ is positive for both values.

6. Linearising $y=a{x}^n$ and $y=a{b}^x$ relationships via logs

Many real-world experimental datasets follow non-linear power or exponential relationships. Taking logs of both sides converts these relationships to linear $y=mx+c$ form, making it easy to calculate unknown constants from plotted data.

• For power models $y=a{x}^n$: Take log of both sides to get $\log y=n\log x+\log a$. Plot $\log y$ on the y-axis against $\log x$ on the x-axis: the gradient is $n$, and the y-intercept is $\log a$, so $a=10^{\text{intercept}}$.
• For exponential models $y=a{b}^x$: Take log of both sides to get $\log y=(\log b)x+\log a$. Plot $\log y$ against $x$: the gradient is $\log b$, and the y-intercept is $\log a$, so $b=10^{\text{gradient}}$.

Worked Example: The relationship between x and y is $y=a{b}^x$. Experimental data gives points (2, 15.2) and (4, 57.8). Find a and b to 2 decimal places.

1. Log both sides: $\log 15.2=\log a+2\log b$ and $\log 57.8=\log a+4\log b$
2. Subtract the first equation from the second: $\log 57.8-\log 15.2=2\log b$
3. Calculate: $\log \left(\frac{57.8}{15.2}\right)=2\log b⟹0.580=2\log b⟹\log b=0.290⟹b=10^{0.290}\approx 1.95$
4. Substitute back: $\log a=\log 15.2-2(0.290)=1.1818-0.58=0.6018⟹a=10^{0.6018}\approx 4.00$

7. Modelling exponential growth and decay

Exponential models describe processes where the rate of change of a quantity is proportional to the quantity itself, with the general form: $$N(t)=N_0{e}^{kt}$$ Where $N_0$ is the initial quantity at $t=0$, $k$ is the rate constant, and $t$ is time. If $k>0$, the model describes growth (e.g., bacterial populations, compound interest), and if $k<0$, it describes decay (e.g., radioactive decay, cooling objects).

Half-life is the time taken for a decaying quantity to reduce to half its initial value, calculated as: $$t_{1/2}=\frac{\ln 2}{|k|}$$

Worked Example: A radioactive isotope has a half-life of 12 years. Calculate the time taken for 75% of the initial sample to decay, to 3 significant figures.

1. First find $k$: $12=\frac{\ln 2}{|k|}⟹|k|=\frac{\ln 2}{12}\approx 0.05776$, so $k=-0.05776$
2. 75% decayed means 25% remains, so $N(t)=0.25N_0$
3. Substitute into model: $0.25=e^{-0.05776t}$
4. Take ln of both sides: $\ln 0.25=-0.05776t⟹t=\frac{-1.3863}{-0.05776}\approx 24.0$ years

8. Common Pitfalls (and how to avoid them)

• Wrong move: Splitting $\log (a+b)$ into $\log a+\log b$. Why: Students mix up addition inside a log with multiplication of arguments. Correct move: Only split logs for products or quotients inside the argument, never sums or differences.
• Wrong move: Including solutions where the argument of $\ln x$ is negative. Why: Rushing through solutions without checking domain constraints. Correct move: Always substitute solutions back into log equations to confirm the argument is positive, discard any invalid solutions to avoid losing 1 mark per error.
• Wrong move: Mixing up linear forms for power and exponential models, plotting $\log y$ against $\log x$ for $y=a{b}^x$. Why: Confusing power and exponential relationship structure. Correct move: Only log the x-axis for power models; for exponential models, plot $\log y$ against raw x values.
• Wrong move: Rounding intermediate values to 3 sig figs, leading to incorrect final answers. Why: Students round early to save time. Correct move: Keep 4+ sig figs for all intermediate calculations, only round the final answer to 3 sig figs as required by A-Level.
• Wrong move: Using a positive k value for decay models, leading to negative half-life. Why: Forgetting that decreasing quantities have negative rate constants. Correct move: Always confirm k sign matches the process (positive for growth, negative for decay) before substituting values.

9. Practice Questions (Paper 3 Style)

Question 1

Solve the equation $2\cdot 9^x-9\cdot 3^x+4=0$, giving all solutions correct to 3 significant figures.

Worked Solution

1. Recognise that $9^x=(3^2{)}^x=(3^x{)}^2$, so the equation is a hidden quadratic in $u=3^x$.
2. Substitute $u=3^x$ (note $u>0$ for all real $x$):$$2u^2-9u+4=0.$$
3. Factorise: $(2u-1)(u-4)=0$, so $u=\frac{1}{2}$ or $u=4$. Both are positive, so both are admissible.
4. Back-substitute and take logs of base 3:

• $3^x=\frac{1}{2}⟹x={\log }_3\frac{1}{2}=-\frac{\ln 2}{\ln 3}\approx -\mathrm{0.631.}$
• $3^x=4⟹x={\log }_{3}4=\frac{2\ln 2}{\ln 3}\approx \mathrm{1.26.}$

5. Both solutions satisfy the original equation, so $x\approx -0.631$ or $x\approx 1.26$ (3 s.f.).

Question 2

The relationship between x and y is given by $y=k{x}^n$, where k and n are constants. A plot of ${\log }_{10}y$ against ${\log }_{10}x$ gives a straight line with gradient 1.5 and y-intercept 0.301. (a) Find the values of k and n. (b) Calculate the value of y when x = 4, to 3 significant figures.

Worked Solution

(a) Linear form is $\log y=n\log x+\log k$. Gradient = n = 1.5, intercept = $\log k=0.301$, so $k=10^{0.301}\approx 2.00$. (b) Substitute $x=4$: $y=2.00\times (4{)}^{1.5}=2\times 8=16.0$

Question 3

A colony of bacteria grows according to $P(t)=500e^{0.03t}$, where t is time in hours. (a) State the initial population. (b) Find the time taken for the population to triple, to the nearest hour. (c) Calculate the rate of change of population at t=20 hours, to 2 significant figures.

Worked Solution

(a) Initial population at t=0: $P(0)=500e^0=500$ (b) Triple population = 1500: $1500=500e^{0.03t}⟹3=e^{0.03t}⟹\ln 3=0.03t⟹t=\frac{1.0986}{0.03}\approx 37$ hours (c) Rate of change = $\frac{dP}{dt}=500\times 0.03e^{0.03t}=15e^{0.03t}$. At t=20: $15e^{0.6}\approx 15\times 1.822=27$ bacteria per hour (2 sig figs)

10. Quick Reference Cheatsheet

11. What's Next

Mastery of logarithms and exponentials is essential for upcoming Paper 3 topics, including first-order differential equations (where solutions often take exponential form), further calculus (integrating $\frac{1}{x}$ gives $\ln x$, a core integration rule), and numerical methods (where logarithms simplify iterative equation solving). You will also encounter these functions in complex number modulus calculations and mechanics problems involving resistive forces proportional to velocity, so strong foundational knowledge will reduce revision time later in the course.

If you need extra support with any of the rules, examples, or practice questions in this guide, you can ask Ollie, our AI tutor, for personalized explanations, additional practice problems, or feedback on your working. Head to the homepage to access more A-Level Mathematics Paper 3 study guides, topic quizzes, and full past paper walkthroughs to prepare for your exam.

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