Edexcel A-Level Maths: Exponentials and Logarithms — Complete Revision Guide (9MA0)

Exponentials and logarithms is one of the most distinctive sections of Edexcel A-Level Maths (9MA0). It is where the algebra you have been building since GCSE meets a genuinely new family of functions — the exponential $e^x$ex and its inverse $\ln x$lnx — and where the language of growth, decay, and modelling becomes part of your mathematical toolkit. The section is examinable on either Paper 1 or Paper 2 of the pure papers, and the same techniques resurface inside calculus, differential equations, and statistical modelling later in the course.

This guide is a topic-by-topic walkthrough of the exponentials and logarithms content in the 9MA0 specification. It covers the laws of indices in their full A-Level form, exponential functions and their graphs, the special role of $e$e and the natural exponential, logarithm laws, solving exponential and logarithmic equations, natural logarithms, exponential growth and decay, logarithmic graphs and linearisation, change of base, and modelling with exponentials. For each topic you will see the core skills, the typical pitfalls, a short worked example, and a link to the full lesson on the LearningBro course.

The aim is not to replace working through problems. Logarithm questions reward fluency, and the only way to develop fluency is to work many of them under time pressure. The aim of this guide is to give you a clear map of what Edexcel expects, in the order the specification teaches it, so your revision is targeted rather than scattered.

What the Edexcel 9MA0 Specification Covers

The Edexcel A-Level Maths qualification (9MA0) is assessed through three two-hour papers, each worth 100 marks. Papers 1 and 2 cover pure mathematics, and Paper 3 covers statistics and mechanics. The exponentials and logarithms content sits in Section 6 of the pure specification and can appear on either pure paper. There is no choice of questions and no coursework, so every mark must be earned in the exam.

Exponentials and logarithms is a high-frequency area of pure. Almost every cohort sees a substantial logs question worth 6-10 marks, and the techniques quietly underpin questions in calculus (differentiating $e^{kx}$ekx, integrating $1/x$1/x), in differential equations on the second pure paper, and in modelling questions across the course. The table below shows the sub-topics, where they sit in the specification, and a realistic estimate of how many marks across a Paper 1 / Paper 2 sitting come from each.

Topic	Spec Section	Typical Paper 1/2 marks weight
Laws of indices (A-Level extension)	6.1	2-4 marks
Exponential functions $y = a^x$y=ax	6.2	3-5 marks
The natural exponential $e^x$ex	6.3	3-5 marks
Logarithms and laws of logarithms	6.4	4-6 marks
Solving exponential equations	6.5	4-6 marks
Natural logarithms $\ln x$lnx	6.6	3-5 marks
Exponential growth and decay	6.7	5-8 marks
Logarithmic graphs and linearisation	6.8	4-6 marks
Change of base	6.9	2-3 marks
Modelling with exponentials	6.10	5-8 marks

These weights are estimates based on the spread of typical 9MA0 papers — not guarantees for any single year. What is reliable, however, is that exponentials and logarithms is a near-certainty on at least one of the pure papers, that modelling and linearisation questions tend to carry the most marks per question, and that the same techniques resurface in calculus and statistical contexts. Mastering this section is high-leverage revision.

Laws of Indices

The laws of indices are GCSE territory in their basic form, but the A-Level pushes them into fractional, negative, and zero powers and assumes you can switch between forms instantly. Every later topic in this section uses indices, so any rust here gets expensive fast.

The core laws are $x^a \cdot x^b = x^{a+b}$xa⋅xb=xa+b, $x^a / x^b = x^{a-b}$xa/xb=xa−b, $(x^a)^b = x^{ab}$(xa)b=xab, $x^0 = 1$x0=1 for $x \neq 0$x=0, $x^{-n} = 1/x^n$x−n=1/xn, and $x^{1/n} = \sqrt[n]{x}$x1/n=nx​. Combinations like $x^{3/2} = \sqrt{x^3} = (\sqrt{x})^3$x3/2=x3​=(x​)3 and $x^{-2/3} = 1/\sqrt[3]{x^2}$x−2/3=1/3x2​ appear constantly. You also need to be able to write expressions like $\sqrt{x^5}/x$x5​/x in the form $x^k$xk — a routine but examinable conversion.

Common pitfalls include misreading $x^{-1/2}$x−1/2 as $-\sqrt{x}$−x​ instead of $1/\sqrt{x}$1/x​, applying $(x^a)^b$(xa)b as $x^{a+b}$xa+b instead of $x^{ab}$xab, and forgetting that the laws only apply to powers with the same base. Expressions like $2^x \cdot 3^x$2x⋅3x cannot be combined as a single power of one base; they must be written as $(2 \cdot 3)^x = 6^x$(2⋅3)x=6x if you need a single exponential.

A short worked example. Simplify $\dfrac{8x^{1/2} \cdot \sqrt[3]{x^2}}{2x^{-1/6}}$2x−1/68x1/2⋅3x2​​ as a single power of $x$x with a numerical coefficient. The numerator gives $8 x^{1/2 + 2/3} = 8 x^{7/6}$8x1/2+2/3=8x7/6. Dividing by $2 x^{-1/6}$2x−1/6 gives $4 x^{7/6 - (-1/6)} = 4 x^{8/6} = 4 x^{4/3}$4x7/6−(−1/6)=4x8/6=4x4/3. Notice how the negative exponent in the denominator becomes positive after subtraction — a frequent slip.

For full coverage with practice questions and worked solutions, see the Laws of Indices lesson.

Exponential Functions

An exponential function has the form $y = a^x$y=ax, where $a > 0$a>0 and $a \neq 1$a=1. The variable is in the exponent, not in the base — this is what distinguishes it from a polynomial like $x^2$x2. Edexcel expects you to recognise the shape of $y = a^x$y=ax and its key features without plotting points.

For $a > 1$a>1, the graph passes through $(0, 1)$(0,1), rises slowly for negative $x$x, and grows without bound as $x$x increases. The $x$x-axis is a horizontal asymptote on the left. For $0 < a < 1$0<a<1, the graph still passes through $(0, 1)$(0,1) but falls instead — large for negative $x$x and approaching the $x$x-axis on the right. Either way, $a^x > 0$ax>0 for all real $x$x, so the curve never crosses the $x$x-axis. The $y$y-intercept is always $1$1 because $a^0 = 1$a0=1.

The defining feature of exponential growth is constant proportional change: increasing $x$x by $1$1 multiplies $y$y by $a$a, regardless of where you start. This is what makes exponentials the right model for populations, bank balances, and radioactive decay — situations where the rate of change is proportional to the current amount.

A common pitfall is confusing $y = a^x$y=ax with $y = x^a$y=xa. The former is exponential and grows much faster than any polynomial; the latter is a power function and grows at a polynomial rate. Another is forgetting the asymptote when sketching, which costs a method mark even if the rest of the curve is right.

For sketching practice and the standard transformations of $y = a^x$y=ax, see the Exponential Functions lesson.

The Natural Exponential

Among all possible bases, one is special. The number $e \approx 2.71828$e≈2.71828 is the unique base for which the gradient of $y = e^x$y=ex at every point equals the value of the function itself: $\dfrac{d}{dx}(e^x) = e^x$dxd​(ex)=ex. This property is what makes $e$e central to calculus and to every model where rate of change is proportional to current value.

The graph of $y = e^x$y=ex behaves like any other exponential with $a > 1$a>1: it passes through $(0, 1)$(0,1), rises rapidly as $x$x increases, and approaches the $x$x-axis as $x$x decreases. What sets it apart is its derivative. From $\dfrac{d}{dx}(e^x) = e^x$dxd​(ex)=ex, the chain rule gives $\dfrac{d}{dx}(e^{kx}) = k e^{kx}$dxd​(ekx)=kekx, so the family $y = A e^{kx}$y=Aekx has the property $\dfrac{dy}{dx} = ky$dxdy​=ky. This is the differential equation of exponential growth, and you will see it again in mechanics, biology, and finance.

You also need to handle transformations. $y = e^{-x}$y=e−x is the reflection of $y = e^x$y=ex in the $y$y-axis and represents exponential decay. $y = A e^{kx} + c$y=Aekx+c shifts the asymptote up to $y = c$y=c and scales by $A$A. Sketches should show the asymptote, the $y$y-intercept, and the direction of growth or decay clearly.

A common pitfall is treating $e$e as if it were a variable. It is a fixed irrational constant, and expressions like $e^2$e2 are numbers (about $7.389$7.389), not algebraic objects. Another is forgetting the chain-rule factor when differentiating $e^{kx}$ekx — the $k$k matters.

For worked examples on differentiating and sketching natural exponentials, see the Natural Exponential lesson.

Logarithms and Laws

A logarithm is the inverse of an exponential. The statement $\log_a x = y$loga​x=y means exactly the same thing as $a^y = x$ay=x. You should be able to switch between the two forms instantly. The base $a$a must satisfy $a > 0$a>0 and $a \neq 1$a=1, and the argument $x$x must be strictly positive — you cannot take the log of zero or a negative number.

The three laws of logarithms are the working tools of the topic. For any valid base $a$a and positive arguments $x$x and $y$y:

Law	Statement
Product law	$\log_a(xy) = \log_a x + \log_a y$loga​(xy)=loga​x+loga​y
Quotient law	$\log_a(x/y) = \log_a x - \log_a y$loga​(x/y)=loga​x−loga​y
Power law	$\log_a(x^k) = k \log_a x$loga​(xk)=kloga​x

Two further identities follow directly from the definition: $\log_a 1 = 0$loga​1=0 (because $a^0 = 1$a0=1) and $\log_a a = 1$loga​a=1 (because $a^1 = a$a1=a). These are easy marks if you remember them and easy marks lost if you do not.

A common pitfall is "distributing" the log over a sum — $\log(x + y)$log(x+y) is not $\log x + \log y$logx+logy. The product law applies to a product inside the log, not a sum. Another is sign errors with the quotient law, particularly when the fraction is written in a long expression and the numerator and denominator are not visually obvious.

A short worked example. Write $2 \log 3 + \log 5 - \log 9$2log3+log5−log9 as a single logarithm. By the power law $2 \log 3 = \log 9$2log3=log9. The sum becomes $\log 9 + \log 5 - \log 9 = \log 5$log9+log5−log9=log5. Notice how the power law lets you absorb coefficients into arguments — a constant move in Edexcel logs questions.

For the laws in detail with practice on combining and splitting logarithmic expressions, see the Logarithms and Laws lesson.

Solving Exponential Equations

Solving an equation like $5^x = 17$5x=17 requires logarithms because $17$17 is not a tidy power of $5$5. The standard move is to take logs of both sides and use the power law to bring the exponent down: $x \log 5 = \log 17$xlog5=log17, so $x = \log 17 / \log 5$x=log17/log5. Either common logs or natural logs work — the answer is the same number — and the calculator gives a decimal value when you need one.

For equations with the unknown in more than one exponent, like $3^{2x} - 4 \cdot 3^x + 3 = 0$32x−4⋅3x+3=0, the trick is substitution. Let $u = 3^x$u=3x. Then $3^{2x} = u^2$32x=u2 and the equation becomes $u^2 - 4u + 3 = 0$u2−4u+3=0, which factorises as $(u - 1)(u - 3) = 0$(u−1)(u−3)=0. So $u = 1$u=1 or $u = 3$u=3, giving $3^x = 1$3x=1 (so $x = 0$x=0) or $3^x = 3$3x=3 (so $x = 1$x=1). The substitution converts an exponential equation into a quadratic — a high-frequency Edexcel pattern.

For mixed equations like $2^{x+1} = 5^{x-2}$2x+1=5x−2, take logs of both sides at once: $(x + 1) \log 2 = (x - 2) \log 5$(x+1)log2=(x−2)log5. Expand, collect $x$x-terms on one side, and divide. The algebra is routine but unforgiving — sign errors here propagate to the final answer.

A common pitfall is forgetting to check that solutions are valid. After the substitution $u = 3^x$u=3x, the requirement $u > 0$u>0 rules out any negative roots of the quadratic. Always check that each candidate solution gives a positive value of the substituted variable.

For worked examples of all three patterns and a clean substitution workflow, see the Solving Exponential Equations lesson.

Natural Logarithms

The natural logarithm $\ln x$lnx is the logarithm with base $e$e. So $\ln x = \log_e x$lnx=loge​x, and $\ln x = y$lnx=y means exactly $e^y = x$ey=x. The natural log obeys all the standard log laws — product, quotient, power — and has the same domain restriction: $x > 0$x>0.

Two key identities link $\ln$ln and $e^x$ex directly: $\ln(e^x) = x$ln(ex)=x and $e^{\ln x} = x$elnx=x for $x > 0$x>0. These are the inverse-function identities, and they are the tools you reach for when you need to "cancel" an exponential or a log. Solving $e^{2x} = 7$e2x=7 becomes $2x = \ln 7$2x=ln7, so $x = \tfrac{1}{2} \ln 7$x=21​ln7. Solving $\ln(3x - 1) = 4$ln(3x−1)=4 becomes $3x - 1 = e^4$3x−1=e4, so $x = (e^4 + 1)/3$x=(e4+1)/3.

You also need to handle the calculus identities $\dfrac{d}{dx}(\ln x) = 1/x$dxd​(lnx)=1/x and $\int 1/x \, dx = \ln |x| + C$∫1/xdx=ln∣x∣+C. These appear repeatedly in calculus questions on Paper 2, and the modulus inside the log on the integral is essential — without it, you would lose the negative half of the domain.

A common pitfall is treating $\ln$ln as a variable. It is an operator, not a multiplier — $\ln(2x)$ln(2x) is not $2 \ln x$2lnx unless you apply the product law correctly: $\ln(2x) = \ln 2 + \ln x$ln(2x)=ln2+lnx. Another is forgetting that $\ln 1 = 0$ln1=0 and $\ln e = 1$lne=1, which are routine substitutions in many problems.

For practice on the inverse identities and on calculus involving $\ln$ln, see the Natural Logarithms lesson.

Exponential Growth and Decay

Exponential growth is described by $P = P_0 e^{kt}$P=P0​ekt with $k > 0$k>0, where $P_0$P0​ is the initial value and $k$k is the growth rate. Exponential decay uses the same equation with $k < 0$k<0, or equivalently $P = P_0 e^{-\lambda t}$P=P0​e−λt with $\lambda > 0$λ>0. These two forms cover almost every Edexcel modelling question on this topic.

The standard tasks are: find $P$P at a given time, find the time at which $P$P reaches a given value, and find the rate of change at a given time. The first uses substitution. The second uses logs to invert the exponential: from $P_0 e^{kt} = P$P0​ekt=P you get $t = \tfrac{1}{k} \ln(P/P_0)$t=k1​ln(P/P0​). The third uses calculus: $\dfrac{dP}{dt} = k P_0 e^{kt} = kP$dtdP​=kP0​ekt=kP, which says the rate of change is proportional to the current value. This is the defining property of exponential models and the reason they appear so widely.

For decay problems, the half-life is the time taken for the quantity to halve. From $P_0 e^{-\lambda t_{1/2}} = P_0 / 2$P0​e−λt1/2​=P0​/2 you get $t_{1/2} = \ln 2 / \lambda$t1/2​=ln2/λ. The half-life depends only on $\lambda$λ, not on the initial amount — a defining feature of exponential decay and a common short-answer question.

A short worked example. A bacterial culture starts at $500$500 cells and triples every $4$4 hours. Find the model and the population after $10$10 hours. Triple every $4$4 hours means $P(4) = 1500 = 500 e^{4k}$P(4)=1500=500e4k, so $e^{4k} = 3$e4k=3 and $k = \tfrac{1}{4} \ln 3$k=41​ln3. Then $P(10) = 500 e^{10 \cdot \tfrac{1}{4} \ln 3} = 500 \cdot 3^{10/4} = 500 \cdot 3^{2.5} \approx 7794$P(10)=500e10⋅41​ln3=500⋅310/4=500⋅32.5≈7794 cells.

A common pitfall is mixing up the sign of $k$k for decay versus growth, or forgetting to convert units of time consistently between the data and the question. Another is rounding $k$k early — keep it exact ($\tfrac{1}{4} \ln 3$41​ln3, not $0.2747$0.2747) and round only the final answer.

For worked growth and decay problems with realistic contexts, see the Exponential Growth and Decay lesson.

Logarithmic Graphs

A core Edexcel skill is using logs to linearise an exponential or power relationship — that is, to convert it into a straight-line form so that its parameters can be read from a graph. There are two standard cases.

For an exponential model $y = A b^x$y=Abx, take logs of both sides: $\log y = \log A + x \log b$logy=logA+xlogb. Plotting $\log y$logy against $x$x gives a straight line with gradient $\log b$logb and intercept $\log A$logA. So if a candidate is given a table of $(x, y)$(x,y) data, they should plot $\log y$logy on the vertical axis against $x$x on the horizontal axis, find the gradient and intercept, and recover $A$A and $b$b from $A = 10^{\text{intercept}}$A=10intercept and $b = 10^{\text{gradient}}$b=10gradient (using natural logs gives the same procedure with $e$e in place of $10$10).

For a power model $y = A x^n$y=Axn, take logs of both sides: $\log y = \log A + n \log x$logy=logA+nlogx. Now plotting $\log y$logy against $\log x$logx gives a straight line with gradient $n$n and intercept $\log A$logA. The choice of which variable to take logs of is the discriminator: a model of the form $y = Ab^x$y=Abx needs logs only on $y$y, while a model of the form $y = Ax^n$y=Axn needs logs on both.

A common pitfall is choosing the wrong linearisation — taking logs of both variables when the model is exponential, or only of $y$y when the model is power. Another is reading the gradient as $b$b or $n$n directly without remembering that the gradient is $\log b$logb in the exponential case. A third is mixing up natural and common logs without committing to one — pick one consistently.

For worked linearisation examples and the gradient-intercept recovery technique, see the Logarithmic Graphs lesson.

Change of Base

The change-of-base formula lets you rewrite a logarithm in any base in terms of logarithms in another base:

$\log_a x = \frac{\log_b x}{\log_b a}$loga​x=logb​alogb​x​

The most common use is to evaluate logs in non-standard bases on a calculator — many calculators have buttons for $\log$log (base $10$10) and $\ln$ln (base $e$e) but not for arbitrary bases. So $\log_3 17 = \log 17 / \log 3 \approx 2.579$log3​17=log17/log3≈2.579 via common logs, or equivalently $\ln 17 / \ln 3$ln17/ln3, which gives the same answer.

Change of base also lets you compare logs in different bases, simplify expressions where the base depends on context, and solve equations like $\log_2 x = \log_4(x + 6)$log2​x=log4​(x+6) by converting both to a common base. From the formula, $\log_4(x + 6) = \log_2(x + 6) / \log_2 4 = \tfrac{1}{2} \log_2(x + 6)$log4​(x+6)=log2​(x+6)/log2​4=21​log2​(x+6). The equation becomes $\log_2 x = \tfrac{1}{2} \log_2(x + 6)$log2​x=21​log2​(x+6), which simplifies to $2 \log_2 x = \log_2(x + 6)$2log2​x=log2​(x+6) and then $x^2 = x + 6$x2=x+6, giving $x = 3$x=3 (rejecting $x = -2$x=−2 since the original log requires $x > 0$x>0).

A common pitfall is misremembering which way round the formula goes — putting $\log_b a$logb​a on top instead of $\log_b x$logb​x. A simple sense-check is that $\log_a a = 1$loga​a=1, so substituting $x = a$x=a should give $\log_b a / \log_b a = 1$logb​a/logb​a=1. If your formula does not, you have it the wrong way round.

For worked examples and a clean memory hook for the formula, see the Change of Base lesson.

Modelling with Exponentials

Modelling questions are where exponentials and logarithms earn their spec section. Edexcel routinely sets a 6-10 mark question that gives a real-world context — population, cooling, drug concentration, savings — and asks the candidate to interpret a given model, find parameters from data, predict future values, and comment on the limitations of the model.

A typical structure: the question states $P = a e^{kt}$P=aekt and gives two data points. Use the data to solve for $a$a and $k$k. Use the resulting model to predict a value at a new time, or to find the time at which $P$P reaches a target. Finally, the question often asks "explain one limitation of this model in this context" — and this is where many candidates lose easy marks by leaving the answer blank or writing something vague.

The standard limitations to mention are: exponential models predict unbounded growth, which is unrealistic for any finite system (a population cannot exceed available resources, a savings account is capped by its principal and interest rules); exponential decay never reaches zero, but real quantities (bacteria killed by an antibiotic, drug levels in the bloodstream) eventually do; and the model assumes constant growth or decay rate, which ignores changes in conditions over time.

A short worked example. A cooling cup of coffee is modelled by $T = 20 + 70 e^{-0.05 t}$T=20+70e−0.05t where $T$T is in degrees Celsius and $t$t is in minutes. The room temperature is $20$20 (the asymptote). The initial temperature is $T(0) = 20 + 70 = 90$T(0)=20+70=90. The temperature after $30$30 minutes is $T(30) = 20 + 70 e^{-1.5} \approx 20 + 15.6 = 35.6$T(30)=20+70e−1.5≈20+15.6=35.6. The time to reach $50$50 comes from $50 = 20 + 70 e^{-0.05 t}$50=20+70e−0.05t, so $e^{-0.05 t} = 30/70 = 3/7$e−0.05t=30/70=3/7, giving $t = -\tfrac{1}{0.05} \ln(3/7) \approx 16.9$t=−0.051​ln(3/7)≈16.9 minutes. A reasonable limitation: the model assumes constant room temperature, which is unrealistic if the window is open or the room warms during the day.

A common pitfall is treating modelling questions as algebra-only and not engaging with the context. The interpretation marks are usually 1-2 marks per part, and they add up over a question. Another is forgetting to round sensibly — quoting a population of $7793.6$7793.6 bacteria looks careless, even if the calculation is correct.

For worked modelling questions across several contexts, see the Modelling with Exponentials lesson.

Common Mark-Loss Patterns Across Exponentials and Logarithms

Across the whole exponentials and logarithms section, a small set of habits accounts for a disproportionate share of lost marks. None of these are about content you do not know. They are all about content you do know, applied carelessly.

• Distributing log over a sum. $\log(x + y)$log(x+y) is not $\log x + \log y$logx+logy. The product law applies to products, not sums, and this slip is one of the most common in Edexcel logs questions.
• Forgetting the domain restriction. $\log x$logx requires $x > 0$x>0. After solving a log equation, always check whether each candidate solution gives a positive argument; reject any that do not.
• Sign errors with negative exponents. $x^{-1/2}$x−1/2 is $1/\sqrt{x}$1/x​, not $-\sqrt{x}$−x​. The minus sign moves the term to the denominator and does not change its sign.
• Forgetting the chain rule when differentiating $e^{kx}$ekx. The derivative is $k e^{kx}$kekx, not $e^{kx}$ekx.
• Choosing the wrong linearisation. Exponential models $y = Ab^x$y=Abx need $\log y$logy against $x$x. Power models $y = Ax^n$y=Axn need $\log y$logy against $\log x$logx. Mixing these up costs the whole question.
• Reading the gradient as the parameter directly in a linearised graph. The gradient is $\log b$logb in the exponential case, not $b$b. Always invert with the appropriate exponential to recover the original parameter.
• Rounding too early in growth and decay problems. Keep $k$k in exact form ($\tfrac{1}{4} \ln 3$41​ln3, not $0.2747$0.2747) until the final answer.
• Misremembering the change-of-base formula. $\log_a x = \log_b x / \log_b a$loga​x=logb​x/logb​a. The argument is on top, the base is on the bottom. A quick sense-check is that $\log_a a$loga​a must equal $1$1.
• Leaving the limitation question blank in modelling problems. There are easy marks here for any sensible comment about unbounded growth, irreversible decay, or constant-rate assumptions.
• Not showing enough working. Edexcel mark schemes award method marks generously when the working is clear. A correct final answer with no working can score fewer marks than an incorrect final answer with clean method.

Many candidates lose marks here every series. A revision plan that explicitly drills these habits — not just the content — will move your grade more than another pass through the textbook.

Recommended Six-Week Revision Plan

This plan is designed for a candidate who has covered the exponentials and logarithms content in lessons but wants to revise it cleanly before the exam. It assumes about 5-6 hours per week on this section. Adjust pace if you are starting earlier or later.

Week	Topics	Practice
1	Laws of indices (A-Level form); exponential functions $y = a^x$y=ax and their graphs	30-40 indices manipulation problems; 15 sketching tasks for $y = a^x$y=ax in transformed forms
2	The natural exponential $e^x$ex; logarithms and laws of logarithms	20 short problems on $e^x$ex and its derivative; 25 problems combining and splitting logarithmic expressions using the three laws
3	Solving exponential equations (single, quadratic-in-$a^x$ax, and mixed-base); natural logarithms	20 exponential equations across all three patterns; 15 problems using $\ln$ln and the inverse identities
4	Exponential growth and decay; half-life and proportional rate	15 modelling problems including 5 with half-life; 5 problems involving rate of change at a given time
5	Logarithmic graphs and linearisation; change of base	10 linearisation problems for both exponential and power models; 10 change-of-base manipulations and equations
6	Mixed practice; targeted review of weakest topics; full exponentials-and-logarithms question sets	One full mixed problem set per day; review marking-scheme working for any question scoring below 60%

The point of the plan is to keep moving forward while maintaining contact with earlier topics. Do not spend three weeks on the laws of logarithms and run out of time before modelling. By the end of week 5, every topic in the section should have had focused contact and a practice round. Week 6 is consolidation and weakness-targeting.

A useful discipline through the whole plan is to treat any question you got wrong as a diagnostic. Was it a content gap, a method error, or a careless slip? Logging the cause means your next review session targets the right thing.

How LearningBro's Edexcel A-Level Maths Exponentials and Logarithms Course Helps

LearningBro's Edexcel A-Level Maths: Exponentials and Logarithms course is built around the structure of this guide. Each of the ten lessons covers one section of the 9MA0 specification, in the order Edexcel teaches it, with worked examples, practice questions, and full mark-scheme-style solutions. Lessons end with a short review and quick-recall questions designed for spaced revisits.

The course is designed to be used in two ways. As a first pass, you can work through the lessons in order, building each topic on the last — indices into exponentials, exponentials into the natural exponential, logs into solving equations, and so on. As a revision tool, you can drop into any lesson and work the practice independently — for example, drilling linearisation problems for a week before mocks. The AI tutor is available throughout to give targeted hints when you get stuck, without giving away full solutions, and to mark your written working with structured feedback.

If you want one place to revise this section of the spec well, with realistic practice and clean explanations of every topic, the full course is the right next step. Start with the Edexcel A-Level Maths: Exponentials and Logarithms course.

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