Exponential Graphs, the Laws of Logarithms and Solving aˣ = b

Before you read on

Two or three questions on exactly what this lesson teaches. Being wrong here is fine — it's the fastest way to find out what to pay attention to next.

What y = aˣ is, and what a logarithm undoes

$y = a^x$ , for a constant $a$ with $a > 0$ and $a \neq 1$ (spec 5.1), is an : the variable sits in the exponent, not the base — the opposite arrangement to every function met so far in this course. Both restrictions on $a$ matter for a concrete reason, not as fine print. If $a$ were allowed to be negative, $a^x$ would be undefined for infinitely many real $x$ — $(-4)^{0.5}$ asks for a square root of a negative number, which has no real value — so the function would have gaps in its domain rather than being defined everywhere. If $a$ were allowed to equal $1$ , $1^x = 1$ for every $x$ : a horizontal line, not a growth or decay curve, and none of the graph facts below mean anything for it.

For any valid $a$ , $a^0 = 1$ — true whatever $a$ is, so every exponential curve in this family passes through the same point, $(0, 1)$ . And $a^x$ , built from repeated multiplication of a positive number by itself (and its reciprocal, for negative $x$ ), is always strictly positive: it can shrink arbitrarily close to zero but it can never reach zero and it can never go negative. That single fact — $a^x > 0$ for every real $x$ — has a graphical consequence worth stating plainly, because it is easy to forget under exam pressure: the curve $y = a^x$ has a horizontal at $y = 0$ and no x-intercept at all. 'Where does the curve cross the x-axis' is a reflex question carried over from quadratics; for an exponential the honest answer is nowhere.

The direction of the curve depends only on whether $a$ is bigger or smaller than $1$ . For $a > 1$ , each step of $+1$ in $x$ multiplies the output by $a$ — a growing quantity — so the function increases everywhere; for $0 < a < 1$ , each step of $+1$ in $x$ multiplies the output by a number less than $1$ , so the function decreases everywhere. There is no third case: $a > 0$ and $a \neq 1$ partitions every valid base into exactly these two behaviours.

Because $y = a^x$ is strictly increasing (for $a > 1$ ) or strictly decreasing (for $0 < a < 1$ ) across its whole domain, it never takes the same output value twice — so the question 'what power of $a$ produces this particular positive number?' always has exactly one answer, for any positive number asked about. That answer is called $\log_a$ , defined by nothing more than undoing the exponential: $\log_a b = c$ means precisely that $a^c = b$ , for $a > 0$ , $a \neq 1$ , $b > 0$ . ( $b$ has to be positive, because $a^c$ is always positive whatever $c$ is — there is no power of $a$ that produces a negative number or zero, so $\log_a$ of a negative number or zero is a question with no answer, not a hard question.) Every law in the next section, and the whole method for solving $a^x = b$ , is built on nothing more than this one definition, applied to specific equations.

Diagram — y = aˣ for a > 1 and for 0 < a < 1 — two shapes, one shared point

x-axis: x · y-axis: y = aˣ

C1 · a > 1 (e.g. a = 3): Increasing for all x. As x → −∞ the curve flattens towards the x-axis from above but never reaches it (aˣ > 0 always); as x → +∞ it rises without bound. The larger a is, the faster it climbs to the right of x = 0.
C2 · 0 < a < 1 (e.g. a = 1/3): Decreasing for all x — a mirror image of C1 in the y-axis, since (1/3)ˣ = 3⁻ˣ. As x → +∞ it flattens towards the x-axis from above without reaching it; as x → −∞ it rises without bound.

(0, 1) on every curve: aº = 1 whatever a is (a > 0), so every exponential graph in this family passes through the same point — the one feature that never moves as a changes.
Asymptote y = 0 (the x-axis): aˣ is a product of positive factors, so it is positive for every real x. The curve approaches the axis without ever touching or crossing it, whichever direction a points.
No x-intercept, ever: Since aˣ > 0 for all real x, the equation aˣ = 0 has no solution. 'Where does it cross the x-axis' — a reflex question from quadratics — has no answer here; do not go looking for one.

Common error: Drawing the 0 < a < 1 branch as a reflection of the a > 1 branch in the x-axis — a curve that dips below the axis for large x.

Correct: It is a reflection in the y-axis, not the x-axis. The 0 < a < 1 curve stays entirely above the x-axis too, just decreasing instead of increasing. Check with a substitution: (1/3)^(−2) = 9, the same value 3² gives, confirming the two curves are mirror images across x = 0.

The laws of logarithms — five statements, before the mechanism behind them

For $a, x, y > 0$ and $a \neq 1$ , the specification requires five (spec 5.2): $\log_a(xy) \equiv \log_a x + \log_a y$ (the addition law — a log of a PRODUCT becomes a SUM of logs); $\log_a(x/y) \equiv \log_a x - \log_a y$ (the subtraction law — a log of a QUOTIENT becomes a DIFFERENCE of logs); $\log_a(x^k) \equiv k\log_a x$ (the power law — an exponent inside the argument moves out in front as a multiplier); $\log_a(1/x) \equiv -\log_a x$ (a special case of the subtraction law); and $\log_a a = 1$ (true for every valid base, since $a^1 = a$ ).

Read left-to-right, each law COMBINES two logs into one, or removes an exponent from inside a log. Read right-to-left, each law SPLITS a single log apart. Both directions are genuinely used on this paper: combining is the standard way to solve an equation containing several separate log terms — get everything to one log, then remove it — and splitting is the standard way to express one log in terms of others whose values are already given.

Mechanism

Why the laws hold — the laws of indices, read through one definition

Let $\log_a x = p$ and $\log_a y = q$ — by the definition above, this means precisely that $a^p = x$ and $a^q = y$ , nothing more. Every law of logarithms is what happens to that pair of equations when you multiply, divide or raise them to a power, translated back into log notation. Multiply the two equations: $x \times y = a^p \times a^q$ . The laws of indices already say $a^p \times a^q = a^{p+q}$ — indices add when you multiply powers of the same base — so $xy = a^{p+q}$ . Taking $\log_a$ of both sides (asking 'what power of $a$ gives this?') recovers $p + q$ exactly, because $a^{p+q}$ passed back through $\log_a$ just returns its own exponent: $\log_a(xy) = p + q = \log_a x + \log_a y$ . The subtraction law is the same argument run on division: $x/y = a^p/a^q = a^{p-q}$ by the quotient rule for indices, so $\log_a(x/y) = p - q = \log_a x - \log_a y$ . The power law comes from raising the first equation to the power $k$ : $x^k = (a^p)^k = a^{pk}$ by the laws of indices (a power raised to a power multiplies the exponents), so $\log_a(x^k) = pk = k\log_a x$ — the exponent $k$ moves from inside the log's argument to a multiplier in front, which is the entire content of the law, and the reason $(\log_a y)^2$ is a genuinely different expression from $\log_a(y^2)$ : squaring the NUMBER $\log_a y$ is not the same operation as squaring the ARGUMENT $y$ before taking its log. The remaining two laws are special cases already covered: $\log_a(1/x) = \log_a 1 - \log_a x$ by the subtraction law, and $\log_a 1 = 0$ always (since $a^0 = 1$ for any valid base), so $\log_a(1/x) = -\log_a x$ ; and $\log_a a = 1$ is immediate from the definition itself, since $a^1 = a$ . None of these five statements is a separate fact to memorise — they are the laws of indices, read backwards through the one definition that connects a power to a log.

Mechanism

Where the change-of-base formula comes from, and why it is the only log law in the booklet

Suppose you want $\log_a x$ but your calculator only gives logs in one particular base, $b$ (almost always base 10 or base $e$ , written $\ln$ ). Start from the definition: let $c = \log_a x$ , so $a^c = x$ . Take $\log_b$ of both sides — a legal move, since applying the same function to two equal quantities keeps them equal: $\log_b(a^c) = \log_b x$ . The left-hand side is exactly the power law derived above, applied with base $b$ instead of $a$ : $\log_b(a^c) = c\log_b a$ . So $c\log_b a = \log_b x$ , and dividing through by $\log_b a$ (never zero, since $a \neq 1$ ) gives $c = \dfrac{\log_b x}{\log_b a}$ . Substituting back $c = \log_a x$ : $\log_a x = \dfrac{\log_b x}{\log_b a}$ — the , spec 5.3's named permitted method ('students may use the change of base formula'), and confirmed as the ONLY logarithm law actually printed on the P2 formula-booklet page: the addition, subtraction and power laws derived above all have to be recalled from memory, but this one identity is given, because it is the identity that turns 'find $\log_a x$ ' into an instruction a calculator with only a $\log$ or $\ln$ button can actually carry out. It is nothing structurally new either — it is the power law again, run through the definition of a logarithm a second time.

Worked, in full

Solving 5 × 3^(2x−1) = 60 — including the rearrangement step scripts most often leave unfinished

01
Isolate the exponential term algebraically, before touching a logarithm at all: divide both sides by 5. $5 \times 3^{2x-1} = 60 \Rightarrow 3^{2x-1} = 12$ .
Earns: M1 — rearranges into the form $a^{(\ldots)} = b$ before taking logs. This is pure algebra, no log law involved yet, which is exactly why it is the mark most safely banked.
02
Take logs of both sides and apply the power law to bring the exponent down: $\log(3^{2x-1}) = \log 12 \Rightarrow (2x-1)\log 3 = \log 12$ .
Earns: dM1 — takes logs and correctly applies $\log_a(x^k) = k\log_a x$ to bring the exponent out. Any consistent base is accepted — a real, checked instance of the mark scheme's 'oe' (or equivalent) convention in use on a log-law step: 'States or uses $\log_2 4x^2 = 2\log_2 2x$ (oe)' (Jan 2020).
03
Solve for the bracket: $2x - 1 = \dfrac{\log 12}{\log 3} = \log_3 12$ .
Earns: A1 — the correct exact expression for $2x - 1$ .
04
Finish the rearrangement — add 1, then divide by 2: $x = \dfrac{\log_3 12 + 1}{2}$ . The $+1$ combines with the log first: $1 = \log_3 3$ (since $3^1 = 3$ ), so $\log_3 12 + 1 = \log_3 12 + \log_3 3 = \log_3(12 \times 3) = \log_3 36$ by the addition law. Dividing by 2 is then the power law read in reverse — halving an exponent is the same operation as taking a square root — so $\dfrac{\log_3 36}{2} = \log_3\left(36^{1/2}\right) = \log_3 6$ . That exact form equals $1.63$ to 3 s.f.
Earns: A1 — the final rearrangement completed and the value given in the form asked for. Finishing the algebra that isolates $x$ (this stage) and combining an exact expression into a single required log form are two distinct skills — the second is its own trap, with verified real-paper evidence, in the trap taxonomy below ('single-log-form-left-uncombined').
05
Sanity check by substitution: $5 \times 3^{2(1.6309)-1} = 5 \times 3^{2.2619} \approx 60.0$ ✓.
Earns: Nothing — the mark scheme has run out by this point. It is on the list anyway because it costs one line and catches a dropped sign or a mis-simplified exact form before the paper is handed in.

Source — Examiner report, Jan 2021

"it is not always advisable to take logs first when the equation is presented in index form"

Complete it yourself

Complete the chain — solve 4 × 2^(3x+1) = 100, giving x to 3 s.f.

01
Isolate the exponential term first: divide both sides by 4. $4 \times 2^{3x+1} = 100 \Rightarrow 2^{3x+1} = 25$ .
02
Take logs of both sides and bring the exponent down using the power law: $(3x+1)\log 2 = \log 25$ , so $3x + 1 = \log_2 25$ .

Same question, every valid method

Solve $6 \times 4^{x+1} = 50$ , giving your answer as an exact logarithm and to 3 significant figures. (VERIDIAN-original question, not a reproduction of any past-paper question; the per-line mark allocations are modelled on verified mark-scheme conventions rather than transcribed from a real scheme.)

2 valid methods · every one reaches $x = \log_4\dfrac{25}{3} - \log_4 4 = \log_4\dfrac{25}{12} \approx 0.529$ (3 s.f.) · 3 marks available

01
Divide both sides by 6: $4^{x+1} = \dfrac{50}{6} = \dfrac{25}{3}$ .
Method mark for rearranging algebraically into the form $a^{(\ldots)} = b$ before any logarithm is taken. Pure rearrangement, no log law involved — the mark most safely banked, and the one real scripts show the highest early completion rate on: 'A fall off from nearly 90% to just over 70% between the first two marks' (Jan 2021, Q3(i)).
M1
02
Take logs and apply the power law: $(x+1)\log 4 = \log\dfrac{25}{3}$ , so $x + 1 = \log_4\dfrac{25}{3}$ .
Method mark, dependent on the rearrangement above, for taking logs and correctly bringing the exponent down. Any consistent base is accepted, per the mark scheme's oe (or equivalent) convention.
dM1
03
$x = \log_4\dfrac{25}{3} - 1$ . Combine into the single exact logarithm the question asks for: $1 = \log_4 4$ , so $x = \log_4\dfrac{25}{3} - \log_4 4 = \log_4\dfrac{25}{12} \approx 0.529$ (3 s.f.).
Accuracy mark for finishing the rearrangement AND combining it into the single exact-logarithm form the question specifically asks for ('giving your answer as an exact logarithm') — the same M1/A1/A1 distinction the real anchor draws: reaching an unsimplified exact expression such as $\log_7 3 - 2$ earns the intermediate A1 in full, but only the further step of combining it into a single log, here $\log_4\frac{25}{12}$ , earns the final one (Jan 2021, Q3(i)).
A1

The default whenever a constant multiplies or divides the exponential term — which is most equations of this shape. Isolating first means the left-hand side is never a log of a product that has to be expanded, so there is nothing to get wrong except the rearrangement itself, and the rearrangement is checkable in your head. Examiners record this exact ordering as the safer one: 'it is not always advisable to take logs first when the equation is presented in index form' (Jan 2021, Q3(i)).

Marked, line by line

$\log_a 2 = p$ and $\log_a 5 = q$ . (a) Express $\log_a 20$ in terms of $p$ and $q$ . (2) (b) Solve $7^{2x+1} = 90$ , giving your answer to 3 significant figures. (3) (c) Solve $\log_2(x+3) + \log_2(x-3) = 4$ . (4) — VERIDIAN-original question, structured to pair a log-law manipulation part with a solving-aˣ=b part and a combined-log equation whose solution has to be checked against the original (spec 5.2 and 5.3 in one question, mirroring the pilot's practice of testing several sub-items together). Not a reproduction of any past-paper question, and the per-line mark allocations below are modelled on verified mark-scheme conventions rather than copied from a real scheme.

9 marks available

(a) — 2 marks

01
$20 = 2^2 \times 5$ , so $\log_a 20 = \log_a(2^2) + \log_a 5 = 2\log_a 2 + \log_a 5$ .
Method mark for expressing 20 as a product built from the two given values (2 and 5) and applying the power and addition laws to it. Any equivalent split that reaches the same two given logs is credited equally — the mark is for the structure, not the specific route to it.
M1
02
$= 2p + q$ .
Accuracy mark, correct answer only, in terms of $p$ and $q$ as the question specifies. A numeric value computed for a particular $a$ is not what is asked for and would not earn this mark even if arithmetically correct.
A1

(b) — 3 marks

101
Take logs of both sides: $(2x+1)\log 7 = \log 90$ .
Method mark for taking logs of both sides and applying the power law to bring the exponent down. The equation is already in clean index form, $7^{(\cdot)} = 90$ , so there is no product to rearrange first, unlike part (c) below.
M1
102
$2x + 1 = \log_7 90$ , so $x = \dfrac{\log_7 90 - 1}{2}$ .
Accuracy mark for any correct expression for $x$ , reached by dividing by $\log 7$ and completing the rearrangement (subtracting 1, then dividing by 2). The real anchor's Alt(i) credits this identical completed rearrangement within a single A1 rather than a separate dependent method mark (Jan 2021, Q3(i)), and the lesson's own worked chain above credits the equivalent finishing-the-rearrangement step as A1, not dM1.
A1
103
$x = 0.656$ (3 s.f.).
Accuracy mark, correct answer only. $\log_7 90 = 2.3124...$ , so $x = (2.3124 - 1)/2 = 0.6562...$ , awrt 0.656. The exact-answer rule from the general marking guidance does not bind here, because the question specifies 3 s.f. — the one place in this lesson where rounding to the requested precision is exactly what is wanted, not an error.
A1

(c) — 4 marks

201
Combine the two logs using the addition law: $\log_2[(x+3)(x-3)] = 4$ .
Method mark for correctly combining two logs of the same base into a single log of a product — the step where the subtraction-as-division misconception most often replaces the correct addition-as-multiplication law. Both terms here are ADDED, so a PRODUCT form has to appear, not a quotient.
M1
202
$(x+3)(x-3) = 2^4 = 16$ , so $x^2 - 9 = 16$ , giving $x^2 = 25$ .
Accuracy mark for removing the logarithm correctly and reaching the correct equation with no logarithm left in it. Real WMA12 mark schemes award this as its own separate line every time a combined-log equation is turned into a polynomial — confirmed across four independent series testing this exact question shape, all with the equation-forming step marked distinctly from both the combining step before it and the solving step after it: 'Correct equation (not involving logs)' (Jun 2024, Q3(i)); 'Correct equation not involving logs' (Jun 2025, Q6(i)); ' $2x^2+x-6=0$ or equivalent 3TQ' (Jun 2023, Q5); ' $2y^2+9y-5=0$ oe 3TQ' (Jan 2021, Q3(ii)) — this lesson's own worked chain and chain-drill already model this same principle correctly for a^x=b equations (splitting 'reach the expression' from 'finish it' into two separate accuracy marks); this line brings the marked solution's log-to-quadratic part into line with it.
A1
203
$x^2 = 25 \Rightarrow x = \pm 5$ .
Method mark for attempting to solve the resulting equation, dependent on the M1 for combining the logs above — real schemes attach this dependency to the earlier method mark, not to the equation-forming accuracy mark immediately before it.
dM1
204
$x = -5$ is rejected: the original equation requires $x + 3 > 0$ and $x - 3 > 0$ for both logarithms to be defined, i.e. $x > 3$ . So $x = 5$ .
Accuracy mark, correct answer only, and it is not earned by the algebra alone. A logarithm is only defined for a positive argument, so any value of $x$ that makes a bracket zero or negative has to be checked against the ORIGINAL equation and discarded if it fails — the same discipline as checking a discriminant question's non-zero condition separately from the inequality itself.
A1

In your own words

In one sentence: why does solving $\log_2(x+3) + \log_2(x-3) = 4$ produce a value of $x$ that has to be checked against the original equation, when solving $\log_2(x+3) = 4$ alone would not?

Named traps

log-subtraction-rewritten-as-division: Confirmed directly on a real WMA12 log-manipulation question: "A fairly common mistake was to rewrite log₃(5x+7) − log₃4x as log₃(5x+7)/log₃4x. Some then crossed out log₃ in the numerator and denominator as though simplifying a fraction" (Jan 2023, Q4(i)). The subtraction law says $\log_a x - \log_a y = \log_a(x/y)$ — the two ARGUMENTS are divided, the two log SYMBOLS never are. Writing $\log_3(5x+7)/\log_34x$ treats "log₃" as a common factor that can be cancelled, which it is not: $\log_a$ is a function name, not a multiplier.
log-of-a-sum-expanded-as-sum-of-logs: Confirmed on the same question: "Another mistake, common for this type of question, was to expand the brackets within the log function, writing log₃(5x+7) as log₃5x + log₃7" (Jan 2023, Q4(i)). No law distributes a log across a sum or difference inside its own argument — the addition law only COMBINES two separate logs into one, it never SPLITS a single log open. $\log_a(x+y)$ has no simpler form; treating it like $\log_a(xy)$ applies the multiplication law to an expression that is added, not multiplied.
log-squared-confused-with-log-of-the-square: Confirmed on a harder question, described as "one of the most challenging parts on the paper": "understanding the difference between logₐ(y²) and (logₐy)²... rewriting (logₐy)² as 2logₐy was a very common mistake" (Jan 2023, Q4(ii)). The power law moves an exponent sitting INSIDE the log — it says nothing about an exponent sitting OUTSIDE the whole expression. $(\log_a y)^2$ means "square the number $\log_a y$ produces," and no law turns that squaring into a multiplication by 2 — the two expressions are built from the same symbols in a different order and are not generally equal.
exponent-rearrangement-left-unfinished: Confirmed on a solving-aˣ=b question with a measured mark-by-mark drop-off: "A fall off from nearly 90% to just over 70% between the first two marks is revealing" (Jan 2021, Q3(i)) — the gap between the M1 for setting up $(x+2)\log 7 = \log 3$ and the first A1 for reaching ANY correct exact expression for $x$ . Taking logs correctly is not the same as finishing the algebra — once the exponent has been isolated as an expression like $2x-1$ or $x+1$ , every remaining operation (adding, subtracting, dividing) still has to be carried out on both sides before an exact expression for $x$ exists at all. (A separate, later-stage error — reaching a correct exact expression for $x$ but leaving it uncombined into a single required log form — is a distinct trap: see "single-log-form-left-uncombined" below.)
single-log-form-left-uncombined: Confirmed on the same question, describing the steeper of its two mark-by-mark drops — "with only 50% scoring all three. For example, [many] achieved x = log₇3 − 2 but then did not reach the required form, failing to deal with the −2" (Jan 2021, Q3(i)). This is NOT an unfinished-rearrangement error: $x = \log_7 3 - 2$ is already a fully isolated, fully-credited exact expression for $x$ — it earns the A1 the trap above describes. The mark actually lost here is a further, distinct accuracy mark for combining that expression into the single logarithm the question specifically demanded ("giving your answer in the form $x = \log_7 a$ , $a$ rational"): $\log_7 3 - 2 = \log_7 3 - \log_7 49 = \log_7\dfrac{3}{49}$ , using $\log_7 49 = 2$ since $7^2 = 49$ — i.e. rewriting the constant $2$ via $\log_a a = 1$ , here $2 = 2\log_7 7 = \log_7(7^2) = \log_7 49$ . Whenever a question specifies the answer form this precisely, a correct but unsimplified expression like $\log_a b - c$ earns only the intermediate accuracy mark, never the final one.
logs-taken-before-the-equation-is-rearranged: A direct piece of technique advice from the same report, given as general guidance rather than a specific error: "Although the question asked for an answer in log form, it is not always advisable to take logs first when the equation is presented in index form" (Jan 2021, Q3(i)). Where a constant multiplies or divides the exponential term, isolating it algebraically first — before any logarithm is taken — removes an entire log law (usually the addition or subtraction law) from the working. Taking logs immediately is not wrong, but it adds a step, and every added step is an added chance for one of the other traps on this list to strike.

Retrieval — with feedback on every choice

Question 1

2 marks

$\log_4(x+9) - \log_4 x$ can be written as a single logarithm. Which is correct?

Question 2

2 marks

Which of these is a correct simplification of $\log_5(x+2)$ ?

Question 3

2 marks

Given that $\log_a y = 3$ , what is $(\log_a y)^2$ ?

Question 4

2 marks

A candidate solving $6^{3x+2} = 50$ correctly reaches $3x + 2 = \log_6 50$ . Which is the correctly finished answer for $x$ ?

Question 5

2 marks

A candidate solving $9 \times 2^{x-1} = 200$ takes logs of both sides immediately, without dividing by 9 first. Which statement about this approach is correct?

Question 6

1 mark

For which of these is the graph of $y = a^x$ decreasing for every real $x$ ?

Reference — not a study method, a lookup

logₐb = c ⇔ aᶜ = b (a>0, a≠1, b>0). y=aˣ passes through (0,1), asymptote y=0, never negative.
logₐ(xy)=logₐx+logₐy; logₐ(x/y)=logₐx−logₐy; logₐ(xᵏ)=k·logₐx; logₐ(1/x)=−logₐx; logₐa=1.
Change of base (in the booklet): logₐx = log_b x / log_b a. The other four laws are not — memorise them.
Solving aˣ=b: isolate the exponential first, then take logs — finish the rearrangement, not just the bracket.
If a question names the required form (e.g. x = logₐk), an unsimplified logₐb − c earns only the intermediate mark — combine it into the single required log to earn the final one (Jan 2021, Q3(i)).
A log of a sum does not split apart. Subtracting logs never divides the logs themselves, only their arguments.
A combined-log equation can gain a root the original excludes — check each argument is positive before accepting it.

Not affiliated with or endorsed by Pearson Edexcel. Every quotation and figure attributed to a mark scheme or examiner report in this lesson was independently verified against the primary Pearson document, not carried over from prior course material.

Question 12 marks

$\log_4(x+9) - \log_4 x$ can be written as a single logarithm. Which is correct?

$\log_4\left(\dfrac{x+9}{x}\right)$
Correct. The subtraction law, $\log_a x - \log_a y \equiv \log_a(x/y)$ , divides the two ARGUMENTS, giving a single log of a quotient.
B $\dfrac{\log_4(x+9)}{\log_4 x}$
This divides the two LOGS themselves rather than their arguments — the exact error real scripts show on this manipulation: 'a fairly common mistake was to rewrite log₃(5x+7) − log₃4x as log₃(5x+7)/log₃4x' (Jan 2023, Q4(i)). $\log_4$ is a function name, not a factor that can sit in a fraction.
C $\log_4 9$
This subtracts the ARGUMENTS directly, $(x+9) - x = 9$ , as if the log applied to the difference of the two expressions rather than to their quotient. There is no law that subtracts inside a single log's argument this way.
DIt cannot be written as a single logarithm without knowing the value of $x$
The subtraction law applies to any two logs of the same base, symbolically, with no numeric value of $x$ required — it is an identity, true for every valid $x$ simultaneously, not a numerical shortcut that needs $x$ plugged in first.

Traps tested: Log subtraction rewritten as division · Arguments subtracted directly instead of divided · Log laws believed to require a numeric value

Question 22 marks

Which of these is a correct simplification of $\log_5(x+2)$ ?

It cannot be simplified further without more information about $x$
Correct. There is no law of logarithms that splits a log of a SUM apart — the addition law only combines two separate logs into one, it never expands a single log the other way. $\log_5(x+2)$ is already in its simplest form.
B $\log_5 x + \log_5 2$
This distributes the log across the sum inside the bracket, as if $\log_5(x+2)$ behaved like $\log_5(x \times 2)$ . Real scripts record exactly this substitution: 'writing log₃(5x+7) as log₃5x + log₃7' (Jan 2023, Q4(i)). The addition law works in the other direction — two SEPARATE logs combine into one — it does not split a single log's SUM apart.
C $\log_5 x + 2$
This drops the log from the ' $+2$ ' entirely, as though only part of the argument were inside $\log_5$ . Whatever operation is applied to $x+2$ , it applies to the whole bracket — a log cannot selectively apply to one term of a sum and not the other.
D $\log_5(2x)$
This misreads the addition $x + 2$ as the multiplication $x \times 2$ inside the bracket — a different expression entirely, and one that would only be reached by starting from $\log_5 x + \log_5 2$ and (correctly) reversing the addition law on the wrong pair of terms.

Traps tested: Log of a sum expanded as sum of logs · Constant inside log argument treated as outside · Sum inside log misread as product

Question 32 marks

Given that $\log_a y = 3$ , what is $(\log_a y)^2$ ?

$9$
Correct. $(\log_a y)^2$ means square the NUMBER $\log_a y$ produces: $3^2 = 9$ .
B $6$
This doubles instead of squaring — treating $(\log_a y)^2$ as if it meant $2\log_a y$ . Real scripts record this exact confusion: 'rewriting $(\log_a y)^2$ as $2\log_a y$ was a very common mistake' (Jan 2023, Q4(ii)). The power law moves an exponent from INSIDE a log's argument to a multiplier out front; it says nothing about an exponent applied to the log's own output.
C $\log_a 9$
This applies the power law backwards to the wrong quantity — turning the squaring of the LOG VALUE into a log of a squared number. $\log_a(y^2)$ would equal $2\log_a y = 6$ (a different expression from the one asked for); neither of those equals $(\log_a y)^2$ .
DCannot be determined without knowing $a$ and $y$ individually
$\log_a y$ is given directly as $3$ — the question does not require unpacking $a$ and $y$ separately, because $(\log_a y)^2$ is just that one given number, squared. Overclaiming missing information here loses a mark available for free.

Traps tested: Log squared confused with log of the square · Power law misapplied to the log value itself · Overclaims uncertainty

Question 42 marks

A candidate solving $6^{3x+2} = 50$ correctly reaches $3x + 2 = \log_6 50$ . Which is the correctly finished answer for $x$ ?

$x = \dfrac{\log_6 50 - 2}{3}$
Correct. Both remaining operations are carried out on both sides in the right order: subtract 2 from both sides of $3x + 2 = \log_6 50$ , then divide the whole thing by 3.
B $x = \log_6 50 - 2$
The subtraction is done but the division by 3 never happens — the exact shape of error real scripts record on this question type: reaching a correct intermediate expression and stopping before the rearrangement is complete.
C $x = \dfrac{\log_6 50}{3} - 2$
This divides only the $\log_6 50$ term by 3 and subtracts the 2 afterwards, rather than dividing the ENTIRE left-hand side (both the 3 $x$ and the $+2$ ) by 3 before isolating $x$ . The 2 has to come off before the division, not after it.
D $x = \dfrac{\log_6 50 + 2}{3}$
The 2 is added instead of subtracted. $3x + 2 = \log_6 50$ rearranges to $3x = \log_6 50 - 2$ — moving a term across the equals sign flips its sign, and this response has not flipped it.

Traps tested: Exponent rearrangement left unfinished · Order of operations error in final rearrangement · Sign error in final rearrangement

Question 52 marks

A candidate solving $9 \times 2^{x-1} = 200$ takes logs of both sides immediately, without dividing by 9 first. Which statement about this approach is correct?

It is still valid — it just requires an extra step splitting $\log(9 \times 2^{x-1})$ into $\log 9 + (x-1)\log 2$ before reaching the same equation the divide-first method reaches in one line
Correct. Both orderings are mathematically valid and both are creditable in full; the mark scheme does not require a specific one. Taking logs first is simply less efficient here, exactly as examiners record: 'it is not always advisable to take logs first when the equation is presented in index form' (Jan 2021, Q3(i)) — 'not always advisable' is a statement about efficiency, not validity.
BIt is invalid — logs cannot be taken until the equation is in the form $a^x = b$
Logs can legally be taken of both sides of ANY true equation, in any form — it is a function applied equally to two equal quantities. Isolating the exponential first is a matter of efficiency and safety, not a precondition for taking logs at all.
CIt is invalid — the 9 must be square-rooted, not divided, before logs can be taken
There is no operation here that calls for a square root; 9 is a constant multiplying the exponential term, and the operation that removes it is division, not a square root of any kind.
DIt is the required method — dividing by 9 first would lose the constant from the final answer
The 9 is not lost by dividing early — it reappears inside the log, as $\log 9$ , whichever order the working takes. Both methods account for the 9 in full; they just account for it at different points in the working.

Traps tested: Believes logs require pre isolated form · Irrelevant operation applied to the constant · Believes isolating the constant loses information

Question 61 mark

For which of these is the graph of $y = a^x$ decreasing for every real $x$ ?

$y = 0.4^x$
Correct. $0 < 0.4 < 1$ , so each step of $+1$ in $x$ multiplies the output by a number less than 1: the curve decreases everywhere.
B $y = 1.4^x$
$a = 1.4 > 1$ , which makes this curve increasing everywhere, not decreasing — the two cases are opposite behaviours either side of $a = 1$ , and this value sits on the increasing side.
C $y = (-0.4)^x$
A negative base is excluded by the spec's own condition on $a$ ( $a > 0$ ), and for a concrete reason: $(-0.4)^{0.5}$ has no real value, so this is not even a function defined for every real $x$ , let alone one with a clear increasing/decreasing direction.
D $y = 1^x$
$a = 1$ is excluded by the spec's own condition ( $a \neq 1$ ): $1^x = 1$ for every $x$ , a horizontal line that is neither increasing nor decreasing.

Traps tested: Increasing and decreasing cases confused · Condition a greater than zero not checked · Condition a not equal one not checked

Practice this for real

This site teaches the mechanism; the exam is sat on Pearson's own real questions. Go find and attempt these yourself — nothing here substitutes for actually sitting a timed paper.

Examiner report: Jan 2021 · Q3(i) — cited directly in this lesson

Pearson's official past-papers portal

Select International Advanced Level → Mathematics → any series, then look for WMA12.

Exponential Graphs, the Laws of Logarithms and Solving aˣ = b

What y = aˣ is, and what a logarithm undoes

The laws of logarithms — five statements, before the mechanism behind them

Why the laws hold — the laws of indices, read through one definition

Where the change-of-base formula comes from, and why it is the only log law in the booklet

Solving 5 × 3^(2x−1) = 60 — including the rearrangement step scripts most often leave unfinished

Complete the chain — solve 4 × 2^(3x+1) = 100, giving x to 3 s.f.

Deep dive

Practice this for real