Exponentials and Logarithms

This lesson covers Exponentials and Logarithms as required by the A-Level Mathematics Pure 1 specification. Exponential functions model growth and decay in real-world contexts, while logarithms are the inverse of exponentials. Understanding the laws of logarithms and the natural logarithm ln is essential for solving a wide range of A-Level problems.

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Exponential Functions

An exponential function has the form y = aˣ where a > 0 and a ≠ 1.

Key Properties of y = aˣ

Property	Description
Domain	All real numbers
Range	y > 0 (always positive)
y-intercept	(0, 1) since a⁰ = 1
Asymptote	The x-axis (y = 0) is a horizontal asymptote
Growth/Decay	If a > 1, the function grows; if 0 < a < 1, it decays

The Natural Exponential Function

The number e (Euler's number) is approximately 2.71828. The function y = eˣ is the most important exponential function because its derivative is itself: d/dx(eˣ) = eˣ.

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Logarithms

A logarithm is the inverse of an exponential. If aˣ = b, then logₐ(b) = x.

aˣ = b  ⟺  logₐ(b) = x

Special Logarithms

Notation	Base	Name
log x or log₁₀ x	10	Common logarithm
ln x or logₑ x	e	Natural logarithm

Key Results

Statement	Meaning
logₐ 1 = 0	a⁰ = 1
logₐ a = 1	a¹ = a
ln e = 1	e¹ = e
ln 1 = 0	e⁰ = 1

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Laws of Logarithms

Law	Rule
Product	logₐ(xy) = logₐ(x) + logₐ(y)
Quotient	logₐ(x/y) = logₐ(x) − logₐ(y)
Power	logₐ(xⁿ) = n logₐ(x)
Change of base	logₐ b = logc b / logc a
Inverse	a^(logₐ x) = x and logₐ(aˣ) = x

Example: Simplify log₂(8) + log₂(4).

log₂(8) + log₂(4) = log₂(8 × 4) = log₂(32) = log₂(2⁵) = 5

Example: Write 2 log 3 − log 6 + log 2 as a single logarithm.

= log 9 − log 6 + log 2
= log(9 × 2 / 6)
= log 3

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The Relationship Between eˣ and ln x

The natural logarithm ln x is the inverse of eˣ:

e^(ln x) = x    (for x > 0)
ln(eˣ) = x      (for all x)

The graph of y = ln x is the reflection of y = eˣ in the line y = x.

Property	y = eˣ	y = ln x
Domain	x ∈ ℝ	x > 0
Range	y > 0	y ∈ ℝ
Asymptote	y = 0	x = 0
Key point	(0, 1)	(1, 0)

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Solving Exponential Equations

Method 1: Using Logarithms

Take logarithms of both sides when the variable is in the exponent.

Example: Solve 5ˣ = 13.

x = log 13 / log 5 ≈ 1.594 (3 d.p.)

Method 2: Substitution for Quadratic Exponentials

Example: Solve e²ˣ − 5eˣ + 6 = 0.

Let u = eˣ: u² − 5u + 6 = 0 → (u − 2)(u − 3) = 0.

So eˣ = 2 giving x = ln 2, or eˣ = 3 giving x = ln 3.

Method 3: Equations with Different Bases

Example: Solve 3^(2x+1) = 7ˣ.

(2x + 1) ln 3 = x ln 7
2x ln 3 + ln 3 = x ln 7
x(2 ln 3 − ln 7) = −ln 3
x = −ln 3 / (2 ln 3 − ln 7) ≈ −4.371

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Solving Logarithmic Equations

Example: Solve ln(x + 3) + ln(x − 1) = ln 5.

ln((x + 3)(x − 1)) = ln 5
(x + 3)(x − 1) = 5
x² + 2x − 3 = 5
x² + 2x − 8 = 0
(x + 4)(x − 2) = 0
x = −4 or x = 2

Check: x = −4 gives ln(−1) which is undefined. So x = 2 is the only solution.

Exam Tip: Always check that your solutions give positive arguments for logarithms. Reject any solutions where the argument would be zero or negative.

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Exponential Growth and Decay

Many real-world processes are modelled by: N = N₀eᵏᵗ

Model	Condition	Examples
Growth	k > 0	Population growth, compound interest
Decay	k < 0	Radioactive decay, cooling

Example: A population P = 25000e^(0.03t) where t is years after 2020. When will P reach 40000?

40000 = 25000e^(0.03t)
e^(0.03t) = 1.6
0.03t = ln 1.6
t = ln 1.6 / 0.03 ≈ 15.65 years (during 2035)

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Summary

• Exponential functions y = aˣ always pass through (0, 1) and have the x-axis as an asymptote.
• Logarithms are the inverse of exponentials: logₐ(b) = x ⟺ aˣ = b.
• Know and apply the laws of logarithms (product, quotient, power, change of base).
• ln x and eˣ are inverses: ln(eˣ) = x and e^(ln x) = x.
• Solve exponential equations by taking logarithms; solve logarithmic equations by converting to exponential form.
• Always check solutions of logarithmic equations for valid (positive) arguments.

Exam Tip: When solving exponential or logarithmic equations, clearly show which law of logarithms you are using at each step. Always give exact answers (e.g., x = ln 3) unless told to give a decimal approximation. Remember: e^(ln x) = x and ln(eˣ) = x — these inverse relationships are frequently tested.

A-Level Deep Dive: Exponentials and Logarithms

Spec mapping

AQA 7357 specification, Pure section F — Exponentials and logarithms: "Know and use the function $a^x$ax and its graph, where $a$a is positive; know and use the function $e^x$ex and its graph; know that the gradient of $e^{kx}$ekx is equal to $ke^{kx}$kekx and hence understand why the exponential model is suitable in many applications. Know and use the definition of $\log_a x$loga​x as the inverse of $a^x$ax, where $a$a is positive and $x \geq 0$x≥0; know and use the function $\ln x$lnx and its graph; know and use $\ln x$lnx as the inverse function of $e^x$ex. Understand and use the laws of logarithms; solve equations of the form $a^x = b$ax=b; use logarithmic graphs to estimate parameters in relationships of the form $y = ax^n$y=axn and $y = kb^x$y=kbx, given data for $x$x and $y$y; understand and use exponential growth and decay." Section F is examined predominantly in Paper 1 (Pure), but appears across Paper 2 through differential-equation contexts and in Paper 3 statistics where $e^{-\lambda}$e−λ governs Poisson-style modelling at the boundary of the syllabus. Synoptic links are heavy: differentiation ($\frac{d}{dx} e^x = e^x$dxd​ex=ex, $\frac{d}{dx} \ln x = \frac{1}{x}$dxd​lnx=x1​), integration ($\int \frac{1}{x}\,dx = \ln|x| + C$∫x1​dx=ln∣x∣+C), and the differential equation $\frac{dy}{dt} = ky$dtdy​=ky whose solution $y = Ae^{kt}$y=Aekt underpins every growth/decay model. The AQA formula booklet lists $\frac{d}{dx} e^{kx} = ke^{kx}$dxd​ekx=kekx and $\frac{d}{dx} \ln x = \frac{1}{x}$dxd​lnx=x1​ but does not list the log laws — these must be memorised.

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Worked example with full mark scheme

Question (8 marks): Solve the equation $5e^{2x} - 3e^x - 2 = 0$5e2x−3ex−2=0, giving the exact value of any solution and rejecting any extraneous root with reasoning.

Solution with mark scheme:

Step 1 — recognise the hidden quadratic.

The structure "term in $e^{2x}$e2x + term in $e^x$ex + constant" is the signature of a quadratic in disguise. Note that $e^{2x} = (e^x)^2$e2x=(ex)2.

M1 — recognising the structure and proposing a substitution. Candidates who attempt to take logarithms term-by-term — for instance writing $\ln(5e^{2x}) - \ln(3e^x) = \ln 2$ln(5e2x)−ln(3ex)=ln2 — earn nothing here, because $\ln$ln does not distribute across addition.

Step 2 — substitute $u = e^x$u=ex.

Let $u = e^x$u=ex, so $e^{2x} = u^2$e2x=u2. The equation becomes:

$5u^2 - 3u - 2 = 0$5u2−3u−2=0

M1 — correct substitution and resulting quadratic.

Step 3 — factorise the quadratic.

$5u^2 - 3u - 2 = (5u + 2)(u - 1) = 0$5u2−3u−2=(5u+2)(u−1)=0.

M1 — correct factorisation. Candidates may also use the quadratic formula here; the AQA mark scheme typically accepts either route provided the factorisation or formula is shown. A bare answer with no working forfeits this M1.

A1 — $u = 1$u=1 or $u = -\dfrac{2}{5}$u=−52​.

Step 4 — back-substitute and reject extraneous roots.

For $u = 1$u=1: $e^x = 1 \implies x = \ln 1 = 0$ex=1⟹x=ln1=0.

For $u = -\dfrac{2}{5}$u=−52​: $e^x = -\dfrac{2}{5}$ex=−52​ has no real solution, since $e^x > 0$ex>0 for all real $x$x.

M1 — back-substituting and stating both candidate values of $u$u.

A1 — explicit rejection of the negative $u$u root with the phrase "since $e^x > 0$ex>0" or equivalent. Silently dropping the root costs this mark even though the final numerical answer is the same.

A1 — exact answer $x = 0$x=0.

B1 — communication mark for a fully reasoned solution: stating the substitution, showing the factorisation, and giving a justified rejection. AQA awards this only when all three are present.

Total: 8 marks (M4 A3 B1).

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Specimen question modelled on the AQA 7357 Paper 1 format

Question (6 marks): A scientist models the temperature $T$T degrees Celsius of a cooling liquid at time $t$t minutes after pouring as

$T = 20 + 60e^{-kt}, \quad k > 0$T=20+60e−kt,k>0

(a) State the initial temperature and the long-term temperature predicted by the model. (2)

(b) Given that $T = 35$T=35 when $t = 8$t=8, find the exact value of $k$k. (4)

Mark scheme decomposition by AO:

(a)

• B1 (AO3.3) — initial temperature: substitute $t = 0$t=0 to give $T = 20 + 60 = 80$T=20+60=80 °C.
• B1 (AO3.4) — long-term temperature: as $t \to \infty$t→∞, $e^{-kt} \to 0$e−kt→0, so $T \to 20$T→20 °C.

(b)

• M1 (AO1.1a) — substituting $T = 35$T=35, $t = 8$t=8: $35 = 20 + 60e^{-8k}$35=20+60e−8k.
• M1 (AO1.1b) — rearranging to isolate the exponential: $e^{-8k} = \dfrac{15}{60} = \dfrac{1}{4}$e−8k=6015​=41​.
• M1 (AO1.1b) — taking natural logarithms: $-8k = \ln\left(\dfrac{1}{4}\right) = -\ln 4$−8k=ln(41​)=−ln4.
• A1 (AO2.5) — exact form: $k = \dfrac{\ln 4}{8} = \dfrac{\ln 2}{4}$k=8ln4​=4ln2​.

Total: 6 marks split AO1 = 3, AO2 = 1, AO3 = 2. AQA uses cooling/growth contexts to load AO3 (modelling) marks onto the interpretation of constants. The "long-term" mark depends on understanding what $e^{-kt} \to 0$e−kt→0 means physically — this is precisely the modelling literacy that the spec singles out.

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Synoptic links

1. Differentiation (section G): the chain rule applied to $y = e^{f(x)}$y=ef(x) gives $\frac{dy}{dx} = f'(x) e^{f(x)}$dxdy​=f′(x)ef(x), and to $y = \ln(f(x))$y=ln(f(x)) gives $\frac{dy}{dx} = \frac{f'(x)}{f(x)}$dxdy​=f(x)f′(x)​. Every growth/decay derivative collapses to these two patterns.

2. Integration (section H): $\int e^{kx}\,dx = \frac{1}{k}e^{kx} + C$∫ekxdx=k1​ekx+C and $\int \frac{1}{x}\,dx = \ln|x| + C$∫x1​dx=ln∣x∣+C. The modulus is non-negotiable — examiners deduct A marks for omitting it on indefinite integrals.

3. Differential equations (section M): the equation $\frac{dy}{dt} = ky$dtdy​=ky (population growth, radioactive decay, Newton's cooling about an asymptote) has general solution $y = Ae^{kt}$y=Aekt. This single ODE template appears in every applied modelling question that uses an exponential.

4. Sequences (section D): geometric sequences $u_n = ar^{n-1}$un​=arn−1 become exponential when $n$n is treated as continuous: $u(t) = a r^t = a e^{t \ln r}$u(t)=art=aetlnr. The bridge $r^t = e^{t \ln r}$rt=etlnr is the conversion every modelling question asks you to do without saying so.

5. Trigonometry (section E, harder synoptic): $e^{ix} = \cos x + i \sin x$eix=cosx+isinx is beyond A-Level but is the bridge to Further Maths; even at A-Level the $\ln$ln and exponential identities mirror the structure of $\sinh$sinh and $\cosh$cosh definitions, which is why $\frac{d}{dx} \sinh x = \cosh x$dxd​sinhx=coshx has the same shape as $\frac{d}{dx} e^x = e^x$dxd​ex=ex.

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Mark-scheme literacy

Exponential/log questions on AQA 7357 split AO marks across all three objectives:

AO	Typical share	Earned by
AO1 (knowledge / procedure)	50–60%	Applying log laws, taking $\ln$ln of both sides, executing substitutions $u = e^x$u=ex, manipulating exponential forms
AO2 (reasoning / interpretation)	20–30%	Justifying rejection of extraneous roots, presenting answers in exact form ($\ln 2$ln2 rather than $0.693$0.693), recognising hidden quadratics
AO3 (problem-solving / modelling)	15–25%	Interpreting the meaning of constants in growth/decay models, reading off long-term behaviour, log-linearising data to estimate parameters

Examiner-rewarded phrasing: "since $e^x > 0$ex>0 for all real $x$x, we reject $u = -\dots$u=−…"; "as $t \to \infty$t→∞, $e^{-kt} \to 0$e−kt→0, so $T \to 20$T→20"; "taking natural logarithms of both sides". Phrases that lose marks: writing decimal approximations when "exact" is demanded; omitting the modulus in $\int \frac{1}{x}\,dx$∫x1​dx; treating $\ln(x + y)$ln(x+y) as $\ln x + \ln y$lnx+lny; failing to state the domain restriction $x > 0$x>0 when $\ln x$lnx appears.

A specific AQA pattern: questions in modelling contexts ask "interpret the value of $k$k in the context of the model". This is an AO3 mark that demands a sentence in plain English ("$k$k is the rate constant; large $k$k means rapid cooling") not a re-statement of the algebra.

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Grade-band model answers

3-mark question