Natural Logarithms ln(x)

This lesson covers the natural logarithm function, written ln(x), which is the logarithm with base e. Since eˣ is the most important exponential function in mathematics, its inverse ln(x) is the most important logarithm. You need to be confident with ln for differentiation, integration, and solving equations at A-Level.

Definition

The natural logarithm ln(x) is defined as:

ln(x) = log_e(x)

In words: ln(x) is the power to which you must raise e to get x.

The key relationship is:

eˡⁿ⁽ˣ⁾ = x and ln(eˣ) = x

These express the fact that eˣ and ln(x) are inverse functions of each other.

The Graph of y = ln(x)

Since y = ln(x) is the inverse of y = eˣ, its graph is the reflection of y = eˣ in the line y = x.

Property	Value
Domain	x > 0 (you can only take the log of a positive number)
Range	All real numbers
x-intercept	(1, 0) — because ln(1) = 0, since e⁰ = 1
Asymptote	x = 0 (the y-axis) is a vertical asymptote
Shape	Increasing, concave down, grows slowly
Key point	(e, 1) — because ln(e) = 1

The graph passes through (1, 0) and (e, 1), rises slowly for large x, and goes to negative infinity as x approaches 0 from the right.

Key Values of ln

x	ln(x)
1	0
e ≈ 2.718	1
e² ≈ 7.389	2
e³ ≈ 20.086	3
1/e ≈ 0.368	-1
√e ≈ 1.649	1/2

Laws of Natural Logarithms

The same three logarithm laws apply to ln:

Law	Statement
Addition	ln(a) + ln(b) = ln(ab)
Subtraction	ln(a) - ln(b) = ln(a/b)
Power	ln(aⁿ) = n × ln(a)

Plus the special identities:

ln(1) = 0
ln(e) = 1
eˡⁿ⁽ˣ⁾ = x
ln(eˣ) = x

Solving Equations with ln

Type 1: eˣ = k

Take ln of both sides: x = ln(k)

Example: Solve eˣ = 10

x = ln(10) ≈ 2.303

Type 2: eᵃˣ⁺ᵇ = k

Take ln of both sides, then solve for x.

Example: Solve e^(3x-1) = 7

3x - 1 = ln(7)
3x = ln(7) + 1
x = (ln(7) + 1)/3 ≈ (1.946 + 1)/3 ≈ 0.982

Type 3: ln(x) = k

Exponentiate both sides: x = eᵏ

Example: Solve ln(x) = 4

x = e⁴ ≈ 54.60

Type 4: ln(ax + b) = k

Example: Solve ln(2x - 3) = 1

2x - 3 = e¹ = e
2x = e + 3
x = (e + 3)/2 ≈ 2.859

Combining ln with Other Techniques

Example: Solve e²ˣ + 3eˣ = 10

Let y = eˣ:

y² + 3y - 10 = 0
(y + 5)(y - 2) = 0
y = -5 (rejected, since eˣ > 0) or y = 2
eˣ = 2
x = ln(2) ≈ 0.693

The Relationship Between ln and log₁₀

Both ln and log₁₀ are logarithms, but with different bases:

ln uses base e ≈ 2.718
log₁₀ uses base 10

They are related by: ln(x) = log₁₀(x) / log₁₀(e) ≈ log₁₀(x) / 0.4343

Or equivalently: ln(x) ≈ 2.303 × log₁₀(x)

On your calculator, ln is the natural logarithm button and log is the common (base 10) logarithm button.

Why ln Is Preferred in Calculus

The natural logarithm has the simplest derivative of any logarithm:

d/dx [ln(x)] = 1/x

For any other base: d/dx [log_a(x)] = 1/(x × ln(a))

This simplicity is why ln is the standard logarithm in university-level mathematics, physics, and engineering.

Summary

ln(x) = log_e(x) is the inverse of eˣ.
The graph passes through (1, 0) and has the y-axis as a vertical asymptote.
The same three log laws apply: addition, subtraction, and power.
To solve eˣ = k, take ln: x = ln(k). To solve ln(x) = k, exponentiate: x = eᵏ.
ln is preferred in calculus because d/dx [ln(x)] = 1/x.

A-Level Deep Dive: Natural Logarithms ln(x)

Spec mapping

Edexcel 9MA0 specification section 6 — Exponentials and logarithms, natural logarithm sub-strand covers the function $\ln x$ and its graph; understand $\ln x$ as the inverse function of $e^x$ (refer to the official specification document for exact wording). The natural logarithm is the logarithm to base $e$ , written $\ln x = \log_e x$ . It satisfies $\ln(e^x) = x$ for all real $x$ and $e^{\ln x} = x$ for all $x > 0$ . Although introduced in section 6, $\ln x$ is examined as a synoptic backbone across section 7 (Differentiation, where $\frac{d}{dx}\ln x = \frac{1}{x}$ is a non-negotiable standard result), section 8 (Integration, where $\int \frac{1}{x}\,dx = \ln|x| + C$ closes a glaring gap in the power-rule integration table), section 10 (Differential equations, where separation of variables routinely produces $\ln$ -form solutions) and section 4 (Sequences and modelling, where exponential growth/decay models require $\ln$ to invert and solve for time). The Edexcel formula booklet lists $\frac{d}{dx}\ln x = \frac{1}{x}$ and $\int \frac{1}{x}\,dx = \ln|x| + C$ , but does not list the laws of logarithms — these must be memorised and applied confidently.

Worked example with full mark scheme

Question (8 marks):

A population of bacteria is modelled by $P(t) = 200 e^{0.45 t}$ , where $P$ is the population at time $t$ hours after the start of an experiment.

(a) Find, to 3 significant figures, the time at which the population first reaches 5000. (4)

(b) Show that the time $t$ at which the population is exactly $k$ times its initial value is given by $t = \dfrac{\ln k}{0.45}$ , and hence find the doubling time of the population to 3 significant figures. (4)

Solution with mark scheme:

(a) Step 1 — set up the equation.

$200 e^{0.45 t} = 5000$

M1 — correct equation formed from the model and the target population. The most common slip is omitting the factor $200$ and writing $e^{0.45 t} = 5000$ , which loses M1 immediately.

Step 2 — isolate the exponential.

$e^{0.45 t} = \dfrac{5000}{200} = 25$

M1 — division by 200 to isolate $e^{0.45 t}$ . This is the standard "divide first, then $\ln$ " sequence.

Step 3 — apply natural logarithm to both sides.

$0.45 t = \ln 25$

M1 — taking $\ln$ of both sides and using $\ln(e^{0.45 t}) = 0.45 t$ . Candidates who write $0.45 t = \log 25$ (using $\log_{10}$ with no further conversion) lose this mark unless the base is corrected later.

Step 4 — solve and round.

$t = \dfrac{\ln 25}{0.45} = \dfrac{3.2189...}{0.45} = 7.153...$

So $t = 7.15$ hours (3 s.f.).

A1 — answer correct to 3 significant figures. Writing $7.2$ (2 s.f.) or $7.1531$ (4 s.f.) loses the accuracy mark; the question demanded 3 s.f.

(b) Step 1 — generalise.

The initial population is $P(0) = 200$ . Setting $P(t) = 200 k$ :

$200 e^{0.45 t} = 200 k$

M1 — interpreting " $k$ times its initial value" correctly. A common error is writing $P(t) = k$ and treating $k$ as the absolute population.

Step 2 — isolate and take logs.

$e^{0.45 t} = k \implies 0.45 t = \ln k \implies t = \dfrac{\ln k}{0.45}$

M1 — taking $\ln$ correctly to obtain the printed result. A1 — printed result reached, with at least one intermediate line of working shown (for "show that" credit).

Step 3 — doubling time.

Substitute $k = 2$ :

$t = \dfrac{\ln 2}{0.45} = \dfrac{0.6931...}{0.45} = 1.540...$

So the doubling time is $1.54$ hours (3 s.f.).

A1 — exact substitution and correctly rounded answer.

Total: 8 marks (M5 A3, split as shown).

Specimen question modelled on the Edexcel 9MA0 Paper 1 format

Question (6 marks): Solve the equation $\ln(2x + 1) + \ln(x - 2) = \ln 14$ , giving your answer in exact form.

Mark scheme decomposition by AO:

M1 (AO1.1a) — applying the product law of logarithms to combine the LHS: $\ln[(2x + 1)(x - 2)] = \ln 14$ .
M1 (AO1.1b) — using the one-to-one property of $\ln$ (or " $\ln$ both sides cancel") to remove logarithms: $(2x + 1)(x - 2) = 14$ .
M1 (AO1.1b) — expanding the bracket and forming a quadratic in standard form: $2x^2 - 3x - 2 = 14$ , hence $2x^2 - 3x - 16 = 0$ .
M1 (AO1.1b) — applying the quadratic formula or factorisation: $x = \dfrac{3 \pm \sqrt{9 + 128}}{4} = \dfrac{3 \pm \sqrt{137}}{4}$ .
A1 (AO2.5) — rejecting the negative root with explicit justification: since $x > 2$ is required for $\ln(x - 2)$ to be defined, only $x = \dfrac{3 + \sqrt{137}}{4}$ is valid.
A1 (AO2.5) — final answer in exact form $x = \dfrac{3 + \sqrt{137}}{4}$ .

Total: 6 marks split AO1 = 4, AO2 = 2. The AO2 marks are reserved for the domain check — Edexcel routinely punishes candidates who solve the algebra correctly but fail to reject roots that violate the implicit domain of the original logarithmic expression. Writing "since the original equation requires $2x + 1 > 0$ and $x - 2 > 0$ , hence $x > 2$ " before stating the final answer is the examiner-rewarded phrasing.

Synoptic links

Connects to:

Section 7 — Differentiation: $\frac{d}{dx}\ln x = \frac{1}{x}$ for $x > 0$ is the standard result. Combined with the chain rule, $\frac{d}{dx}\ln(f(x)) = \dfrac{f'(x)}{f(x)}$ — the so-called logarithmic derivative. This unlocks differentiation of products and quotients via logarithmic differentiation: $y = x^x \implies \ln y = x \ln x \implies \dfrac{1}{y}\dfrac{dy}{dx} = \ln x + 1$ .
Section 8 — Integration: $\int \dfrac{1}{x}\,dx = \ln|x| + C$ fills the $n = -1$ gap in the power rule $\int x^n\,dx = \dfrac{x^{n+1}}{n + 1} + C$ . The absolute value is essential for negative $x$ . More generally $\int \dfrac{f'(x)}{f(x)}\,dx = \ln|f(x)| + C$ — the reverse-chain integral that examiners love because it tests recognition rather than mechanical rule application.
Section 4 — Modelling with exponential functions: every "exponential growth/decay" model $P(t) = P_0 e^{kt}$ requires $\ln$ to invert. Half-lives, doubling times, and "time to reach threshold" questions all reduce to $t = \dfrac{1}{k}\ln\left(\dfrac{P}{P_0}\right)$ .
Section 10 — Differential equations: separating variables in $\dfrac{dy}{dx} = ky$ gives $\int \dfrac{1}{y}\,dy = \int k\,dx$ , hence $\ln|y| = kx + C$ and $y = A e^{kx}$ . The natural logarithm is the bridge between the differential and the closed-form solution.
Section 2 — Algebra: the substitution method $u = \ln x$ converts equations of the form $(\ln x)^2 - 5\ln x + 6 = 0$ into routine quadratics. Recognising "polynomial in $\ln x$ " as a hidden quadratic is exactly the kind of synoptic pattern-spotting Edexcel rewards on long-form questions.

Mark-scheme literacy

Natural logarithm questions on 9MA0 split AO marks across procedure, reasoning and synoptic application:

AO	Typical share	Earned by
AO1 (knowledge / procedure)	50–60%	Applying $\ln$ correctly, using log laws (product, quotient, power), differentiating and integrating standard $\ln$ forms
AO2 (reasoning / interpretation)	25–35%	Domain checks (rejecting $\ln$ of non-positives), justifying use of one-to-one property, presenting answers in requested exact form ( $\ln 2$ rather than $0.693$ )
AO3 (problem-solving)	10–20%	Modelling: setting up exponential models from worded contexts, interpreting the meaning of $\ln$ -derived parameters, reverse-chain integration recognition

Examiner-rewarded phrasing: "since $\ln$ is one-to-one, we may equate arguments"; "the domain requires $x - 2 > 0$ , so we reject $x = ...$ "; "in exact form, $t = \dfrac{\ln 5}{k}$ (no decimals required)"; "applying the product law $\ln a + \ln b = \ln(ab)$ ". Phrases that lose marks: " $\ln(x + y) = \ln x + \ln y$ " (this is the cardinal misconception, not a law); " $\ln(-3)$ is small and negative" (it is undefined, not small); rounding mid-calculation to 2 d.p. and propagating; writing " $\log$ " without specifying base when $\ln$ is required.

A specific Edexcel pattern: when a question asks for an answer "in exact form" or "in terms of $\ln 2$ ", the final mark is awarded only for an unevaluated logarithmic expression. A decimal — however accurate — forfeits the A1.

Grade-band model answers

3-mark question

Question: Solve $e^{2x + 1} = 7$ , giving your answer in exact form.

Grade C response (~210 words):

Take $\ln$ of both sides:

$\ln(e^{2x + 1}) = \ln 7$ , so $2x + 1 = \ln 7$ .

Subtract 1: $2x = \ln 7 - 1$ .

Divide by 2: $x = \dfrac{\ln 7 - 1}{2}$ .

Examiner commentary: Full marks (3/3). The candidate identifies that $\ln$ is the inverse of $e^x$ , applies it correctly, and rearranges cleanly. Crucially the answer is left as an exact logarithmic expression — $x = \dfrac{\ln 7 - 1}{2}$ — rather than evaluated as $x \approx 0.473$ . The "exact form" instruction is honoured. Working is brief but each operation is justified by the equation that follows it. This is the standard Grade C answer for a procedural $\ln$ inversion question — examiners want the inversion seen, the rearrangement seen, and the final form matched to the question stem. Many candidates lose a mark here by computing the decimal and presenting it alongside the exact form, which examiners read as the candidate not trusting the exact answer.

Grade A response (~270 words):*

The equation $e^{2x + 1} = 7$ has $e$ on the LHS, so the natural logarithm is the appropriate inverse. Apply $\ln$ to both sides and use the property $\ln(e^y) = y$ for all real $y$ :

$\ln(e^{2x + 1}) = \ln 7 \implies 2x + 1 = \ln 7$

Now solve linearly for $x$ :

$2x = \ln 7 - 1 \implies x = \dfrac{\ln 7 - 1}{2}$

This is the exact solution. Numerically, $\ln 7 \approx 1.9459$ , giving $x \approx 0.473$ , but the exact form is preferred per the question.

Natural Logarithms ln(x)

Natural Logarithms ln(x)

Definition

The Graph of y = ln(x)

Key Values of ln

Laws of Natural Logarithms

Solving Equations with ln

Type 1: eˣ = k

Type 2: eᵃˣ⁺ᵇ = k

Type 3: ln(x) = k

Type 4: ln(ax + b) = k

Combining ln with Other Techniques

The Relationship Between ln and log₁₀

Why ln Is Preferred in Calculus

Summary

A-Level Deep Dive: Natural Logarithms ln(x)

Spec mapping

Worked example with full mark scheme

Specimen question modelled on the Edexcel 9MA0 Paper 1 format

Synoptic links

Mark-scheme literacy

Grade-band model answers

3-mark question

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