Logarithms & Exponentials | FractionRush A-Level

Welcome to Logarithms & Exponentials

Logarithms and exponentials are among the most powerful tools in mathematics. They appear everywhere — in compound interest and finance, radioactive decay in physics, population growth in biology, pH in chemistry, and the decibel scale in acoustics. Mastering this topic unlocks an entire world of real-world modelling.

log_a(x) = y ⟺ a^y = x | ln x = log_e x | e ≈ 2.71828

Why do logs matter? They let us solve equations where the unknown is in the exponent (e.g. 3^x = 20), and convert multiplicative relationships into additive ones — making analysis and graphing much simpler.

Learning Objectives

Define logarithms and convert between log and index form
Apply the product, quotient and power laws of logarithms
Use the change of base formula
Understand the natural logarithm ln x and the number e
Differentiate and integrate e^x and ln x
Solve exponential equations by taking logarithms
Solve logarithmic equations using log laws
Apply exponential growth and decay models Q = Q₀e^kt
Linearise data using y = axⁿ and y = ae^bx models
Sketch and transform graphs of y = e^x and y = ln x

Topic Overview

Log Laws

Product, quotient, power rules and change of base

Natural Log & e

The special number e ≈ 2.718 and its inverse ln x

Exponential Equations

Solving a^x = b by taking logs of both sides

Log Equations

Combining log laws to isolate the unknown

Exponential Models

Growth/decay, half-life, doubling time

Graphs & Transformations

Shapes, reflections, translations of e^x and ln x

Learn 1 — Logarithm Definition & Laws

Definition

The logarithm answers the question: "What power must I raise a to, to get x?"

log_a(x) = y ⟺ a^y = x

Conditions: a > 0, a ≠ 1, x > 0. You cannot take the log of a negative number or zero.

Special Values

log_a(1) = 0 | log_a(a) = 1 | log_a(aⁿ) = n

Why? Because a⁰ = 1, a¹ = a, and aⁿ = aⁿ.

The Three Log Laws

Product Law: log_a(AB) = log_a(A) + log_a(B)

Example: log₂(4 × 8) = log₂(4) + log₂(8) = 2 + 3 = 5 ✓ (since 2⁵ = 32 = 4×8)

Quotient Law: log_a(A/B) = log_a(A) − log_a(B)

Example: log₃(27/3) = log₃(27) − log₃(3) = 3 − 1 = 2 ✓ (since 3² = 9 = 27/3)

Power Law: log_a(Aⁿ) = n · log_a(A)

Example: log₂(8⁴) = 4 · log₂(8) = 4 × 3 = 12

Change of Base Formula

log_a(x) = ln(x) / ln(a) = log(x) / log(a)

This is essential when your calculator only has log₁₀ and ln buttons.

Example: log₅(17) = ln(17)/ln(5) = 2.833/1.609 = 1.760 (to 4sf)

Worked Examples with Log Laws

Simplify: log₂(12) − log₂(3)
= log₂(12/3) = log₂(4) = 2

Simplify: 2·log₃(6) − log₃(4)
= log₃(6²) − log₃(4) = log₃(36) − log₃(4) = log₃(36/4) = log₃(9) = 2

Express log₂(5) in terms of ln:
log₂(5) = ln(5)/ln(2) = 1.609/0.693 ≈ 2.322

Tip: When combining logs, always make sure all terms have the same base before applying the laws.

Learn 2 — The Natural Logarithm & e

The Number e

The number e is one of the most important constants in mathematics, approximately equal to 2.71828... It arises naturally when studying continuous growth and is defined as:

e = lim_n→∞ (1 + 1/n)ⁿ ≈ 2.71828182845...

Just like π appears naturally in circles, e appears naturally in exponential growth, compound interest, and calculus.

The Natural Logarithm

ln x = log_e(x) — logarithm with base e

The natural logarithm is the inverse of the exponential function e^x. Key values:

ln(1) = 0 | ln(e) = 1 | ln(e^x) = x | e^{ln x} = x

Graphs of y = e^x and y = ln x

y = e^x: Passes through (0, 1). Always positive. Increasing. As x → −∞, y → 0 (horizontal asymptote y = 0). As x → +∞, y → +∞.

y = ln x: Passes through (1, 0) and (e, 1). Defined only for x > 0. Increasing. Vertical asymptote x = 0.

Key fact: y = ln x is the reflection of y = e^x in the line y = x.

Derivatives

d/dx (e^x) = e^x | d/dx (e^kx) = ke^kx

d/dx (ln x) = 1/x | d/dx (ln(kx)) = 1/x

Example: d/dx(e^3x+1) = 3e^3x+1 (chain rule: multiply by derivative of 3x+1)

Example: d/dx(ln(x²+1)) = 2x/(x²+1) (chain rule: f'(x)/f(x))

Integrals

∫ e^x dx = e^x + C | ∫ e^kx dx = (1/k)e^kx + C

∫ (1/x) dx = ln|x| + C (note: absolute value!)

Example: ∫ e^4x dx = (1/4)e^4x + C

Example: ∫ (3/x) dx = 3 ln|x| + C

Remember: ln|x| includes the absolute value so the formula works for negative x too. In A-Level, you usually write ln x when the domain is clearly x > 0.

Learn 3 — Solving Exponential Equations

Key Strategy: Take logs of both sides

When the unknown is in the exponent, use logarithms to bring it down.

a^x = b ⟹ x·ln a = ln b ⟹ x = ln(b)/ln(a)

Type 1: Simple base

Solve 3^x = 20:
Take ln both sides: x · ln 3 = ln 20
x = ln 20 / ln 3 = 2.996/1.099 = 2.727 (to 4sf)

Type 2: Different bases on both sides

Solve 2^x+1 = 5^x−1:
Take ln: (x+1)ln2 = (x−1)ln5
x·ln2 + ln2 = x·ln5 − ln5
x(ln2 − ln5) = −ln5 − ln2
x = (−ln5 − ln2)/(ln2 − ln5) = (ln2 + ln5)/(ln5 − ln2)
x = ln10 / ln(5/2) = 2.303/0.916 = 2.513 (to 4sf)

Type 3: e^f(x) = k — take ln directly

Solve e^2x−1 = 8:
Take ln: 2x − 1 = ln 8
2x = 1 + ln 8
x = (1 + ln 8)/2 = (1 + 2.079)/2 = 1.540 (to 4sf)

Type 4: Hidden quadratic — substitution

Solve e^2x − 5e^x + 6 = 0:
Let u = e^x. Note e^2x = (e^x)² = u²
u² − 5u + 6 = 0
(u − 2)(u − 3) = 0
u = 2 or u = 3
e^x = 2 → x = ln 2 OR e^x = 3 → x = ln 3
x = ln 2 or x = ln 3

Check signs! Since e^x > 0 always, if your substitution gives u = negative number, discard that solution (e^x can never be negative).

Exam tip: If asked for exact answers, leave your answer as ln(2) or (ln 5 − ln 2)/ln 3 etc. Only convert to decimals if the question says "to 3sf" or similar.

Learn 4 — Solving Logarithmic Equations

Key Strategy: Convert to index form

log_a(x) = b ⟹ x = a^b

Type 1: Single log equals a number

Solve log₃(2x+1) = 4:
Convert: 2x + 1 = 3⁴ = 81
2x = 80
x = 40
Check: log₃(81) = 4 ✓ and 2(40)+1 = 81 > 0 ✓

Type 2: Logs on both sides

Solve ln(x) + ln(x+1) = ln(6):
Apply product law: ln(x(x+1)) = ln(6)
x(x+1) = 6
x² + x − 6 = 0
(x+3)(x−2) = 0
x = −3 or x = 2
Check domain: ln(x) requires x > 0, so x = −3 is rejected
x = 2

Type 3: Multiple logs combined

Solve 2·log x − log(x−1) = 1:
Using base 10; apply power law: log(x²) − log(x−1) = 1
Quotient law: log(x²/(x−1)) = 1
Convert: x²/(x−1) = 10¹ = 10
x² = 10(x−1) = 10x − 10
x² − 10x + 10 = 0
x = (10 ± √60)/2 = 5 ± √15
x = 5 + √15 ≈ 8.87 or x = 5 − √15 ≈ 1.13
Check: both give x > 0 and x−1 > 0, so both solutions are valid

Domain Checking — Critical!

Always check your answers! After solving, verify that every expression inside a log is strictly positive. Any solution that makes a log argument ≤ 0 must be rejected.

Combining Log and Exponential

Solve ln(x+3) = 2 + ln(x−1):
Rearrange: ln(x+3) − ln(x−1) = 2
Quotient: ln((x+3)/(x−1)) = 2
Convert: (x+3)/(x−1) = e²
x + 3 = e²(x−1) = e²x − e²
x − e²x = −e² − 3
x(1 − e²) = −(e² + 3)
x = (e² + 3)/(e² − 1) ≈ (7.389 + 3)/(7.389 − 1) = 10.389/6.389 ≈ 1.627

Learn 5 — Exponential Models & Linearisation

The General Exponential Model

Q = Q₀ · e^kt

Where Q₀ is the initial quantity (at t = 0), k is the growth rate constant, and t is time. If k > 0: growth. If k < 0: decay.

Doubling Time

For Q = Q₀e^kt, find time T when Q = 2Q₀:
2Q₀ = Q₀e^kT → 2 = e^kT → kT = ln 2 → T = ln 2 / k

Half-Life

For Q = Q₀e^−kt (decay), find t_½ when Q = Q₀/2:
Q₀/2 = Q₀e^−kt½ → 1/2 = e^−kt½ → −kt½ = ln(1/2) = −ln 2 → t½ = ln 2 / k

Linearisation: y = ae^bx

To check if data fits an exponential model, take ln of both sides:

y = ae^bx ⟹ ln y = ln a + bx

Plot ln y vs x. If linear, the data fits the model. The gradient = b and the y-intercept = ln a (so a = e^intercept).

Linearisation: y = axⁿ

y = axⁿ ⟹ log y = log a + n·log x

Plot log y vs log x. Gradient = n, y-intercept = log a (so a = 10^intercept).

Example: Population Model

A population grows according to P = 500e^0.03t (t in years).
(i) Initial population: t = 0 → P = 500e⁰ = 500
(ii) When does population reach 1000?
1000 = 500e^0.03t → 2 = e^0.03t → 0.03t = ln 2 → t = ln2/0.03 = 23.1 years
(iii) Population after 10 years:
P = 500e^0.3 = 500 × 1.3499 ≈ 675

Example: Reading from a ln y vs x graph

A ln y vs x graph has gradient 0.4 and y-intercept 2.1.
So: ln y = 2.1 + 0.4x
Model: y = e^2.1 · e^0.4x = 8.17e^0.4x (a ≈ 8.17, b = 0.4)

A-Level exam tip: When asked to "find a and b" from a linearised graph, read off gradient and intercept carefully — they give b and ln a respectively, not a and b directly. Don't forget to exponentiate!

Worked Examples

Example 1 — Applying Log Laws

Simplify: log₂(8) + log₂(4) − log₂(2)

M1 Evaluate each term: log₂(8) = 3, log₂(4) = 2, log₂(2) = 1

A1 3 + 2 − 1 = 4

Alternative: log₂(8 × 4 / 2) = log₂(16) = 4 ✓

Example 2 — Solving 5^x = 30

M1 Take ln of both sides: x · ln 5 = ln 30

M1 x = ln 30 / ln 5

A1 x = 3.4012 / 1.6094 = 2.113 ≈ 2.11 (to 3sf)

Example 3 — Hidden Quadratic: e^2x − 7e^x + 12 = 0

M1 Let u = e^x, so e^2x = u². Equation becomes: u² − 7u + 12 = 0

M1 Factorise: (u − 3)(u − 4) = 0 → u = 3 or u = 4

M1 e^x = 3 → x = ln 3 OR e^x = 4 → x = ln 4

A1 x = ln 3 or x = ln 4 (exact answers)

Example 4 — Log Equation with Two Logs

Solve: log₃(x+2) + log₃(x−2) = 2

M1 Product law: log₃((x+2)(x−2)) = 2

M1 log₃(x²−4) = 2 → x² − 4 = 3² = 9

M1 x² = 13 → x = ±√13

A1 Check domain: x+2 > 0 and x−2 > 0 requires x > 2. So x = −√13 is rejected. x = √13

Example 5 — Different Bases: 2^x+1 = 3^x−1

M1 Take ln: (x+1)ln 2 = (x−1)ln 3

M1 Expand: x·ln2 + ln2 = x·ln3 − ln3

M1 x(ln2 − ln3) = −ln3 − ln2 = −ln6

A1 x = −ln6 / (ln2 − ln3) = ln6 / (ln3 − ln2) = ln6 / ln(3/2) ≈ 1.792/0.405 ≈ 4.42 (to 3sf)

Example 6 — Population Model

P = 500·e^0.03t. Find t when P = 1000.

M1 1000 = 500·e^0.03t → 2 = e^0.03t

M1 Take ln: 0.03t = ln 2

A1 t = ln 2 / 0.03 = 0.6931/0.03 = 23.1 years (to 3sf)

Example 7 — Interpreting Linearised Form

Given ln y = 2.3 + 0.5x, find y in the form ae^bx.

M1 Compare with ln y = ln a + bx → ln a = 2.3, b = 0.5

M1 a = e^2.3 ≈ 9.974

A1 y = e^2.3·e^0.5x ≈ 9.97e^0.5x

Example 8 — Combined Log-Exponential

Solve: ln(x+3) = 2 + ln(x−1)

M1 Rearrange: ln(x+3) − ln(x−1) = 2

M1 Quotient: ln((x+3)/(x−1)) = 2

M1 Exponentiate: (x+3)/(x−1) = e²

A1 x + 3 = e²(x−1) → x + 3 = e²x − e² → x(1−e²) = −e² − 3 → x = (e²+3)/(e²−1)

Numerically: x ≈ 10.389/6.389 ≈ 1.63. Check: x > 1 ✓

Common Mistakes

Mistake 1: Splitting log of a sum

✗ Wrong: log(A + B) = log A + log B

✓ Right: log(A · B) = log A + log B (product law applies to multiplication, not addition)

Example: log(5 + 3) = log(8) ≠ log5 + log3 = 1.398

Mistake 2: Dividing logs equals log of quotient?

✗ Wrong: log(A/B) = log A / log B

✓ Right: log(A/B) = log A − log B (subtraction, not division)

The division form is change of base: log_a(b) = log b / log a (different bases!)

Mistake 3: Forgetting domain restrictions

✗ Wrong: accepting x = −3 from solving ln(x+4) = ln(1) without checking

✓ Right: always verify every log argument is > 0 after solving. Reject invalid solutions.

Mistake 4: Inverting the change of base formula

✗ Wrong: log_a(b) = log a / log b

✓ Right: log_a(b) = log b / log a (what you're finding goes on top)

Memory aid: log_a(b) means "log of b, base a" → log(b) ÷ log(a)

Mistake 5: Power law applied to coefficient

✗ Wrong: 2·ln x = ln(2x)

✓ Right: 2·ln x = ln(x²) — the coefficient becomes the exponent of x, not a multiplier

Mistake 6: Thinking e^x can be negative

✗ Wrong: e^x = −5 → x = ln(−5) (not defined!)

✓ Right: e^x > 0 for all real x. If your quadratic substitution gives a negative value for e^x, reject it.

Mistake 7: Forgetting t = 0 gives Q₀

✗ Wrong: setting up Q = Ae^kt but forgetting that the given initial value is exactly A

✓ Right: at t = 0, Q = A·e⁰ = A. So the initial value directly gives you the constant A = Q₀.

Key Formulas

Category	Formula	Notes
Definition	log_a(x) = y ⟺ a^y = x	a > 0, a ≠ 1, x > 0
Special values	log_a(1) = 0, log_a(a) = 1	log_a(aⁿ) = n
Product law	log_a(AB) = log_aA + log_aB	Multiplication → addition
Quotient law	log_a(A/B) = log_aA − log_aB	Division → subtraction
Power law	log_a(Aⁿ) = n · log_aA	Power → multiplier
Change of base	log_a(x) = ln(x)/ln(a) = log(x)/log(a)	Works with any common base
Natural log	ln x = log_e(x)	e ≈ 2.71828
ln rules	ln(e) = 1, ln(1) = 0, ln(e^x) = x	e^{ln x} = x
Derivative of e^x	d/dx(e^x) = e^x	d/dx(e^kx) = ke^kx
Derivative of ln x	d/dx(ln x) = 1/x	d/dx(ln(f(x))) = f'(x)/f(x)
Integral of e^x	∫e^x dx = e^x + C	∫e^kx dx = (1/k)e^kx + C
Integral of 1/x	∫(1/x) dx = ln\|x\| + C	Note absolute value
Exponential model	Q = Q₀e^kt	k > 0 growth; k < 0 decay
Doubling time	T = ln 2 / k	When Q doubles
Half-life	t½ = ln 2 / k	Decay model Q = Q₀e^−kt
Linearise y = ae^bx	ln y = ln a + bx	Plot ln y vs x; gradient = b, intercept = ln a
Linearise y = axⁿ	log y = log a + n log x	Plot log y vs log x; gradient = n, intercept = log a

Proof Bank

Proof 1 — Product Law of Logarithms

To prove: log_a(AB) = log_a(A) + log_a(B)

Proof:
Let log_a(A) = p, so A = a^p
Let log_a(B) = q, so B = a^q

Then AB = a^p · a^q = a^p+q (using index law)

Taking log_a of both sides:
log_a(AB) = log_a(a^p+q) = p + q (by definition)

Therefore log_a(AB) = p + q = log_a(A) + log_a(B) QED

Proof 2 — Change of Base Formula

To prove: log_a(x) = ln(x) / ln(a)

Proof:
Let log_a(x) = y, so a^y = x (by definition)

Take the natural log (ln) of both sides:
ln(a^y) = ln(x)

Apply the power law of logarithms:
y · ln(a) = ln(x)

Divide both sides by ln(a) (valid since a ≠ 1, so ln(a) ≠ 0):
y = ln(x) / ln(a)

Therefore log_a(x) = ln(x) / ln(a) QED

Note: the same proof with log₁₀ instead of ln gives log_a(x) = log(x)/log(a).

Proof 3 — d/dx(ln x) = 1/x

Using implicit differentiation of y = ln x

Proof:
Let y = ln x, so by definition x = e^y

Differentiate both sides with respect to x:
1 = e^y · dy/dx (chain rule on the right)

Solve for dy/dx:
dy/dx = 1 / e^y

Since e^y = x:
dy/dx = 1/x

Therefore d/dx(ln x) = 1/x QED

This proof assumes the result d/dx(e^x) = e^x, which follows from the definition of e as the base of the natural exponential function.

Interactive Graph Plotter

Explore the shapes of exponential and logarithmic functions. Adjust the parameters and see the graph update in real time.

a = 1

b = 2.72 (base)

y-intercept: (0, a)

Base b: 2.72

Growth: b > 1

Domain: all reals

Showing y = 1 · b^x. The y-intercept is always (0, a). When b > 1 the curve grows; when 0 < b < 1 it decays. The natural base e ≈ 2.718 is special because it is its own derivative.

Exercise 1 — Log Laws (10 Questions)

Exercise 2 — Solving Exponential Equations (10 Questions)

Exercise 3 — Solving Log Equations (10 Questions)

Exercise 4 — Natural Logarithm & e (10 Questions)

Exercise 5 — Exponential Models (10 Questions)

Practice (30 Mixed Questions)

Challenge (15 Harder Questions)

Exam Style Questions

Question 1 [3 marks] — Exponential equation

Solve e^2x − 4e^x + 3 = 0, giving exact answers.

Let u = e^x: u² − 4u + 3 = 0 M1
(u−1)(u−3) = 0 → u = 1 or u = 3 M1
e^x = 1 → x = 0 e^x = 3 → x = ln 3 A1
x = 0 or x = ln 3

Question 2 [4 marks] — Log equation

Solve ln(x+2) − ln(x−1) = 1, giving your answer in exact form.

Quotient law: ln((x+2)/(x−1)) = 1 M1
Exponentiate: (x+2)/(x−1) = e M1
x + 2 = e(x−1) = ex − e M1
x − ex = −e − 2 → x(1−e) = −(e+2)
x = (e+2)/(e−1) A1
Check: x > 1 ✓ (since e+2 > e−1, so x > 1)

Question 3 [3 marks] — Linearisation

Given y = 3·2^x, express ln y in terms of ln 2 and x.

ln y = ln(3·2^x) M1
ln y = ln 3 + ln(2^x) M1
ln y = ln 3 + x ln 2 A1
(This is linear in x; gradient = ln 2, y-intercept = ln 3)

Question 4 [5 marks] — Radioactive decay model

The mass M grams of a radioactive substance satisfies M = M₀·e^−0.04t, where t is time in years. Find (i) the half-life, (ii) the percentage remaining after 50 years.

(i) Set M = M₀/2: M₀/2 = M₀e^−0.04t M1
1/2 = e^−0.04t → −0.04t = ln(1/2) = −ln2 M1
t = ln2/0.04 = 17.3 years A1

(ii) At t = 50: M = M₀·e^−0.04×50 = M₀e⁻² M1
Percentage = e⁻² × 100 = 13.5% A1

Question 5 [4 marks] — Different bases

Solve 5^x = 3^x+2, giving your answer to 3 significant figures.

Take ln: x ln5 = (x+2)ln3 M1
x ln5 = x ln3 + 2ln3 M1
x(ln5 − ln3) = 2ln3 M1
x = 2ln3 / (ln5 − ln3) = 2(1.0986) / (1.6094 − 1.0986) = 2.197/0.5108
x = 4.30 (to 3sf) A1

Question 6 [5 marks] — Graph of ln y vs x

Given that ln y = a + bx and the graph of ln y against x passes through (2, 5) and (6, 13), find a and b. Hence write y in terms of x.

Gradient b = (13−5)/(6−2) = 8/4 = 2 M1 A1
Using (2, 5): 5 = a + 2(2) → a = 5 − 4 = 1 M1 A1
So ln y = 1 + 2x
y = e^1+2x = e·e^2x
y = e · e^2x (or y = e^1+2x) A1

Question 7 [4 marks] — Log equation

Solve log₂(x) + log₂(x−6) = 4.

Product law: log₂(x(x−6)) = 4 M1
x(x−6) = 2⁴ = 16 M1
x² − 6x − 16 = 0
(x−8)(x+2) = 0 → x = 8 or x = −2 M1
Check: x > 0 and x−6 > 0 requires x > 6, so x = −2 is rejected
x = 8 A1

Question 8 [5 marks] — Inverse function

f(x) = 2e^x − 3. (i) State the range of f. (ii) Find f⁻¹(x). (iii) State the domain of f⁻¹.

(i) e^x > 0 always, so 2e^x > 0, so 2e^x − 3 > −3
Range of f: f(x) > −3 (or (−3, ∞)) B1

(ii) Let y = 2e^x − 3. Swap x and y to find inverse:
x = 2e^y − 3 M1
e^y = (x+3)/2 M1
y = ln((x+3)/2)
f⁻¹(x) = ln((x+3)/2) A1

(iii) Domain of f⁻¹ = Range of f: x > −3 B1

Past Paper Questions

These questions are adapted from Cambridge A-Level 9709 past papers. Always show full working in exams.

Q1 — 9709/22/O/N/19 Q1 [3 marks]

Solve the equation 2·ln(x) − ln(x+1) = ln(5).

Power law: ln(x²) − ln(x+1) = ln5 M1
Quotient: ln(x²/(x+1)) = ln5 M1
x²/(x+1) = 5 → x² = 5x + 5 → x² − 5x − 5 = 0
x = (5 ± √45)/2 = (5 ± 3√5)/2
x = (5 + 3√5)/2 ≈ 5.85 (reject negative since ln(x) requires x > 0) A1

Q2 — 9709/21/M/J/20 Q3 [4 marks]

A population P at time t years satisfies P = 2000/(1 + 3e^−0.2t). Find (i) the initial population, (ii) the value of t when P = 1500.

(i) At t = 0: P = 2000/(1 + 3e⁰) = 2000/(1+3) = 2000/4 = 500 B1

(ii) 1500 = 2000/(1 + 3e^−0.2t) M1
1 + 3e^−0.2t = 2000/1500 = 4/3
3e^−0.2t = 1/3 → e^−0.2t = 1/9 M1
−0.2t = ln(1/9) = −ln9 → t = ln9/0.2 = 11.0 years (to 3sf) A1

Q3 — 9709/23/M/J/18 Q2 [3 marks]

Solve 3^2x+1 = 4^x, giving your answer to 3 significant figures.

Take ln: (2x+1)ln3 = x·ln4 M1
2x·ln3 + ln3 = x·ln4
x(2ln3 − ln4) = −ln3 M1
x = −ln3 / (2ln3 − ln4) = −1.0986 / (2.1972 − 1.3863) = −1.0986/0.8109
x = −1.35 (to 3sf) A1

Q4 — 9709/22/O/N/17 Q4 [5 marks]

The variables x and y satisfy the relation y = ab^x. A straight line graph of ln y against x passes through (1, 3.4) and (4, 7.6). Find a and b.

Linearise: ln y = ln a + x·ln b B1
Gradient = ln b = (7.6 − 3.4)/(4 − 1) = 4.2/3 = 1.4 M1
b = e^1.4 ≈ 4.05 (to 3sf) A1
Using (1, 3.4): 3.4 = ln a + 1.4 → ln a = 2.0 → a = e² ≈ 7.39 M1 A1

Q5 — 9709/21/M/J/21 Q1 [4 marks]

Solve the equation ln(2x+3) + ln(x+1) = 2ln(x+2).

RHS: 2ln(x+2) = ln(x+2)² M1
Product law LHS: ln((2x+3)(x+1)) = ln(x+2)² M1
(2x+3)(x+1) = (x+2)²
2x² + 5x + 3 = x² + 4x + 4 M1
x² + x − 1 = 0
x = (−1 ± √5)/2
Check domain: 2x+3 > 0 and x+1 > 0 and x+2 > 0 → x > −1
x = (−1+√5)/2 ≈ 0.618 ✓ x = (−1−√5)/2 ≈ −1.618 ✗
x = (√5 − 1)/2 A1

Logarithms & Exponentials A-Level Pure 2