Lesson 4: Logarithms & Exponentials | Additional Mathematics

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1. Exponential Functions

Exponential Function: A function of the form f(x) = aˣ where a > 0 and a ≠ 1. The variable x is in the exponent (power). These functions model growth and decay in finance, biology, and physics.

Key Properties of y = aˣ

Growth: a > 1

Graph rises steeply left to right
Always passes through (0, 1)
Asymptote: y = 0 (x-axis) as x → −∞
Domain: all real x. Range: y > 0
Example: y = 2ˣ, y = 10ˣ, y = eˣ

Decay: 0 < a < 1

Graph falls left to right
Always passes through (0, 1)
Asymptote: y = 0 as x → +∞
Domain: all real x. Range: y > 0
Example: y = (½)ˣ = 2⁻ˣ

The Natural Exponential: e

Euler's Number e ≈ 2.71828... The most important base in mathematics. The function y = eˣ is unique because its gradient at every point equals its y-value — it is its own derivative. This property makes it essential in calculus (Lessons 6 and 7).

2. Logarithms — Definition and Basics

Logarithm Definition: The logarithm is the inverse operation of exponentiation.

log_a x = y ⟺ aʸ = x

Read as: "log base a of x equals y" means "a raised to the power y gives x".
Example: log₂ 8 = 3 because 2³ = 8. log₁₀ 1000 = 3 because 10³ = 1000.

Conditions and Special Values:
• Base a must satisfy: a > 0 and a ≠ 1
• Argument x must satisfy: x > 0 (cannot take log of zero or a negative number)
• log_a 1 = 0 for any base a (since a⁰ = 1)
• log_a a = 1 for any base a (since a¹ = a)
• log_a aˣ = x and a^(log_a x) = x (inverse relationship)

Common Logarithms and Natural Logarithms

Common Log: log₁₀ x = log x

Base 10. Written as "log x" without a base shown. Used in pH scales, decibels, Richter scale.

log 100 = 2 | log 0.01 = −2 | log 1 = 0

Natural Log: log_e x = ln x

Base e ≈ 2.718. Written as "ln x". Used in calculus, compound interest, population growth.

ln e = 1 | ln 1 = 0 | ln eˣ = x | e^(ln x) = x

3. Laws of Logarithms

The Three Laws of Logarithms (same base throughout)

Law 1 (Product): log_a(xy) = log_a x + log_a y

Law 2 (Quotient): log_a(x/y) = log_a x − log_a y

Law 3 (Power): log_a(xⁿ) = n log_a x

These mirror the index laws: multiplication ↔ addition of logs; division ↔ subtraction; power ↔ multiplication.

Derived Results (from the three laws):
log_a(1/x) = −log_a x     (from Law 2: log(1)−log(x) = 0−log x)
log_a(√x) = ½ log_a x    (from Law 3: x^(½))
Change of base: log_a x = log_b x / log_b a    (especially: log_a x = ln x / ln a)

📐 Worked Example 1 — Using Log Laws to Simplify

Simplify: (a) log 6 + log 5 − log 3 (b) 2 log 3 + log 4 (c) ½ log 16 − 2 log 3

(a) log 6 + log 5 − log 3 = log(6×5/3) = log 10 = 1

(b) 2 log 3 + log 4 = log 3² + log 4 = log 9 + log 4 = log 36 = log 36

(c) ½ log 16 − 2 log 3 = log 16^(½) − log 3² = log 4 − log 9 = log(4/9)

📐 Worked Example 2 — Expressing in Terms of log p and log q

Given that log p = a and log q = b, express in terms of a and b: (a) log(p²q³) (b) log(p/√q) (c) log(10p²)

(a) log(p²q³) = log p² + log q³ = 2 log p + 3 log q = 2a + 3b

(b) log(p/√q) = log p − log q^(½) = log p − ½ log q = a − b/2

(c) log(10p²) = log 10 + log p² = 1 + 2 log p = 1 + 2a

4. Solving Exponential and Logarithmic Equations

Solving Exponential Equations

Two Methods for Solving aˣ = b:
Method 1 — Same Base: If both sides can be written with the same base, equate exponents.
e.g. 8ˣ = 32 → 2³ˣ = 2⁵ → 3x = 5 → x = 5/3

Method 2 — Take Logarithms: Take log (or ln) of both sides, then use Law 3.
e.g. 3ˣ = 20 → x log 3 = log 20 → x = log 20 / log 3

📐 Worked Example 3 — Exponential Equations

Solve: (a) 5²ˣ⁻¹ = 125 (b) 3ˣ = 7 (c) 2ˣ⁺¹ = 5ˣ⁻²

(a) Same base: 125 = 5³. So 5²ˣ⁻¹ = 5³
2x−1 = 3 → x = 2

(b) Take log: x log 3 = log 7
x = log 7 / log 3 = 1.771 (3 d.p.)

(c) Take log of both sides:
(x+1) log 2 = (x−2) log 5
x log 2 + log 2 = x log 5 − 2 log 5
x log 2 − x log 5 = −2 log 5 − log 2
x(log 2 − log 5) = −2 log 5 − log 2
x = (−2 log 5 − log 2)/(log 2 − log 5) = (2 log 5 + log 2)/(log 5 − log 2)
= (log 25 + log 2)/log(5/2) = log 50 / log 2.5 ≈ 4.824

Solving Logarithmic Equations

📐 Worked Example 4 — Logarithmic Equations

Solve: (a) log₃(2x−1) = 3 (b) log x + log(x−3) = 1 (c) ln(3x+1) = 2

(a) Convert to exponential form: 2x−1 = 3³ = 27
x = 14

(b) log x + log(x−3) = 1 → log[x(x−3)] = 1
x(x−3) = 10¹ = 10 (base 10)
x²−3x−10 = 0 → (x−5)(x+2) = 0
x = 5 or x = −2. But log(−2) is undefined → x = 5 only

(c) Convert: 3x+1 = e²
x = (e²−1)/3 ≈ 2.128

⚠ Always Check Solutions! When solving logarithmic equations, always substitute back to verify the argument of every logarithm is positive. Negative or zero values of the argument must be rejected even if they satisfy the algebra.

Equations of the Form aˣ = bˣ⁺ᶜ — Substitution Method

📐 Worked Example 5 — Substitution for Exponential Equations

Solve: 4ˣ − 6(2ˣ) + 8 = 0

Write 4ˣ = (2²)ˣ = (2ˣ)². Let y = 2ˣ:
y² − 6y + 8 = 0

Factorise: (y−2)(y−4) = 0 → y = 2 or y = 4

Substitute back: 2ˣ = 2 → x = 1 or 2ˣ = 4 → x = 2
x = 1 or x = 2

5. Graphs of Logarithmic and Exponential Functions

Function	Shape	Key Points	Asymptote	Domain	Range
y = eˣ	Increasing curve, concave up	(0,1), (1,e), (−1, 1/e)	y = 0 (as x→−∞)	ℝ	y > 0
y = e⁻ˣ	Decreasing curve, concave up	(0,1), (1, 1/e)	y = 0 (as x→+∞)	ℝ	y > 0
y = ln x	Increasing, concave down	(1,0), (e,1), (1/e,−1)	x = 0 (as x→0⁺)	x > 0	ℝ
y = aˣ (a>1)	Increasing	(0,1)	y = 0	ℝ	y > 0
y = log_a x (a>1)	Increasing	(1,0)	x = 0	x > 0	ℝ

Transformations of Exponential and Log Graphs

📐 Worked Example 6 — Describing Transformations

Describe the transformation mapping y = eˣ to: (a) y = e²ˣ (b) y = eˣ + 3 (c) y = 3eˣ (d) y = e⁻ˣ (e) y = ln x

(a) y = e²ˣ: Replace x with 2x → horizontal stretch by factor ½ (towards y-axis).

(b) y = eˣ + 3: Add 3 → translation by vector (0 over 3) — moves up by 3. Asymptote moves from y=0 to y=3.

(c) y = 3eˣ: Multiply by 3 → vertical stretch by factor 3 (from x-axis). y-intercept moves from (0,1) to (0,3).

(d) y = e⁻ˣ: Replace x with −x → reflection in the y-axis.

(e) y = ln x: Inverse of eˣ → reflection in the line y = x.

6. Reducing a Non-Linear Relationship to Linear Form

Linearisation: Many real-world relationships are not linear but can be transformed into a straight-line graph (Y = mX + c) using logarithms. This allows us to determine unknown constants from a graph of the linearised data.

The Two Standard Forms

Form 1: y = axⁿ (Power Law)

Take log of both sides:

log y = log a + n log x

Plot log y against log x.
Gradient = n y-intercept = log a
So a = 10^(y-intercept) [or e^(y-intercept) if using ln]

Form 2: y = ab^x (Exponential)

Take log of both sides:

log y = log a + x log b

Plot log y against x.
Gradient = log b y-intercept = log a
So a = 10^(y-intercept), b = 10^(gradient)

📐 Worked Example 7 — Linearising y = axⁿ

The variables x and y satisfy y = axⁿ. When log y is plotted against log x, a straight line passes through (1, 3.2) and (3, 5.6). Find a and n.

The equation in linear form: log y = n log x + log a
This is Y = mX + c where Y = log y, X = log x, gradient = n, intercept = log a.

Gradient (= n): n = (5.6−3.2)/(3−1) = 2.4/2 = 1.2

Find log a (y-intercept): using point (1, 3.2):
3.2 = 1.2×1 + log a → log a = 2 → a = 100

Equation: y = 100x^1.2

📐 Worked Example 8 — Linearising y = abˣ

The variables x and y satisfy y = abˣ. When ln y is plotted against x, a straight line passes through (0, 1.6) and (4, 5.2). Find a and b.

Take ln: ln y = ln a + x ln b
Linear form: Y = (ln b)x + ln a, where Y = ln y.

Gradient = ln b = (5.2−1.6)/(4−0) = 3.6/4 = 0.9
b = e^0.9 ≈ 2.46

y-intercept = ln a = 1.6 → a = e^1.6 ≈ 4.95

Equation: y = 4.95 × (2.46)ˣ

Reading Values from a Linearised Graph

📐 Worked Example 9 — Using Linearised Graph to Find x or y

From Worked Example 7, y = 100x^1.2. Find (a) y when x = 5 (b) x when y = 500.

(a) y = 100 × 5^1.2 = 100 × 6.899 = 690 (3 s.f.)

(b) 500 = 100x^1.2 → x^1.2 = 5 → x = 5^(1/1.2) = 5^(5/6) = 3.34 (3 s.f.)

Non-Standard Linearisations

📐 Worked Example 10 — Non-Standard Form

The variables x and y are related by y = A·eˢˣ + B. When (y−B) is plotted against x on a graph with ln(y−B) on the vertical axis, a straight line with gradient 3 and vertical intercept ln 5 is obtained. Find A and s.

y − B = Aeˢˣ → ln(y−B) = ln A + sx

Comparing with Y = mx + c where Y = ln(y−B), X = x:
Gradient = s = 3
y-intercept = ln A = ln 5 → A = 5

7. Change of Base and Mixed Equations

Change of Base Formula

log_a x = log_b x / log_b a

Most useful: log_a x = log x / log a = ln x / ln a

Use when solving equations involving different bases, or when evaluating logs that your calculator does not have a direct button for.

📐 Worked Example 11 — Equations Requiring Change of Base

Solve: (a) log₄ x = 3 (b) log₉ x = log₃ 5 (c) 2 log₃ x − log₃(x+6) = 1

(a) x = 4³ = 64

(b) Change log₉ x to base 3: log₉ x = log₃ x / log₃ 9 = log₃ x / 2
So ½ log₃ x = log₃ 5 → log₃ x = 2 log₃ 5 = log₃ 25 → x = 25

(c) 2 log₃ x − log₃(x+6) = 1
log₃ x² − log₃(x+6) = log₃ 3
log₃[x²/(x+6)] = log₃ 3
x²/(x+6) = 3 → x² = 3x+18 → x²−3x−18 = 0
(x−6)(x+3) = 0 → x = 6 or x = −3
Reject x=−3 (log of negative). x = 6

📐 Worked Example 12 — Disguised Quadratic in Logs

Solve: (log₂ x)² − log₂ x⁴ + 3 = 0

Let u = log₂ x. Note: log₂ x⁴ = 4 log₂ x = 4u
u² − 4u + 3 = 0

(u−1)(u−3) = 0 → u = 1 or u = 3

u = log₂ x = 1 → x = 2
u = log₂ x = 3 → x = 8
x = 2 or x = 8

📝 Exam Practice Questions

Q1 [3 marks] — Without using a calculator, evaluate: (a) log₂ 32 (b) log₃(1/27) (c) log₅ 5√5

(a) 2⁵ = 32 → log₂ 32 = 5
(b) 1/27 = 3⁻³ → log₃(1/27) = −3
(c) 5√5 = 5¹×5^(½) = 5^(3/2) → log₅ 5^(3/2) = 3/2

Q2 [3 marks] — Given log p = 2 and log q = −1, find: (a) log(p²/q) (b) log(√p × q²) (c) log(100p)

(a) 2 log p − log q = 4−(−1) = 5
(b) ½ log p + 2 log q = 1 + (−2) = −1
(c) log 100 + log p = 2+2 = 4

Q3 [3 marks] — Solve: (a) 3²ˣ⁻¹ = 243 (b) 5ˣ = 30 (give answer to 3 d.p.) (c) e²ˣ⁻³ = 7

(a) 243 = 3⁵. So 2x−1=5 → x = 3
(b) x log 5 = log 30 → x = log 30/log 5 = 1.4771/0.6990 = 2.113
(c) 2x−3 = ln 7 → 2x = 3+ln 7 → x = (3+ln 7)/2 = (3+1.9459)/2 = 2.473

Q4 [4 marks] — Solve: (a) log(x+1) + log(x−2) = 1 (b) 9ˣ − 4(3ˣ) + 3 = 0

(a) log[(x+1)(x−2)] = 1 → (x+1)(x−2) = 10
x²−x−2 = 10 → x²−x−12 = 0 → (x−4)(x+3) = 0
x=4 or x=−3. Reject x=−3 (log(−3+1)=log(−2) undefined).
x = 4

(b) (3ˣ)²−4(3ˣ)+3=0. Let u=3ˣ:
u²−4u+3=0 → (u−1)(u−3)=0 → u=1 or u=3
3ˣ=1 → x=0 or 3ˣ=3 → x=1
x = 0 or x = 1

Exam Tip: For exponential equations like 9ˣ−4(3ˣ)+3=0, recognise that 9ˣ=(3ˣ)² and substitute u=3ˣ to get a quadratic. This substitution technique is very commonly tested.

Q5 [4 marks] — The variables x and y satisfy y = Axⁿ. The graph of log y against log x is a straight line passing through points (0.5, 2.3) and (2.5, 5.1). Find A and n.

Linear form: log y = n log x + log A
Gradient = n: (5.1−2.3)/(2.5−0.5) = 2.8/2 = n = 1.4
log A (y-intercept): Using (0.5, 2.3): 2.3 = 1.4×0.5 + log A → log A = 2.3−0.7 = 1.6
A = 10^1.6 ≈ 39.8
Equation: y = 39.8 x^1.4

Q6 [4 marks] — Solve the equation (log₃ x)² − 2 log₃ x − 8 = 0.

Let u = log₃ x: u²−2u−8 = 0
(u−4)(u+2) = 0 → u=4 or u=−2
log₃ x = 4 → x = 3⁴ = 81
log₃ x = −2 → x = 3⁻² = 1/9

Exam Tip: Equations of the form (log x)² + a(log x) + b = 0 are disguised quadratics. Always substitute u = log x (or ln x) to convert to a standard quadratic, solve for u, then convert back.

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